electrode potential	

	Name:	
________________________	

Class:	
________________________	

Date:	
________________________	

 

 	
Time:	251 minutes	
Marks:	243 marks	
Comments:		

Q1.Use the data in the table below to answer this question.
 
 	 	E / V	
 	MnO (aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)	+ 1.52	
 	Cr2O(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l)	+ 1.33	
 	Fe3+(aq) + e− → Fe2+(aq)	+ 0.77	
 	Cr3+(aq) + e− → Cr2+(aq)	− 0.41	
 	Zn2+(aq) + 2e− → Zn(s)	− 0.76	

The most powerful oxidising agent in the table is
A       Mn2+(aq)
B       Zn(s)
C       MnO(aq)
D       Zn2+(aq)
(Total 1 mark)

Q2.Use the data in the table below to answer this question.
 
 	 	E / V	
 	MnO (aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)	+ 1.52	
 	Cr2O(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l)	+ 1.33	
 	Fe3+(aq) + e− → Fe2+(aq)	+ 0.77	
 	Cr3+(aq) + e− → Cr2+(aq)	− 0.41	
 	Zn2+(aq) + 2e− → Zn(s)	− 0.76	

Which one of the following statements is not correct?
A       Fe2+(aq) can reduce acidified MnO(aq) to Mn2+(aq)
B       CrO(aq) can oxidise acidified Fe2+(aq) to Fe3+(aq)
C       Zn(s) can reduce acidified Cr2O(aq) to Cr2+(aq)
D       Fe2+(aq) can reduce acidified Cr3+(aq) to Cr2+(aq)
(Total 1 mark)

Q3.The following cell has an EMF of +0.46 V.
Cu  Cu2+  Ag+  Ag
Which statement is correct about the operation of the cell?
A        Metallic copper is oxidised by Ag+ ions.                                                
B        The silver electrode has a negative polarity.                                        
C        The silver electrode gradually dissolves to form Ag+ ions.                   
D        Electrons flow from the silver electrode to the copper electrode          
 via an external circuit.
(Total 1 mark)

Q4.          Large blocks of magnesium are bolted onto the hulls of iron ships in an attempt to prevent the iron being converted into iron(II), one of the steps in the rusting process.
Use the data below, where appropriate, to answer the questions which follow.
E/ V
                Mg2+(aq) + 2e–          Mg(s)                    –2.37
                Fe2+(aq) + 2e–          Fe(s)                    –0.44
     O2(g) + 2H2O(l) + 4e–          4OH–(aq)              +0.40
(a)     Calculate the e.m.f. of the cell represented by Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s) under standard conditions. Write a half-equation for the reaction occurring at the negative electrode of this cell when a current is drawn.
Cell e.m.f. ....................................................................................................
Half-equation ................................................................................................
......................................................................................................................
(2)

(b)     Deduce how the e.m.f. of the cell Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s) changes when the concentration of Mg2+ is decreased. Explain your answer.
Change in e.m.f. ..........................................................................................
Explanation ..................................................................................................
......................................................................................................................
(3)

(c)     Calculate a value for the e.m.f. of the cell represented by
Pt(s)|OH–(aq)|O2(g)||Fe2+(aq)|Fe(s) and use it to explain why iron corrodes when in contact with water which contains dissolved oxygen.
Cell e.m.f. ....................................................................................................
Explanation ..................................................................................................
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(2)
(Total 7 marks)

 

Q5.Use the standard electrode potential data in the table below to answer the questions which follow.
                                                                                                        E/ V
_________________________________________________________
                       Ce4+(aq) + e–          Ce3+(aq)                              +1.70
   MnO–(aq) + 8H+(aq)+ 5e–          Mn2+(aq) + 4H2O(l)               +1.51
                         Cl2(g) + 2e–          2Cl–(aq)                               +1.36
       VO2+(aq) +2H+(aq) + e–          VO2+(aq) + H2O(l)               +1.00
                       Fe3+(aq) + e–          Fe2+(aq)                               +0.77
   SO42–(aq) + 4H+(aq) + 2e–          H2SO3(aq) + H2O(l)            +0.17
_________________________________________________________
(a)     Name the standard reference electrode against which all other electrode potentials are measured.
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(1)

(b)     When the standard electrode potential for Fe3+(aq) / Fe2+(aq) is measured, a platinum electrode is required.
(i)      What is the function of the platinum electrode?
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(ii)     What are the standard conditions which apply to Fe3+(aq)/Fe2+(aq) when measuring this potential?
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(3)

(c)     The cell represented below was set up under standard conditions.
Pt|H2SO3(aq), SO42–(aq)||MnO4–(aq), Mn2+(aq)|Pt
Calculate the e.m.f. of this cell and write an equation for the spontaneous cell reaction.
Cell e.m.f. ….................................................................................................
Equation .......................................................................................................
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(3)

(d)     (i)      Which one of the species given in the table is the strongest oxidising agent?
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(ii)     Which of the species in the table could convert Fe2+(aq) into Fe3+(aq) but could  not convert Mn2+(aq) into MnO4–(aq)?
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(3)

(e)     Use data from the table of standard electrode potentials to deduce the cell which would have a standard e.m.f. of 0.93 V. Represent this cell using the convention shown in part (c).
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(2)
(Total 12 marks)

Q6.An experiment was carried out to measure the e.m.f. of this cell.
Al(s)   Al3+(aq)   H+(aq)   H2(g)   Pt(s)
(a)     The aluminium used as the electrode is rubbed with sandpaper prior to use.
Suggest the reason for this.
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(1)

(b)     Draw a labelled diagram of a suitable apparatus for the right-hand electrode in this cell. You do not need to include the salt bridge or the external electrical circuit.

 

 

(2)

(c)     A simple salt bridge can be prepared by dipping a piece of filter paper into potassium carbonate solution. Explain why such a salt bridge would not be suitable for use in this cell.
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(2)
(Total 5 marks)

Q7.The table shows some electrode half-equations and the associated standard electrode potentials.
 
Equation
number	Electrode half-equation	Eϴ / V	
1	Cd(OH)2(s)  +  2e–   Cd(s)  +  2OH–(aq)	–0.88	
2	Zn2+(aq)  +  2e–   Zn(s)	–0.76	
3	NiO(OH)(s)   + H2O(I) + e–   Ni(OH)2(s) + OH–(aq)	+0.52	
4	MnO2(s)   + H2O(I) + e–   MnO(OH)(s) + OH–(aq)	+0.74	
5	O2(g)   + 4H+(aq) +4e–   2H2O(I)	+1.23	

(a)     In terms of electrons, state the meaning of the term oxidising agent.
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(1)

(b)     Deduce the identity of the weakest oxidising agent in the table.
Explain how Eϴ values can be used to make this deduction.
Weakest oxidising agent ...............................................................................
Explanation ....................................................................................................
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(2)

(c)     The diagram shows a non-rechargeable cell that can be used to power electronic devices. The relevant half-equations for this cell are equations 2 and 4 in the table above.

(i)      Calculate the e.m.f. of this cell.
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(1)

(ii)     Write an equation for the overall reaction that occurs when the cell discharges.
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(1)

(iii)     Deduce one essential property of the non-reactive porous separator labelled in the diagram.
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(1)

(iv)     Suggest the function of the carbon rod in the cell.
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(1)

(v)     The zinc electrode acts as a container for the cell and is protected from external damage. Suggest why a cell often leaks after being used for a long time.
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(1)

(d)      A rechargeable nickel–cadmium cell is an alternative to the cell shown in part (c).
The relevant half-equations for this cell are equations 1 and 3 in the table above.
(i)      Deduce the oxidation state of the nickel in this cell after recharging is complete.
Write an equation for the overall reaction that occurs when the cell is recharged.
Oxidation state ......................................................................................
Equation ................................................................................................
...............................................................................................................
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(3)

(ii)     State one environmental advantage of this rechargeable cell compared with the non-rechargeable cell described in part (c).
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(1)

(e)     An ethanol–oxygen fuel cell may be an alternative to a hydrogen–oxygen fuel cell.
When the cell operates, all of the carbon atoms in the ethanol molecules are converted into carbon dioxide.
(i)      Deduce the equation for the overall reaction that occurs in the ethanol–oxygen fuel cell.
 
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(1)

(ii)     Deduce a half-equation for the reaction at the ethanol electrode.
In this half-equation, ethanol reacts with water to form carbon dioxide and hydrogen ions.
 
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(1)

(iii)     The e.m.f. of an ethanol–oxygen fuel cell is 1.00 V. Use data from the table above to calculate a value for the electrode potential of the ethanol electrode.
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(1)

(iv)     Suggest why ethanol can be considered to be a carbon-neutral fuel.
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(2)
(Total 17 marks)

 

Q8.This table shows some standard electrode potential data.
 
 	Electrode half-equation	Eϴ / V	
 	Au+(aq) + e−  Au(s)	+1.68	
 	O2(g) + 2H+(aq) + 2e−  H2O(l)	+1.23	
 	Ag+(aq) + e−  Ag(s)	+0.80	
 	Fe3+(aq) + e−  Fe2+(aq)	+0.77	
 	Cu2+(aq) + 2e−  Cu(s)	+0.34	
 	Fe2+(aq) + 2e−  Fe(s)	−0.44	

(a)     Draw a labelled diagram of the apparatus that could be connected to a standard hydrogen electrode in order to measure the standard electrode potential of the Fe3+ / Fe2+ electrode.
In your diagram, show how this electrode is connected to the standard hydrogen electrode and to a voltmeter. Do not draw the standard hydrogen electrode.
State the conditions under which this cell should be operated in order to measure the standard electrode potential.

 

Conditions .....................................................................................................
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(5)

(b)     Use data from the table to deduce the equation for the overall cell reaction of a cell that has an e.m.f. of 0.78 V.
Give the conventional cell representation for this cell.
Identify the positive electrode.
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(4)

(c)     Use data from the table to explain why Au+ ions are not normally found in aqueous solution.
Write an equation to show how Au+ ions would react with water.
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(3)

(d)     Use data from the table to predict and explain the redox reactions that occur when iron powder is added to an excess of aqueous silver nitrate.
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(3)
(Total 15 marks)

Q9.In a test, aqueous iron(III) ions are reduced to aqueous iron(II) ions by iodide ions. 
This reaction could be used to provide electrical energy in a cell.
(a)     The standard electrode potential for the reduction of iron(III) ions into iron(II) ions can be measured by connecting a suitable electrode to a standard hydrogen electrode. 
Draw a clearly labelled diagram to show the components and reagents, including their concentrations, in this Fe(III)/Fe(II) electrode. 
Do not draw the salt bridge or the standard hydrogen electrode.

 

 
(3)

(b)     A salt bridge is used to complete the cell. This could be prepared using potassium nitrate solution and filter paper.
State the purpose of the salt bridge. State one essential requirement of the soluble ionic compound used to make the salt bridge.
Purpose of salt bridge ....................................................................................
........................................................................................................................
Requirement ..................................................................................................
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(2)
(Total 5 marks)

Q10.An electrochemical cell is shown in the diagram. In this cell, the amount of copper in the electrodes is much greater than the amount of copper ions in the copper sulfate solutions.

(a)     Explain how the salt bridge D provides an electrical connection between the two electrodes.
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(1)

(b)     Suggest why potassium chloride would not be a suitable salt for the salt bridge in this cell.
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(1)

(c)     In the external circuit of this cell, the electrons flow through the ammeter from right to left.
Suggest why the electrons move in this direction.
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(2)

(d)     Explain why the current in the external circuit of this cell falls to zero after the cell has operated for some time.
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(1)

(e)     The simplified electrode reactions in a rechargeable lithium cell are
Electrode A    Li+ + MnO2 + e−   LiMnO2     E = −0.15 V
Electrode B    Li+ + e−  Li
Electrode B    is the negative electrode.
(i)      The e.m.f. of this cell is 2.90 V.
Use this information to calculate a value for the electrode potential of electrode B.
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(1)

(ii)     Write an equation for the overall reaction that occurs when this lithium cell is being recharged.
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(2)

(iii)    Suggest why the recharging of a lithium cell may lead to release of carbon dioxide into the atmosphere.
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(1)
(Total 9 marks)

Q11.Copper, in the form of nanoparticles of copper(II) hexacyanoferrate(II), has recently been investigated as an efficient method of storing electrical energy in a rechargeable cell.
(a)     Solar cells generate an electric current from sunlight. These cells are often used to provide electrical energy for illuminated road signs.
Explain why rechargeable cells are connected to these solar cells.
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(2)

(b)     Suggest one reason why many waste disposal centres contain a separate section for cells and batteries.
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(1)
(Total 3 marks)

Q12.One cell that has been used to provide electrical energy is the Daniell cell. This cell uses copper and zinc.
(a)     The conventional representation for the Daniell cell is
Zn(s) | Zn2+(aq) | | Cu2+(aq) | Cu(s)
The e.m.f. of this cell under standard conditions is +1.10 V.
Deduce the half-equations for the reactions occurring at the electrodes.
At Zn electrode ..............................................................................................
At Cu electrode ..............................................................................................
(2)

(b)     A Daniell cell was set up using 100 cm3 of a 1.0 mol dm–3 copper(II) sulfate solution. The cell was allowed to produce electricity until the concentration of the copper(II) ions had decreased to 0.50 mol dm–3.
Calculate the decrease in mass of the zinc electrode. Show your working.
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(3)

(c)     You are provided with the Daniell cell referred to in part (b), including a zinc electrode of known mass.
Briefly outline how you would carry out an experiment to confirm your answer to part (b).
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(3)
(Total 8 marks)

Q13.Hydrogen–oxygen fuel cells are used to provide electrical energy for electric motors in vehicles.
(a)     In a hydrogen–oxygen fuel cell, a current is generated that can be used to drive an electric motor.
(i)      Deduce half-equations for the electrode reactions in a hydrogen–oxygen fuel cell.
Half-equation 1 .....................................................................................
Half-equation 2 .....................................................................................
(2)

(ii)     Use these half-equations to explain how an electric current can be generated.
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(2)

(b)     Explain why a fuel cell does not need to be recharged.
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(1)

(c)     To provide energy for a vehicle, hydrogen can be used either in a fuel cell or in an internal combustion engine.
Suggest the main advantage of using hydrogen in a fuel cell rather than in an internal combustion engine.
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(1)

(d)     Identify one major hazard associated with the use of a hydrogen–oxygen fuel cell in a vehicle.
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(1)
(Total 7 marks)

Q14.The table below shows some standard electrode potential data.
 
 	 	E / V	
 	ZnO(s)  +  H2O(I)  +  2e–   Zn(s)  +  2OH–(aq)	–1.25	
 	Fe2+(aq)  +  2e–   Fe(s)                     	–0.44	
 	O2(g)  +  2H2O(I)   +  4e–    4OH–(aq)               	+0.40	
 	2HOCl(aq)  +  2H+(aq)  +  2e–   Cl2(g)  +  2H2O(I)	+1.64	

(a)     Give the conventional representation of the cell that is used to measure the standard electrode potential of iron as shown in the table.
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(2)

(b)     With reference to electrons, give the meaning of the term reducing agent.
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(1)

(c)     Identify the weakest reducing agent from the species in the table.
Explain how you deduced your answer.
Species...........................................................................................................
Explanation...................................................................................................
........................................................................................................................
(2)

(d)     When HOCl acts as an oxidising agent, one of the atoms in the molecule is reduced.
(i)      Place a tick (✓) next to the atom that is reduced.
 
 	Atom that is
reduced	Tick (✓)	
 	 H	 	
 	 O	 	
 	 Cl	 	

(1)

(ii)     Explain your answer to part (i) in terms of the change in the oxidation state of this atom.
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(1)

(e)     Using the information given in the table, deduce an equation for the redox reaction that would occur when hydroxide ions are added to HOCl
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(2)

(f)      The table is repeated to help you answer this question.
 
 	 	E / V	
 	ZnO(s)  +  H2O(I)  +  2e–   Zn(s)  +  2OH–(aq)	–1.25	
 	Fe2+(aq)  +  2e–   Fe(s)                     	–0.44	
 	O2(g)     +  2H2O(I)   +  4e–    4OH–(aq)               	+0.40	
 	2HOCl(aq)  +  2H+(aq)  +  2e–   Cl2(g)  +  2H2O(I)	+1.64	

The half-equations from the table that involve zinc and oxygen are simplified versions of those that occur in hearing aid cells.
A simplified diagram of a hearing aid cell is shown in the following figure.
 
(i)      Use data from the table to calculate the e.m.f. of this cell.
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Answer = .................................
(1)

(ii)     Use half-equations from the table to construct an overall equation for the cell reaction.
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(1)

(iii)    Identify which of A or B, in the figure, is the positive electrode. Give a reason for your answer.
Positive electrode ..................................................................................
Reason ..................................................................................................
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(2)

(iv)    Suggest one reason, other than cost, why this type of cell is not recharged.
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(1)
(Total 14 marks)

Q15.The table contains some standard electrode potential data.
 
 	Electrode half-equation	E / V	
 	F2  +  2e−    2F−	+2.87	
 	Au+  +  e−    Au	+1.68	
 	2HOCl  +  2H+  +  2e−    Cl2  +  2H2O	+1.64	
 	Cl2  +  2e−    2Cl−	+1.36	
 	O2  +  4H+  +  4e−    2H2O	+1.23	
 	Ag+  +  e−    Ag	+0.80	
 	Fe3+  +  e−    Fe2+	+0.77	
 	2H+  +  2e−    H2	0.00	
 	Fe2+  +  2e−    Fe	−0.44	

(a)     In terms of electrons, explain the meaning of the term oxidising agent.
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(1)

(b)     Identify the weakest oxidising agent in the table.
Explain your choice.
Weakest oxidising agent ................................................................................
Explanation ....................................................................................................
........................................................................................................................
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(2)

(c)     Write the conventional representation of the cell used to measure the standard electrode potential for the Ag+ / Ag electrode.
State the conditions necessary when measuring this value.
Conventional representation ..........................................................................
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Conditions ......................................................................................................
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(4)

(d)     Use data from the table to explain, in terms of redox, what happens when a soluble gold(I) compound containing Au+ ions is added to water.
State what you would observe.
Write an equation for the reaction that occurs.
Explanation .....................................................................................................
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Observation ....................................................................................................
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Equation .........................................................................................................
(4)

The table is repeated below to help you answer these questions.
 
 	Electrode half-equation	E / V	
 	F2  +  2e−    2F−	+2.87	
 	Au+  +  e−    Au	+1.68	
 	2HOCl  +  2H+  +  2e−    Cl2  +  2H2O	+1.64	
 	Cl2  +  2e−    2Cl−	+1.36	
 	O2  +  4H+  +  4e−    2H2O	+1.23	
 	Ag+  +  e−    Ag	+0.80	
 	Fe3+  +  e−    Fe2+	+0.77	
 	2H+  +  2e−    H2	0.00	
 	Fe2+  +  2e−    Fe	−0.44	

(e)     A cell is made by connecting Fe2+ / Fe and Ag+ / Ag electrodes with a salt bridge.
(i)      Calculate the e.m.f. of this cell.
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Answer ......................................................................
(1)

(ii)     Suggest why potassium chloride would not be suitable for use in the salt bridge of this cell.
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(1)

(f)     Use data from the table to explain what happens when a solution of iron(II) chloride is exposed to the air.
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(2)
(Total 15 marks)

Q16.          Nickel–cadmium cells are used to power electrical equipment such as drills and shavers.
The electrode reactions are shown below.
                              NiO(OH) + H2O + e–  →  Ni(OH)2 + OH–       Eο = +0.52 V
                                       Cd(OH)2 + 2e–  →  Cd + 2OH–             Eο = –0.88 V
(a)     Calculate the e.m.f. of a nickel–cadmium cell.
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(1)

(b)     Deduce an overall equation for the reaction that occurs in the cell when it is used.
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(2)

(c)     Identify the oxidising agent in the overall cell reaction and give the oxidation state of the metal in this oxidising agent.
Oxidising agent ............................................................................................
Oxidation state .............................................................................................
(2)
(Total 5 marks)

 

Q17.A disproportionation reaction occurs when a species M+ spontaneously undergoes simultaneous oxidation and reduction.
                2M+(aq) → M2+(aq) + M(s)
The table below contains E data for copper and mercury species.
 
 	 	E/ V	
 	Cu2+(aq) + e− → Cu+(aq)	+ 0.15	
 	Cu+(aq) + e− → Cu(s)	+ 0.52	
 	Hg2+(aq) + e− → Hg+(aq)	+ 0.91	
 	Hg+(aq) + e− → Hg(l)	+ 0.80	

Using these data, which one of the following can be predicted?
A       Both Cu(I) and Hg(I) undergo disproportionation.
B       Only Cu(I) undergoes disproportionation.
C       Only Hg(I) undergoes disproportionation.
D       Neither Cu(I) nor Hg(I) undergoes disproportionation.
(Total 1 mark)

Q18.          The electrons transferred in redox reactions can be used by electrochemical cells to provide energy.
Some electrode half-equations and their standard electrode potentials are shown in the table below.
 
Half-equation	Eο/V	
Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l)	+1.33	
Fe3+(aq) + e– → Fe2+(aq)	+0.77	
2H+(aq) + 2e– → H2(g)	0.00	
Fe2+(aq) + 2e– → Fe(s)	–0.44	
Li+(aq) + e– → Li(s)	–3.04	

(a)     Describe a standard hydrogen electrode.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)

(b)     A conventional representation of a lithium cell is given below.
This cell has an e.m.f. of +2.91 V
Li(s) | Li+(aq) || Li+(aq) | MnO2(s) , LiMnO2(s) | Pt(s)
Write a half-equation for the reaction that occurs at the positive electrode of this cell.
Calculate the standard electrode potential of this positive electrode.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)

(c)     Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)

(d)     A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O
(Mr = 277.9) in water and making up to 250 cm3 of solution. The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions. A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid.
Calculate the percentage of iron(II) ions that had been oxidised by the air.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(6)
(Total 14 marks)

 

Q19.In this question consider the data below.
                                                                       E / V
    Ag+(aq) + e− → Ag(s)                                +0.80
  2H+(aq) + 2e− → H2(g)                                  0.00
Pb2+(aq) + 2e− → Pb(s)                                −0.13
The e.m.f. of the cell Ag(s) | Ag+(aq) || Pb2+(aq) | Pb(s) is
A       0.93 V
B       0.67 V
C       −0.67 V
D       −0.93 V
(Total 1 mark)

Q20.In this question consider the data below.
                                                                       E / V
    Ag+(aq) + e− → Ag(s)                                +0.80
  2H+(aq) + 2e− → H2(g)                                  0.00
Pb2+(aq) + 2e− → Pb(s)                                −0.13
The e.m.f. of the cell Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s) would be increased by
A       increasing the concentration of H+(aq).
B       increasing the surface area of the Pt electrode.
C       increasing the concentration of Ag+(aq).
D       decreasing the pressure of H2(g).
(Total 1 mark)

Q21.         Use the data in the table below, where appropriate, to answer the questions which follow.
Standard electrode potentials                                                                            E/ V
Fe3+(aq) + e–		Fe2+(aq)	+0.77	
Cl2(g) + 2e–		2Cl–(aq)	+1.36	
 + 12H+ (aq) + 10e–		Br2(aq) + 6H2O(l)	+1.52	
O3(g) + 2H+(aq) + 2e– 		O2(g) + H2O(l)	+2.08	
F2O(g) + 2H+(aq) + 4e– 		2F–(aq) + H2O(l)	+2.15	

Each of the above can be reversed under suitable conditions.
(a)     (i)      Identify the most powerful reducing agent in the table.
.............................................................................................................

(ii)     Identify the most powerful oxidising agent in the table.
.............................................................................................................

(iii)     Identify all the species in the table which can be oxidised in acidic solution by BrO(aq).
.............................................................................................................
(4)

(b)     The cell represented below was set up.
Pt|Fe2+ (aq), Fe3+ (aq) || BrO (aq), Br2(aq)|Pt
(i)      Deduce the e.m.f. of this cell.
.............................................................................................................

(ii)     Write a half-equation for the reaction occurring at the negative electrode when current is taken from this cell.
.............................................................................................................

(iii)     Deduce what change in the concentration of Fe3+(aq) would cause an increase in the e.m.f. of the cell. Explain your answer.
Change in concentration .....................................................................
Explanation .........................................................................................
.............................................................................................................
.............................................................................................................
(6)
(Total 10 marks)

 

Q22.          Hydrogen–oxygen fuel cells can operate in acidic or in alkaline conditions but commercial cells use porous platinum electrodes in contact with concentrated aqueous potassium hydroxide. The table below shows some standard electrode potentials measured in acidic and in alkaline conditions.
 
Half-equation	Eο /V	
  O2(g) + 4H+(aq) + 4e–  →      2H2O(l)	+1.23	
  O2(g) + 2H2O(l) + 4e–   →      4OH–(aq)	+0.40	
2H+(aq) + 2e–   →      H2(g)	0.00	
2H2O(l) + 2e–   →      2OH–(aq) + H2(g)	– 0.83	

(a)     State why the electrode potential for the standard hydrogen electrode is equal to 0.00V.
......................................................................................................................
(1)

(b)     Use data from the table to calculate the e.m.f. of a hydrogen–oxygen fuel cell operating in alkaline conditions.
......................................................................................................................
(1)

(c)     Write the conventional representation for an alkaline hydrogen–oxygen fuel cell.
 
......................................................................................................................
(2)

(d)     Use the appropriate half-equations to construct an overall equation for the reaction that occurs when an alkaline hydrogen–oxygen fuel cell operates. Show your working.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)

(e)     Give one reason, other than cost, why the platinum electrodes are made by coating a porous ceramic material with platinum rather than by using platinum rods.
......................................................................................................................
......................................................................................................................
(1)

(f)      Suggest why the e.m.f. of a hydrogen–oxygen fuel cell, operating in acidic conditions, is exactly the same as that of an alkaline fuel cell.
......................................................................................................................
(1)

(g)     Other than its lack of pollution, state briefly the main advantage of a fuel cell over a re-chargeable cell such as the nickel–cadmium cell when used to provide power for an electric motor that propels a vehicle.
......................................................................................................................
......................................................................................................................
(1)

(h)     Hydrogen–oxygen fuel cells are sometimes regarded as a source of energy that is carbon neutral. Give one reason why this may not be true.
......................................................................................................................
(1)
(Total 10 marks)

 

Q23.Ethanedioic acid is an important industrial chemical with a number of uses.
Ethanedioate ions, C2O42–, act as bidentate ligands with transition metal ions.
(a)     Write an equation for the ligand substitution reaction of an excess of ethanedioate ions with aqueous cobalt(II) ions.
........................................................................................................................
(1)

(b)     The table below shows some standard electrode potentials.
 
 			E / V	
 	Fe3+(aq) + e–	→ Fe2+(aq)	+0.77	
 	2CO2(g) + 2e–	→ C2O42–(aq)	–0.49	

Use E values from the table to explain why an iron(III) complex is not formed when solutions containing ethanedioate ions and iron(III) ions are mixed.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 3 marks)

Q24.The table below shows some standard electrode potentials.
 
 				E / V	
 	MnO4– + 8H+ + 5e–	 	Mn2+ + 4H2O	+1.51	
 	Cl2(g) + 2e–	 	2Cl–(aq)	+1.36	
 	Cr2O72– + 14H+ + 6e–	 	2Cr3+ + 7H2O	+1.33	

A student determined the concentration of iron(II) ions in a solution of iron(II) chloride by titration with acidified potassium dichromate(VI) solution. A second student titrated the same solution of iron(II) chloride with acidified potassium manganate(VII) solution. 
By reference to the table, explain why the second student obtained a greater value for the concentration of iron(II) ions.
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
(Total 2 marks)

Q25.The diagram below shows a cell that can be used to measure the standard electrode potential for the half-reaction Fe3+(aq) + e– Fe2+(aq). In this cell, the electrode on the right-hand side is positive.

(a)     Identify solution A and give its concentration. State the other essential conditions for the operation of the standard electrode that forms the left-hand side of the cell.
Solution A .......................................................................................................
Conditions ......................................................................................................
........................................................................................................................
........................................................................................................................
(3)

(b)     Identify the material from which electrodes B are made. Give two reasons why this material is suitable for its purpose.
Material ..........................................................................................................
Reason 1 .......................................................................................................
........................................................................................................................
Reason 2 .......................................................................................................
........................................................................................................................
(3)

(c)     Identify a solution that could be used in C to complete the circuit. Give two reasons why this solution is suitable for its purpose.
Solution ..........................................................................................................
Reason 1 .......................................................................................................
........................................................................................................................
Reason 2 .......................................................................................................
........................................................................................................................
(3)

(d)     Write the conventional representation for this cell.
 
........................................................................................................................
(1)

(e)     The voltmeter V shown in the diagram of the cell was replaced by an ammeter.
(i)      Write an equation for the overall cell reaction that would occur.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(ii)     Explain why the ammeter reading would fall to zero after a time.
...............................................................................................................
...............................................................................................................
(1)
(Total 12 marks)

 

Q26.          (a)     Lithium ion cells are used to power cameras and mobile phones.
A simplified representation of a cell is shown below.
Li | Li+ || Li+, CoO2 | LiCoO2 | Pt
The reagents in the cell are absorbed onto powdered graphite that acts as a support medium. The support medium allows the ions to react in the absence of a solvent such as water.
The half-equation for the reaction at the positive electrode can be represented as follows.
Li+ + CoO2 + e–  Li+[CoO2]–
(i)      Identify the element that undergoes a change in oxidation state at the positive electrode and deduce these oxidation states of the element.
Element ..............................................................................................
Oxidation state 1 .................................................................................
Oxidation state 2 .................................................................................
.............................................................................................................
(3)

(ii)     Write a half-equation for the reaction at the negative electrode during operation of the lithium ion cell.
.............................................................................................................
(1)

(iii)     Suggest two properties of platinum that make it suitable for use as an external electrical contact in the cell.
Property 1 ..........................................................................................
Property 2 ...........................................................................................
(2)

(iv)    Suggest one reason why water is not used as a solvent in this cell.
.............................................................................................................
.............................................................................................................
(1)

(b)     The half-equations for two electrodes used to make an electrochemical cell are shown below.
ClO3–(aq) + 6H+(aq) + 6e–  Cl–(aq) + 3H2O(I)        Eο = +1.45 V
SO42–(aq) + 2H+(aq) + 2e–  SO32–(aq) + H2O(I)       Eο = +0.17 V
(i)      Write the conventional representation for the cell using platinum contacts.
.............................................................................................................
(2)

(ii)     Write an overall equation for the cell reaction and identify the oxidising and reducing agents.
Overall equation ..................................................................................
.............................................................................................................
.............................................................................................................
Oxidising agent ....................................................................................
Reducing agent ....................................................................................
(3)
(Total 12 marks)

Q27.Redox reactions occur in the discharge of all electrochemical cells. Some of these cells are of commercial value.
The table below shows some redox half-equations and standard electrode potentials.
 
 	Half-equation	Eο/ V	
 	Zn2+(aq) + 2e– → Zn(s)	–0.76	
 	Ag2O(s) + 2H+(aq) + 2e– → 2Ag(s) + H2O(I)	+0.34	
 	O2(g) + 4H+(aq) + 4e– → 2H2O(I)	+1.23	
 	F2(g) + 2e– → 2F–(aq)	+2.87	

(a)     In terms of electrons, state what happens to a reducing agent in a redox reaction.
......................................................................................................................
(1)

(b)     Use the table above to identify the strongest reducing agent from the species in the table.
Explain how you deduced your answer.
Strongest reducing agent ............................................................................
Explanation ..................................................................................................
......................................................................................................................
(2)

(c)     Use data from the table to explain why fluorine reacts with water.
Write an equation for the reaction that occurs.
Explanation ..................................................................................................
......................................................................................................................
......................................................................................................................
Equation .......................................................................................................
......................................................................................................................
(3)

(d)     An electrochemical cell can be constructed using a zinc electrode and an electrode in which silver is in contact with silver oxide. This cell can be used to power electronic devices.
(i)      Give the conventional representation for this cell.
.............................................................................................................
(2)
(ii)     Calculate the e.m.f. of the cell.
.............................................................................................................
(1)

(iii)     Suggest one reason why the cell cannot be electrically recharged.
.............................................................................................................
.............................................................................................................
(1)

(e)     The electrode half-equations in a lead–acid cell are shown in the table below.
 
 	Half-equation	Eο/ V	
 	PbO2(s) + 3H+(aq) + HSO4–(aq) + 2e– → PbSO4(s) + 2H2O(I)	+1.69	
 	PbSO4(s) + H+(aq) + 2e– → Pb(s) + HSO4–(aq)	to be
calculated	

(i)      The PbO2/PbSO4 electrode is the positive terminal of the cell and the e.m.f. of the cell is 2.15 V.
Use this information to calculate the missing electrode potential for the half-equation shown in the table.
.............................................................................................................
.............................................................................................................
(1)

(ii)     A lead–acid cell can be recharged.
Write an equation for the overall reaction that occurs when the cell is being recharged.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)

(f)      The diagrams below show how the e.m.f. of each of two cells changes with time when each cell is used to provide an electric current.

(i)      Give one reason why the e.m.f. of the lead–acid cell changes after several hours.
.............................................................................................................
.............................................................................................................
(1)

(ii)     Identify the type of cell that behaves like cell X.
.............................................................................................................
(1)

(iii)     Explain why the voltage remains constant in cell X.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(Extra space) ......................................................................................
.............................................................................................................
(2)
(Total 17 marks)

Q28.Some electrode potentials are shown in the table below. These values are not listed in numerical order.
 
                Electrode half-equation	Eϴ / V	
                            Cl2(aq)  +  2e–    2Cl–(aq)	+1.36	
2HOCl(aq)  +  2H+(aq)  +  2e–    Cl2(aq)  +  2H2O(I)	+1.64	
      H2O2(aq)  +  2H+(aq)  +  2e–    2H2O(I)	+1.77	
            O2(g)  +  2H+(aq)  +  2e–    H2O2(aq)	+0.68	
           O2(g)  +  4H+(aq)  +  4e–    2H2O(I)	+1.23	

(a)     Identify the most powerful reducing agent from all the species in the table.
........................................................................................................................
(1)

(b)     Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction.
Explanation ....................................................................................................
........................................................................................................................
........................................................................................................................
Equation .........................................................................................................
........................................................................................................................
........................................................................................................................
(2)

(c)     Suggest one reason why the redox reaction between chlorine and water does not normally occur in the absence of light.
........................................................................................................................
........................................................................................................................
(1)

(d)     Use the appropriate half-equation from the table to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)

(e)     Use data from the table to explain why one molecule of hydrogen peroxide can oxidise another molecule of hydrogen peroxide. Write an equation for the redox reaction that occurs.
Explanation ....................................................................................................
........................................................................................................................
Equation .........................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 8 marks)

 

Q29.The table shows some standard electrode potential data.
 
 	Electrode half-reaction	EΘ/V	
 	Zn2+(aq) + 2e– → Zn(s)	–0.76	
 	Co2+(aq) + 2e– → Co(s)	–0.28	
 	½O2(g) + 2H+ (aq) + 2e– → H2O(l)	+1.23	
 	Au+(aq) + e– → Au(s)	+1.68	
 	Co3+(aq) + e– → Co2+(aq)/p>	+1.82	

(a)     (i)      Identify the weakest oxidising agent in the table.
...................................................................................................................
(1)

(ii)     Give the conditions under which the electrode potential of the Zn2+/Zn electrode
is –0.76 V.
...................................................................................................................
...................................................................................................................
...................................................................................................................
(2)

(b)     Two half-cells, involving species in the table, are connected together to give a cell with an e.m.f. = +0.48 V.
Use data from the table to deduce the conventional representation of this cell. Write the half-equation for the reaction that occurs at the negative electrode.
Conventional representation ...............................................................................
.............................................................................................................................
Half-equation .......................................................................................................
(3)

(c)     Use data from the table to identify a cobalt species that can react with water.
Write an equation for the redox reaction that occurs and identify the oxidation product in the reaction.
Cobalt species .....................................................................................................
Equation ..............................................................................................................
.............................................................................................................................
.............................................................................................................................
Oxidation product ................................................................................................
(3)

(d)     Use data from the table to explain why gold jewellery is unreactive in moist air.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
(2)
(Total 11 marks)

Q30.A representation of a hydrogen–oxygen fuel cell that operates in alkaline conditions is
Pt|H2|H2O||O2|OH–|Pt
(a)     (i)      Write a half-equation for the reaction that occurs at each electrode.
Use the half-equations to deduce an overall equation for the cell.
Half-equation at positive electrode ............................................................
Half-equation at negative electrode ...........................................................
Overall equation .........................................................................................
(3)

(ii)     State and explain the effect, if any, of increasing the pressure of oxygen on the e.m.f of this cell.
Effect on e.m.f. ..........................................................................................
Explanation ................................................................................................
...................................................................................................................
(2)

(b)     Complete the diagram to show how the e.m.f. of a hydrogen–oxygen fuel cell changes with time.
 
(1)

(c)     (i)      Suggest the effect, if any, on the e.m.f. of this cell if the surface area of each platinum electrode is increased.
...................................................................................................................
(1)

(ii)     State the main environmental advantage of using a hydrogen–oxygen fuel cell to power a car.
...................................................................................................................
...................................................................................................................
(1)

(d)     Suggest why the use of a hydrogen–oxygen fuel cell might not be carbon-neutral.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
(1)
(Total 9 marks)

Q31.Fuel cells are an increasingly important energy source for vehicles. Standard electrode potentials are used in understanding some familiar chemical reactions including those in fuel cells.
The following table contains some standard electrode potential data.
 
 	Electrode half-equation	EƟ / V	
 	F2 + 2e−		2F−	+2.87	
 	Cl2 + 2e−		2Cl−	+1.36	
 	O2 + 4H+ + 4e−		2H2O	+1.23	
 	Br2 + 2e−		2Br−	+1.07	
 	I2 + 2e−		2I−	+0.54	
 	O2 + 2H2O + 4e−		4OH−	+0.40	
 	SO42− + 4H+ + 2e−		SO2 + 2H2O	+0.17	
 	2H+ + 2e−		H2	 0.00	
 	4H2O + 4e−		4OH− + 2H2	−0.83	

(a)     A salt bridge was used in a cell to measure electrode potential.
Explain the function of the salt bridge.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)

(b)     Use data from the table above to deduce the halide ion that is the weakest reducing agent.
........................................................................................................................
(1)

(c)     Use data from the table to justify why sulfate ions should not be capable of oxidising bromide ions.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)

(d)     Use data from the table to calculate a value for the EMF of a hydrogen–oxygen fuel cell operating under alkaline conditions.

 
 
EMF = ............................... V
(1)

(e)     There are two ways to use hydrogen as a fuel for cars. One way is in a fuel cell to power an electric motor, the other is as a fuel in an internal combustion engine.
Suggest the major advantage of using the fuel cell.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(Total 6 marks)

 

M1.C
[1]

M2.D
[1]

M3.A
[1]

M4.          (a)     Cell e.m.f.: 1.93 (v) CE if negative value given (1)
Half equation: Mg → Mg2+ + 2 e– (1)
or 
Ignore state symbols
Mark on after an AE
2

(b)     Change in e.m.f.: increases (1)
Mark on even if incorrect
          Explanation: Equilibrium displaced to Mg2+ or to the left (1)
cell reaction or overall reaction goes to the right
Electrode is more negative or E decreases
or gives more electron
or forms more Mg2+ ions
Mark separately
3

(c)     Cell e.m.f. : –0.84 (V) (1)
          Explanation: Fe is giving electrons or forming Fe2+
or reaction goes in the reverse direction (1)
Mark on after AE
N.B. In (a) and (c) mark on if no value given,
but CE in both (a) and (c) if e.m.f. = 0
2
[7]

 

M5.(a)     (Standard) hydrogen (electrode) (1)
1

(b)     (i)      To allow transfer of electrons / provide a reaction surface (1)
(ii)     298 K (1)
Both F3+ (aq) and Fe2+ (aq) have a concentration of 1
mol dm–3 (1) (QoL)
OR [H+] = 1 mol dm–3 
NOT zero current or 100 kPa
3

(c)     +1.34 V (1)
2 MnO4– + 5 H2SO3 → 2 Mn2+ + 5 SO42– + 3 H2O +4 H+
Correct species / order (1)
Balanced and cancelled (1)
Allow one for 2 MnO4– + 5 H2SO3 → 2 Mn2+ + 5 SO42–
3

(d)     (i)      Ce4+ (aq) (1)
(ii)     VO2+ (aq) (1); Cl2 (1)
Penalise additional answers to zero
3

(e)     Pt | Fe2+ (aq), Fe3+ (aq) || Ce4+(aq), Ce3+ (aq) | Pt
Correct species (1)
Correct order (1)
Deduct one mark for each error
2
[12]

M6.(a)     To remove the oxide layer on the aluminium
Do not allow ‘cleaning’ or ‘removal of grease’.
Do not allow ‘removal of impurities’ without qualification.
1

(b)     An appropriate method for delivering H2 gas over a Pt electrode
Need H2 gas and Pt electrode labelled (allow gas delivered directly below the electrode).
1

The Pt electrode must clearly be in contact with a solution of a named acid.
Ignore any concentration or pressure values.
Ignore absence of bubbles.
Allow if electrode is below outer acid level.
1

(c)     The carbonate ion reacts with the acid (in the SHE) / reaction between carbonate and Al3+
Lose this mark if aluminium carbonate formed but mark on.
1

Reaction given (either equation or products specified)
OR H+ / Al3+ concentrations change / cell e.m.f. altered
1
[5]

M7.(a)     Electron acceptor / gains electrons / takes electrons away
Do not allow electron pair acceptor / gain of electrons / definition of redox (QWC)
1

(b)     Cd(OH)2
Do not allow ‘Cd(OH)2/Cd’
1

Species (on LHS) with the least positive/most negative electrode potential / lowest E
/ smallest E
Only allow this mark if M1 answer given correctly or blank
Do not allow negative emf
1

(c)     (i)      1.5 (V) / 1.50
1

(ii)     2MnO2 + 2H2O + Zn →2MnO(OH) + 2OH– + Zn2+
Ignore state symbols
e– must be cancelled
(take care that Zn2+ is on RHS)
1

(iii)    Allows ions to pass (through it) or words to that effect
Penalise passage of electrons
Allow mention of particular ions
1

(iv)     Allows electrons to flow / makes electrical contact / conductor
Allow acts as an (inert) electrode / anode / cathode
1

(v)      Zn is ‘used up’ / has reacted / oxidised
Allow idea that zinc reacts
Do not allow just zinc corrodes
1

(d)     (i)      3 / +3 / III
1

2Ni(OH)2 + Cd(OH)2 → 2NiO(OH) + Cd + 2H2O
For correct nickel and cadmium species in correct order (allow H2O missing and OH– not cancelled)
1

For balanced equation (also scores M2)
Allow max 1 for M2 and M3 if correct balanced equation but reversed.
Ignore state symbols
1

(ii)     Metal / metal compounds are re-used / supplies are not depleted / It (the cell)
can be re-used
Allow does not leak / no landfill problems / less mining / less energy to extract metals / less waste
Do not allow less CO2 unless explained
1

(e)     (i)      C2H5OH + 3O2 → 2CO2 + 3H2O
Allow C2H6O
1

(ii)     C2H5OH + 3H2O → 2CO2 + 12H+ + 12e–
Allow C2H6O
1

(iii)    (+)0.23 (V)
1

(iv)     CO2 released by combustion / fermentation / fuel cell / reaction with water
Can be answered with the aid of equations
1

(atmospheric) CO2 taken up in photosynthesis
1
[17]

M8.(a)    Diagram of an Fe3+ / Fe2+ electrode that includes the following parts labelled:
Solution containing Fe2+ and Fe3+ ions
1

Platinum electrode connected to one terminal of a voltmeter
Must be in the solution of iron ions (one type will suffice)
1

Salt bridge
Do not allow incorrect material for salt bridge and salt bridge must be in the solution (ie it must be shown crossing a meniscus)
1

298 K and 100 kPa / 1 bar
1

all solutions unit / 1 mol dm−3 concentration
Allow zero current / high resistance voltmeter as alternative to M4 or M5
Ignore hydrogen electrode even if incorrect
1

(b)     Cu2+ + Fe → Cu + Fe2+
Ignore state symbols
1

Fe|Fe2+||Cu2+|Cu   correct order
Allow Cu|Cu2+||Fe2+|Fe
1
Phase boundaries and salt bridge correct, no Pt
Allow single / double dashed line for salt bridge
Penalise phase boundary at either electrode end
Can only score M3 if M2 correct
1

Copper electrode
Allow any reference to copper
1

(c)     Eϴ Au+( / Au) > Eϴ O2 ( / H2O)
Allow E cell / e.m.f. = 0.45 V
Allow 1.68 > 1.23
1

So Au+ ions will oxidise water / water reduces Au+
QoL
1

2Au+ + H2O → 2Au + O2 + 2H+
Allow multiples
1

(d)     Eϴ Ag+( / Ag) > Eϴ Fe2+( / Fe)
Allow E cell / e.m.f. = 1.24
Allow 0.80 > −0.44
1

And Eϴ Ag+( / Ag) > Eϴ Fe3+( / Fe2+)
Allow E cell / e.m.f. = 0.03
Allow 0.80 > 0.77
1

So silver ions will oxidise iron (to iron(II) ions) and then oxidise Fe(II) ions (further to Fe(III) ions producing silver metal)
Allow Ag+ ions will oxidise iron to iron(III)
1
[15]

M9.(a)     Platinum electrode
1

Solution in beaker is a mixture of named soluble iron(II) compound and named soluble iron(III) compound
Allow correct formulae for the iron compounds.
1

Concentrations of Fe(II) and Fe(III) ions are both 1 mol dm−3
Ignore any references to temperature.
If eg Fe2(SO4)3 used then concentration must be 0.5
1

(b)     Purpose: Allow movement of ions between electrodes
Allow to maintain an electric circuit.
Do not allow reference to movement of electrons in salt bridge.
1

Requirement: Must not react with the electrolyte / ions in solution
Do not allow ‘must not react’ without further qualification.
1
[5]

M10.(a)    It has mobile ions / ions can move through it / free ions
Do not allow movement of electrons.
Allow specific ions provided they are moving but do not react.
1

(b)     Chloride ions react with copper ions / Cu2+ OR [CuCl4]2− formed
If incorrect chemistry, mark = 0
1

(c)     The Cu2+ ions / CuSO4 in the left-hand electrode more concentrated
Allow converse.
1

So the reaction of Cu2+ with 2e− will occur (in preference at) left-hand electrode / Cu → Cu2+ + electrons at right-hand electrode
Allow left-hand electrode positive / right-hand electrode negative.
Also reduction at left-hand electrode / oxidation at right-hand electrode.
Also left-hand electrode has oxidising agent / right-hand electrode has reducing agent.
Allow E left-hand side > E right-hand side
1

(d)     (Eventually) the copper ions / CuSO4 in each electrode will be at the same concentration
1

(e)     (i)      −3.05 (V)
Must have minus sign.
−3.05 only.
1

(ii)     LiMnO2 → Li + MnO2 correct equation
Allow 1 for reverse equation.
Allow multiples.
1

Correct direction
If Li+ not cancelled but otherwise correct, max = 1
If electrons not cancelled, CE = 0
LiMnO2 → Li + MnO2 scores 2
Li+ + LiMnO2 → Li++ Li + MnO2 scores 1
Li + MnO2 → LiMnO2 scores 1
1

(iii)    Electricity for recharging the cell may come from power stations burning (fossil) fuel
Allow any reference to burning (of carbon-containing) fuels.
Note combustion = burning.
1
[9]

M11.(a)     Solar cells do not supply electrical energy all the time
1

Rechargeable cells can store electrical energy for use when the solar cells are not working
1

(b)     Prevent pollution of the environment by toxic or dangerous substances / recycling of valuable components
Do not allow ‘will not use up landfill sites’.
1
[3]

M12.(a)     Zn(s)  →  Zn2+(aq)  +  2e−
If equations reversed, allow M1 only.
1

Cu2+(aq)  +  2e−  →  Cu(s)
Ignore state symbols.
1

(b)     Moles of copper(II) reacted = (100 / 1000) × 0.5 = 0.05
1

Moles of zinc reacted = 0.05
1

Mass of zinc lost = 0.05 × 65.4 = 3.27 g
Correct final answer without working scores M3 only.
1

(c)     Allow cell to discharge until [Cu2+] is 0.5
Alternative: Allow cell to discharge completely.
1

Confirmed by colorimetric measurement or other suitable method
Solution colourless or use of chemical test to determine absence of copper(II)
1

Weigh the Zn electrode before and after the experiment
Weigh Zn electrodes before and after and halve the mass change.
1
[8]

M13.(a)    (i)      H2 + 2OH- → 2H2O + 2e- / H2 → 2H+ + 2e-
Any order
1

O2 + 4e- + 2H2O → 4OH- / O2 + 4H+ + 4e- → 2H2O
1

(ii)     Hydrogen (electrode) produces electrons
Ignore reference to salt bridge
Do not allow at negative / positive electrode – must identify hydrogen and oxygen
1

Oxygen (electrode) accepts electrons
Allow electrons flow to the oxygen electrode
1

(b)    Hydrogen / the fuel / reactants supplied continuously / fed in
Do not accept oxygen supplied as the only statement
1

(c)    In the fuel cell, a greater proportion of the energy available from the hydrogen–oxygen reaction is converted into useful energy
Allow less energy wasted / more efficient
Do not allow reference to safety
1

(d)    Hydrogen is flammable / H+ corrosive / OH– corrosive / hydrogen explosive
1
[7]

M14.(a)    Pt|H2|H+||Fe2+|Fe
Allow 1 for correct order of symbols but lose second mark for a wrong phase boundary(s) / Pt missing / extra Pt on RHS, additional phase boundary
Note, allow one mark only for correct symbol in reverse:
Fe|Fe2+||H+|H2|Pt
Allow dashed lines for salt bridge
Ignore state symbols
Ignore 2 if used before H+
2

(b)    Electron donor
Allow (species that) loses electrons
Do not allow reference to electron pairs
1

(c)    Cl2 / chlorine
If M1 blank or incorrect cannot score M2
1

(Species on RHS / electron donor) has most positive / largest EΘ / has highest potential
Do not allow reference to e.m.f. or E(cell)
1

(d)    (i)      Cl / chlorine
1

(ii)     Chlorine +1 to chlorine 0
CE if chlorine not identified in part (i)
Allow chlorine +1 to chlorine –1 (in Cl–)
Allow oxidation state decreases by one OR two
Allow oxidation state changes by –1 OR –2
1

(e)    4HOCl + 4H+ + 4OH-→ 2Cl2 + O2 + 6H2O
OR
4HOCl → 2Cl2 + O2 + 2H2O
Allow one mark for any incorrect equation that shows
HOCl → Cl2 + O2
Allow multiples
Ignore state symbols
Penalise one mark for uncancelled or uncombined species (eg H2O + H2O instead of 2H2O)
2

(f)    (i)      e.m.f. = 0.40 – (–1.25) = 1.65 (V) / +1.65 (V)
Allow –1.65 (V)
1

(ii)     2Zn + O2 → 2ZnO
Allow multiples
Ignore state symbols
Do not allow uncancelled species
If more than one equation given, choose the best
1

(iii)    A / stainless lid
If M1 incorrect or blank CE=0
1

O2 (electrode) has a more positive EΘ / oxygen (electrode) requires / gains electrons from external circuit
Or reference to the overall equation and a link to electrons going into A
Allow oxygen is reduced and reduction occurs at the positive electrode
OR Zinc (electrode) has more negative EΘ
Do not allow reference to e.m.f. or E(cell)
1

(iv)    (Cell) reaction(s) cannot be reversed / zinc oxide cannot be reduced to zinc by passing a current through it / zinc cannot be regenerated
Allow danger from production of gas / oxygen produced / hydrogen produced
1
[14]

M15.(a)     Electron acceptor / gains electrons
do not allow electron pair acceptor
1

(b)     Fe2+ ions
1
Fe2+ / Fe or Fe2+ or it has smallest / most negative electrode potential / E°
Do not allow Fe / Fe2+
Cannot score M2 if M1 incorrect
1

(c)     Pt|H2|H+||Ag+|Ag
M1 for H2 H+ Ag+ Ag in correct order
1
allow dashed phase boundaries
2H+ loses one mark (M2)
M2 for Pt correct and correct phase boundaries
Ignore state symbols. M1 must be correct to score M2
If answer correct but all in reverse order allow 1 mark out of two
1

Any two correct conditions
•        298 K / 25 °C
•        100 kPa
•        both solutions of unit concentration
•        zero current
Allow 1 bar
Do not apply list principle, mark correct answers.
2

(d)     E Au+( / Au) > E O2 ( / H2O) OR e.m.f. / Ecell = 0.45 V
If both species in electrode given, must be in correct order i.e. Au+ / Au
1
Au+ (ions) oxidise water OR water reduces Au+ (ions)
Allow water donates electrons to Au+
1
Gold  metal / solid / precipitate OR bubbles / effervescence of (oxygen gas) / gas produced
Penalise incorrect observations
1
2Au+ + H2O → 2Au + 2H+ + O2 OR 4Au+ + 2H2O → 4Au + 4H+ + O2
Allow multiples
1

(e)     (i)      1.24 (V)
Do not allow −1.24
1
(ii)     Chloride ions / Cl− react with / form a precipitate with silver ions / Ag+ / form AgCl
Penalise reaction of chloride ions with iron ions or iron
1

(f)     E O2 ( / H2O) > E Fe3+ ( / Fe2+) (or e.m.f / Ecell = 0.46 V)
Species in electrode if all given must be in correct order
1
Therefore the iron(II) ions are oxidised (or converted) into iron(III) ions (by oxygen)
If chloride ions oxidised to chlorine, lose M2
M2 can be obtained or lost from equation.
Ignore observations.
1
[15]

M16.          (a)     1.4 V
Allow + or –
1

(b)     2NiO(OH) + 2H2O + Cd → 2Ni(OH)2 + Cd(OH)2
Mark for species, Deduct a mark for additional species
(eg OH–) but allow balance mark
1
Balanced
If equation is reversed CE=0
1

(c)     NiO(OH) or Ni(III) or nickel
1
+3
Allow conseq on wrong species
1
[5]

 

M17.B
[1]

M18.          (a)     Hydrogen/H2 gas/bubbles
1
1.0 mol dm–3 HCl/H+
1
At 298K and 100kPa
Allow 1 bar instead of 100 kPa
Do not allow 1 atm
1
Pt (electrode)
1

(b)     Li+ + MnO2 + e– → LiMnO2
Ignore state symbols
1
–0.13(V)
1

(c)     Fe3+ ions reduced to Fe2+
Can score from equation/scheme
1
Because E(Fe3+(/Fe2+)) > E(H+/H2)/E(hydrogen)
Allow emf/Ecell +ve/0.77V
Allow Fe3+ better oxidising agent than H+
Allow H2 better reducing agent than Fe2+
Only award this explanation mark if previous mark given
1

(d)     Moles Cr2O72– =23.7 × 0.01/1000 = 2.37 × 10–4
1
1 mol Cr2O72– reacts with 6 mol Fe2+ so moles
Fe2+ in 25 cm3 = 6 × 2.37 × 10–4 = 1.422 × 10–3
1
M1 × 6
Moles Fe2+ in 250 cm3 = 1.422 × 10–2
M2 × 10 or M4/10
1
Original moles Fe2+ = 10.00/277.9 = 0.0360
Independent mark
1
Moles Fe2+ oxidised = 0.0360 – 0.0142 = 0.0218
M4 – M3
1
% oxidised = (0.0218 × 100)/0.0360 = 60.5%
(M5 × 100)/M4
Allow 60 to 61
Note Max 3 if mol ratio for M2 wrong
eg 1:5 gives 67.1%
1:1 gives 93.4%
Note also, 39.5% (39-40) scores M1, M2, M3 and M4 (4 marks)
1
[14]

 

M19.D
[1]

M20.C
[1]

M21.          (a)     (i)      Fe2+
1
(ii)     F2O
1
(iii)     Fe2+
1
Cl–
1
Use list principle if more than two answers

(b)     (i)      e.m.f. = E(rhs) – E(lhs)
1
= 1.52 – 0.77 = 0.75
(0.75 scores first mark also)
1
(ii)     Fe2+ → Fe3+ + e–
1

(iii)     Decrease
(Increase is CE, no further marks)
1
Equilibrium (or reaction) shifts to R
(or L if refers to half equation in table)
(or in favour of more Fe3+)
(or more Fe3+ formed)
(or more electrons formed)
1
Electrode potential (for Fe3+/Fe2+) less positive (or decreases)
1
[10]

 

M22.          (a)     By definition
allow ‘set to this value’
1

(b)     1.23 V
Allow + or –
1

(c)     Pt|H2(g)|OH–(aq),H2O(l)||O2(g)|H2O(l),OH–(aq)|Pt
H2O not essential, allow reverse order
Correct but with Pt missing
1
Includes Pt with correct representation
1

(d)     Uses O2 + 2H2O + 4e– → 4OH–
And (2×) 2OH– + H2 → 2H2O + 2e–
1
2H2 + O2 → 2H2O
1

(e)     Increases the surface area (so reaction faster)
1

(f)      Overall reaction is the same (2H2 + O2 → 2H2O)
Or shows e.m.f. is the same
1

(g)     Hydrogen and oxygen supplied continuously
OR
Can be operated without stopping to recharge
Or can be refuelled quickly
Allow any one mark
1

(h)     Hydrogen may need to be made using an energy source that is
not ‘carbon neutral’
1
[10]

 

M23.(a)      3C2O42− + [Co(H2O)6]2+ → [Co(C2O4)3]4− + 6H2O
Accept multiples.
Equation must have cobalt(II) hexaaqua ion.
1

(b)     Ethanedioate ion reduces iron(III) ion or
iron(III) ion oxidises ethanedioate ion
Allow answer using equations.
1

(CO2 / C2O42−) more negative than  (Fe3+ / Fe2+) or
 (Fe3+ / Fe2+) >  (CO2 / C2O42−)
or e.m.f. positive or cell voltage = +1.26
1
[3]

M24.MnO4- will oxidise the chloride ion / reaction of MnO4- and Cl- feasible
Accept converse argument with Cr2O72-
Accept calculations of overall E° values.
1

Larger volume needed
1
[2]

M25.(a)    HCl 1.0 mol dm–3
Allow H2SO4 0.5 mol dm–3
Allow HNO3 1.0 mol dm–3
Allow name or formula
Concentration can be given after “conditions”
1

(Hydrogen at) 100kPa / 1 bar
1

298 K
1

(b)     Pt / Platinum
Mark on if no answer for M1
If wrong answer for M1, only mark on if electrode is Au, Ag, Pb or Ti
1

Inert / unreactive / does not create a potential difference
1

Conducts electricity / allows electron flow / conducts / conductor
1

(c)     KCl
Allow NaCl, KNO3, Na2SO4 etc NOT NH4Cl
1

Does not react with either electrode / solution in electrode
Allow unreactive / inert
1

Ions can move
Allow conducts electricity / electrical connection / carries charge
Do not allow just connects / completes the circuit
Do not allow conducts / carries electrons
Mark these independently
1

(d)     Pt|H2|H+||Fe3+,Fe2+|Pt
Ignore state symbols
Order must be correct
| must be correct but allow | instead of , separating Fe3+ from Fe2+
Allow , instead of | separating H2 and H+
1

(e)     (i)     2Fe3+   +   H2      2Fe2+   +   2H+
Allow multiples
1

(ii)     The Fe3+ ions would be used up / reaction completed
Answer must relate to reactants in (e)(i) equation if given
Allow reactant / reactants used up
Do not allow concentration of Fe3+ decreases
Allow concentration of Fe3+ falls to zero
1
[12]

M26.          (a)     (i)      Co/Cobalt
If Co or Cobalt not given CE = 0
ignore case in symbol for Co
1
(+) 4
1
(+) 3
Allow 4 and 3 in either order
1

(ii)     Li → Li+ + e–
Ignore state symbols
Allow e without -ve sign
Do not allow equilibrium sign
1

(iii)     Platinum is a conductor
1
(Platinum is) unreactive/inert
Ignore mention of surface area or catalyst
Allow 2 marks if two properties given on one answer line
Apply list principle to contradictions/wrong answers
Do not allow platinum resists corrosion
1

(iv)    Li reacts with water/forms lithium hydroxide
Allow water breaks down (or is electrolysed) on re-charge
1

(b)     (i)      Pt│SO32– (aq), SO42– (aq)││ClO3– (aq), Cl–(aq)│Pt
State symbols an ‘,’ not necessary
Allow | in place of ‘,’ NOT ‘,’ in place of |
Ignore H+ and H2O
Deduct one mark for each mistake (e.g. Pt missed twice counts as two mistakes)
Allow reverse order for whole cell
Pt | Cl–, ClO3– || SO42–, SO32– | Pt
2

(ii)     ClO3– + 3SO32– → Cl– + 3SO42–
1
Oxidising agent ClO3–
1
Reducing agent SO32–
1
[12]

 

M27.(a)     loses electrons / donates electrons
penalise donates electron pair
1
(b)     Zn
1
(most) negative Eo / lowest Eo / least positive
can only score M2 if M1 correct
do not allow e.m.f instead of Eo
1

(c)     Eo F2 (/F–) > Eo O2 (/H2O)
or e.m.f is positive or e.m.f = 1.64 V
1
Fluorine reacts to form oxygen (can score from equation in
M3 even if equation unbalanced provided no contradiction)
or fluorine oxidises water
or fluorine is a more powerful oxidising agent than oxygen
1
2F2 + 2H2O → 4F– + 4H+ + O2
allow 4HF in equation
balanced equation scores M2 and M3
1

(d)     (i)      order correct Zn Zn2+ Ag2O Ag or reverse of this order
ignore ss , H+ and H2O, no. of moles
1
all phase boundaries correct
allow Zn|Zn2+||Ag2O,Ag
or Zn|Zn2+||Ag2O|H+|Ag for M1 & M2
e.g. Zn|Zn2+||Ag2O|Ag or Ag|Ag2O||Zn2+|Zn scores 2
M2 cannot be gained unless M1 scored
allow H+ either side of Ag2O with comma or |
for M2 penalise
•    wrong phase boundary (allow dashed lines for
     salt bridge)
•    Pt
•    use of + (from half equation)
•    water/H+ outside Ag in Ag electrode
1

(ii)     1.1 (V)
Allow no units, penalise wrong units
allow correct answer even if no answer to (d)(i) or answer to (d)(i) incorrect
allow –1.1 if silver electrode on Left in (d)(i) even if the species are in the wrong order.
1

(iii)     Reaction(s) not reversible or H2O electrolyses
do not allow hard to reverse
mention of primary cell is not enough to show that reaction(s) are irreversible
1

(e)     (i)      –0.46 (V)
Allow no units, penalise wrong units
1
(ii)     2PbSO4 + 2H2O → Pb +PbO2 + 2HSO4– + 2H+
lead species correct on correct sides of equation
1
equation balanced and includes H2O,
HSO4– and H+ (or H2SO4)
allow ions / species must be fully cancelled out or combined
allow 1/2 for balanced reverse equation
1

(f)      (i)      reagents / PbO2 / H2SO4 /acid / ions used up
(or concentration decreases)
1
(ii)     fuel cell
Ignore any other words
1
(iii)     reagents / fuel supplied continuously
1
concentrations (of reagents) remain constant
1
[17]

M28.(a)    H2O2
Ignore state symbols
1

(b)     E Ɵ Cl2/Cl– > E Ɵ O2/H2O
Allow potential for chlorine/Cl2 greater than for oxygen/O2
Allow 1.36 > 1.23 / E cell = 0.13
1

Cl2 + H2O    2Cl– + 1/2O2 + 2H+
Allow multiples
Allow + HCl
1

(c)     Activation energy is high / light/UV provides the activation energy / light breaks
chlorine molecule / Cl–Cl bond
If light used to break Cl–Cl bond award 1 mark and ignore product e.g. Cl–
1

(d)     O (–1) (in H2O2 )
Must give oxidation state of O in H2O2= –1
1

Changes to O(–2) (in water)
Must give oxidation state of O in water = –2
CE = 0/2 if refers to oxidation state of H changing
1

(e)     E Ɵ H2O2/H2O > E Ɵ O2/H2O2
Allow stated in words
Allow 1.77 > 0.68 / E cell = 1.09
1

2H2O2      O2 + 2H2O
Allow multiples
H+ and e– must be cancelled
1
[8]

M29.(a)     (i)      Zn2+
Zn2+(aq)
Apply List
1
(ii)     298 K /25°C
Ignore pressure
Ignore standard conditions
Ignore state symbols
1
(Solutions at) unit concentration / 1 mol dm-3 (of Zn2+)
Ignore references to S.H.E
1

(b)     Identifying it is the Zn/Zn2+ and Co2+/Co half cells
1
 
Correct order with phase boundaries and salt bridge correct, no Pt
If this is correct it scores M1 and M2
Allow double dashed line for salt bridge
Extra phase boundaries loses M2
Ignore state symbols
1
Zn → Zn2+ + 2e(-)
M3 independent
Allow –2e- on LHS
1

(c)     Co3+
Mark independently
1
2Co3+(aq) + H2O(l) 2Co2+(aq)+ ½ O2(g) + 2H+(aq)
Ignore state symbols
Allow multiples
1
Oxygen /O2
Allow ½ O2
1

(d)     EΘ (O2(|H2O)) electrode EΘ (Au+(|Au))
OR EΘ (Au+(|Au)) > EΘ (O2(|H2O))
OR the EΘ (Au+|Au) electrode potential is more positive than the EΘ (O2|H2O) electrode
OR The emf (for the reaction of Au and oxygen) is –0.45 V (and therefore not spontaneous)
Mark independently
1
So oxygen is unable to oxidise gold
Ignore references to water
Allow gold cannot reduce oxygen
1
[11]

M30.(a)     (i)      M1 Positive electrode O2 + 2H2O +4e(-) → 4OH-
M2 Negative electrode H2 + 2OH- → 2H2O + 2e(-)
Allow multiples, ignore state symbols
If equations both correct but at the wrong electrodes allow 1 mark
1
1
2H2 + O2 → 2H2O
Mark independently
Must be this way round
1
(ii)     Increase (emf )
If decrease/no change then CE=0/2; if blank then mark on
1
2H2 + O2 → 2H2O will move to the right
Or overall equation moves to the right
Allow O2 + 2H2O +4e- → 4OH- will move to the right / oxygen half equation moves to the RHS / EΘ O2|OH- half cell moves to the right
1

(b)     
Must start at y-axis
1

(c)     (i)      Unchanged
1
(ii)     Water is the only product / fuel cell does not give out pollutants such as NOx or CO2 or SO2 or C or CO or CxHy or unburnt hydrocarbons
Not fuel cell does not give out pollutants unless pollutant stated
1

(d)     CO2 is released because fossil fuels are burned to produce electricity to generate hydrogen
OR
CO2 is released when methane reacts with steam to produce hydrogen
Allow CO2 is released to produce the hydrogen
1
[9]

M31.(a)     The ions in the ionic substance in the salt bridge move through the salt bridge
1

To maintain charge balance / complete the circuit
1

(b)     F–
1

(c)     EƟ SO42– / SO2 EƟ Br2 / Br–
Allow correct answer expressed in words, eg electrode potential for sulfate ions / sulfur dioxide is less than that for bromine / bromide
1

(d)     1.23 (V)
1

(e)     A fuel cell converts more of the available energy from combustion of hydrogen into kinetic energy of the car / an internal combustion engine wastes more (heat) energy
1
[6]

 

E4.          This proved to be the most challenging question on the paper and candidates appeared unfamiliar with the conventional representation of cells. In part (a) many candidates were able to calculate the e.m.f. of the given cell but rather less gave a correct half-equation for the reaction occurring at the negative electrode. Some gave the overall equation for the reaction whilst others gave incorrect half-equations. Good candidates were able to deduce how a change in the concentration of Mg2+ ions would affect the cell e.m.f. and gave correct explanations for this change. Credit was given to candidates who correctly explained the effect of changing the concentration of Mg2+ ions even though the were unable to link this to the change in the e.m.f of the cell. Part (c) was not well answered. Many candidates failed to use the conventional representation of cells to deduce that, in this case, the cell e.m.f. was negative and that the reaction would proceed spontaneously in the reverse direction with iron being oxidised to Fe2+ ions.

E5.There were many good answers to this question. Virtually all candidates were able to name the standard reference electrode in part (a). In part (b) only the best candidates stated the function of the platinum electrode correctly and many failed to state in section (ii) that, for standard conditions to apply, both Fe3+ (aq) and Fe2+ (aq) would need to be at a concentration of 1 mol dm–3 in the solution. Most deduced correctly the e.m.f. of the cell given in part (c).
Many scored one further mark in part (c) when they deduced the correct molar ratio of the reacting species in the equation but lost one mark when they failed to balance the equation correctly. In part (d) most candidates were able to deduce which of the species in the table of data was the strongest oxidising agent but many only gave one of the two correct answers in section (ii). A pleasing number of candidates gave a correct cell representation in part (e). Almost all were able to select the correct species for the cell but some then used the convention, as shown in part (c), incorrectly.

E6.(a)     This part was usually well answered. The usual mistakes were to make some reference to impurities without specifying aluminium oxide or contradicting an answer by giving an incorrect formula.
(b)     Answers to this part were variable. Students familiar with the standard hydrogen electrode gave good answers, although many omitted to name the acid used.
(c)     In this part, most students realised that a reaction would occur but gave impossible products and lost the mark. The consequences of reaction were not well understood. A number of students gave answers implying that a salt bridge would not be necessary or describing details of one specific use of a standard hydrogen electrode, which did not relate to the question being asked.

E7.          Part (a) was answered well. The most common error was to give the meaning of oxidation rather than an oxidising agent. Answers to part (b) were very often incorrect. All of the species from the table were seen in students’ answers. The most common wrong answer was probably ‘cadmium’. It still seems to be the case that many students do not understand that when redox half–equations are presented in their standard form with electrons on the left, the species on the left are oxidising agents and the species on the right are reducing agents. Parts (c)(i) and (ii) were answered correctly by most students. Part (c)(iii) proved to be more testing and only the best students recognised that the porous partition was acting in the same way as a salt bridge by permitting the transport of ions. Parts (c)(iv) and (v) were answered quite well although in part (c)(v) it was not always realised that the zinc container reacts to form zinc ions whilst the cell is in operation.
Part (d)(i) proved to be a discriminating question. An oxidation state of +2 for nickel in NiO(OH) was a common error and the equation was often unbalanced or not simplified by removing hydroxide ions. Part (d)(ii) was almost always answered correctly. In comparison, parts (e)(i) and (ii) were not answered well, usually because of a lack of attention to detail. Instead of using ethanol, a common mistake was to use CH3OH. In part (e)(i), the equation for the combustion of ethanol was often unbalanced because students seemed to fail to take into account the oxygen in the ethanol. In part (e)(ii), despite being asked to write a half–equation, the electrons were often omitted or the numbers of hydrogen ions and electrons were wrong. Answers to part (e)(iii) were usually correct. In part (e)(iv), it seems that many students are still not clear about the meaning of the term carbon–neutral. The term is too often thought to refer just to carbon rather than to carbon dioxide in the atmosphere. Also, the majority of students did not make it clear that the carbon dioxide, released into the atmosphere by the processes that manufacture and use ethanol, can be balanced by the carbon dioxide taken out of the atmosphere by photosynthesis in plants.

 

E8.This proved to be a difficult question. In part (a), answers were often very scrappy and disappointing. Diagrams showed a half cell that had no chance of working. The salt bridge, if present rarely dipped into a solution and the electrode material was often iron instead of the correct material, platinum. Only very good students showed a solution containing iron(II) and iron(III) ions and only the very highest-scoring students stated clearly that all aqueous species should have a concentration of 1 mol dm-3. Answers to part (b) were more encouraging but only the better students realised that no platinum electrode are necessary with copper and iron electrodes. In parts (c) and (d), most students were unable to use data from the table correctly. For instance, it is correct to state that Eϴ Au + /Au > Eϴ O2/H2O but not correct to state that Eϴ Au + /Au > Eϴ H2O/O2 or that gold ions are better oxidising agents than water. In this reaction, water is the reducing agent. In part (d), only 6% of students recognised that silver ions would oxidise iron to iron(II) ions and then go on to oxidise iron(II) ions to iron(III) ions.

E9.(a)     The diagrams given for the half-cell in this part were occasionally poorly labelled and, too often, leniently marked. Full names or formulae of suitable iron(II) and iron(III) compounds must appear when the term ’reagent’ is requested. In a few cases, students giving iron(III) sulfate as one reagent failed to realise that 1.0 mol dm–3 solutions of this would give double the ion concentration required. When only one reagent was given, this was penalised but the concentration mark could still be awarded and this was sometimes harshly treated by some teachers.
(b)     This part was well understood by many but the use of ’movement of electrons’ appeared occasionally and was not always penalised.

E10.This was a testing question and in common with redox questions in previous years, performance by students was very mixed. In part (a), over 30% of students answered incorrectly that the salt bridge allowed electrons to flow through it. In part (b), students were expected to use their knowledge of copper chemistry to suggest that the chloride ions from the potassium chloride would react with the aqueous copper ions but only a very small percentage of students answered this correctly. Part (c) also proved to be a very demanding question. Only the best students gave a correct answer involving the higher concentration of copper ions in the left hand electrode. It was surprising that many students appeared to consider that the 0.2 mol dm−3 solution of CuSO4 was more concentrated than the 1.0 mol dm−3 solution. Part (d) was very discriminating. The great majority of students answered incorrectly by stating that one of the reagents would be used up or had reacted completely. Answers to part (e) were much better. When students are asked to write an equation for an overall reaction, they should take care to cancel out identical species that appear on both sides of the final equation. In part (e)(ii), many students lost a mark because they failed to cancel out a lithium ion that appeared on both sides of the equation. In part (e)(iii), it was surprising to note that a considerable number of students thought that carbon dioxide would be produced by a reaction with an imaginary carbon electrode in the lithium cell.

E11.In this part (a), there were many examples of full marks being awarded for vague, unclear or confusing answers. Unless the student makes it very clear which type of cell they are referring to in their answer they cannot gain full credit. Lenient marking was often seen here. The ideas in this part (b), however, were well understood.
Despite the long catalogue of points listed above, this was, overall, another very positive session in many schools and colleges. The new administrative arrangements were applied well and schools and colleges are thanked for this. Teachers in schools and colleges have, in the main, successfully and securely delivered the practical assessment and have impressed with the accurate and consistent standard of their marking. In my first report as Principal Moderator, I am grateful for the opportunity to salute their professionalism and thank them for their integrity.

E12.Lots of correct answers were seen for in part (a) and (b) but part (c) caused many more problems with sketchy answers sometimes being credited with full marks. It was not clearly understood that a method for determining the point at which the copper(II) concentration was 0.5 mol dm–3 was needed. Calorimetry was a common error. A novel approach was to wait until the cell potential had fallen to half its original value but this is incorrect chemistry and so should not have been credited. This is a good example of where reference to the Assessment Adviser would be helpful.

E13.Answers to parts of this question were generally disappointing. In part (a)(i), the half-equation for the hydrogen electrode was usually correct but most students gave an incorrect equation for the reaction at the oxygen electrode involving the formation of oxide ions. Part (a)(ii) proved to be difficult and only the highest-scoring students were able to address the question properly by using the half equations to explain how a flow of electrons is generated in the external circuit.
Part (b) was answered well but answers to part (c) were mostly disappointing. Students appeared not to know that the major advantage of a fuel cell over a heat engine is its greater efficiency. It is difficult for examiners to understand why most students thought that the burning of hydrogen in an internal combustion engine would lead to formation of carbon dioxide and hence exacerbate global warming.

E14.Few correct answers were seen to part (a). Some students did not comply with the convention of placing the hydrogen electrode on the left when measuring the standard electrode potential. Some failed to include all the phase boundaries, putting a comma between H2 and H+ instead of a solidus, whilst other students included too many phase boundaries at the extremities or an incorrect, additional platinum connected to iron. Parts (b) and (c) were answered well. All the elements were chosen by students in part (d)(i) and where the correct answer of chlorine was selected, the oxidation states in part (d)(ii) and the equation in part (e) were usually correct. It was pleasing to find that answers to parts (f)(i) and (f)(ii) were usually correct. Most students identified the positive electrode correctly in part (f)(iii) but only the highest-scoring students used an appropriate electrode potential from the table to explain their answer.

E15.Most students gave a correct answer to part (a) but some continued to confuse an oxidising agent with a Lewis acid and incorrectly stated that an oxidising agent is an electron pair acceptor. Inaccurate terminology continues to disadvantage students when they are asked to use electrode potential data to explain the action of oxidising and / or reducing agents. For example, in part (b), the iron (II) ion was the weakest oxidising agent in the Table because the Eθ value for the Fe2+ / Fe electrode is the least positive (or most negative). It is not precise enough to refer to the electrode potential of iron. In a similar way when gold (I) ions are added to water, the water is oxidised because the electrode potential of the Au+ / Au electrode is more positive than that of the O2 / H2O electrode. It is correct in this context to say that Au+ ions are a better oxidising agent than O2 but it is not correct to say that Au+ ions are a better oxidising agent than water. Parts (c) and (e)(i) were answered well by many students. Many answers to part (e)(ii) stated incorrectly that chloride ions would react with iron (II) ions. In part (f) similar imprecise use of the correct Eθ notation led to no marks being awarded for the use of data.

E16.          This question was answered well except for part (b) where only the best candidates were able to write a balanced equation with the oxidising and reducing agent in the correct position.

E18.          This was the most difficult question in the paper. In part (a) most candidates knew that a standard hydrogen electrode uses hydrogen gas and a platinum electrode but only a minority went on to state the correct and necessary conditions of concentration, temperature and pressure for operation of the electrode. In part (b) most candidates gave a correct value for the electrode potential of the positive electrode but very few candidates were able to use the cell representation to deduce a half-equation for the electrode reaction. Answers to part (c) were also disappointing. Despite the given electrode potentials and half-equations, most candidates could not predict and explain the redox reaction between hydrogen and iron(III) ions. Part (d) also proved to be a difficult question. Candidates were usually able to score three marks out of six for a correct calculation of the number of moles of dichromate(VI) ions, a calculation of the original number of moles of iron(II) sulfate and for recognising the involvement of a factor of ten. Only the very best candidates realised that the number of moles of iron ions was six times the number of moles of dichromate(VI) ions and that a subtraction was required to calculate the percentage of iron(II) ions that had been oxidised by air before the titration.

 

E21.          This question discriminated very effectively between candidates. In part (a) weak candidates were confused about reducing and oxidising agents. These agents were confused with the species that could be reduced and oxidised. Thus for part (a)(i) Fe3+ was a very common wrong answer and for part (a)(ii) the F– ion was often suggested. Similarly in part (a)(iii) species were often incorrectly chosen from the left-hand side of the redox half-equations. Parts (b)(i) and (b)(ii) were usually correct but candidates were less successful with part (b) (iii). About half of all candidates gave the incorrect answer, an increase in the concentration of Fe3+ ions. For those candidates who gave a correct change in concentration only the best gained two marks for the explanation. Most omitted to mention that the electrode potential of the Fe3+/Fe2+ electrode would be reduced.

E22.          This question covered a new area of the Specification although the principles of redox that were involved have been tested for many years. Candidates have always found difficulty with the understanding of these principles when applied to electrochemical cells. This question discriminated very effectively between candidates. In part (a) only about 20% of candidates knew that the zero electrode potential of the hydrogen electrode is an arbitrary definition. In part (c) candidates had great difficulty with the conventional representation of the alkaline fuel cell. The most common mistakes were to give the oxidised and reduced forms in the wrong order and to use the hydrogen half-equation in acidic rather than alkaline conditions. Similar mistakes were made in part (d) although 39% of candidates did score full marks. About half of all candidates scored full marks in parts (g) and (h) but only 23% of candidates gained the mark in part (f). Many candidates opted not to answer this part.

E23.(a)     Many candidates failed to score the mark in this part, an incorrect charge for the complex being the usual error.
(b)     Answers to this part were disappointing. Most candidates had some idea of the answer but could not explain themselves clearly. Many centres awarded marks for answers that made little sense.

E24.This Question proved very demanding as expected. Many candidates provided some Eθ calculations without any explanation of the significance of the results, and could not be given any marks. Confusion between chlorine and chloride ion was very common, often rendering the answer meaningless. A fair number of candidates thought that the oxidation of chloride ions took place instead of the oxidation of iron(II) ions. Full marks were quite rare.

E25.In part (a), only about 20% of students scored all three marks. In order to gain full marks it was necessary to identify solution A as, for example, hydrochloric acid not just as H+ ions. Part (b) was answered well. Answers to part (c) very rarely scored all three marks. The most common error was to suggest that the salt bridge allowed transfer of electrons rather than movement of ions. Answers to parts (d), (e)(i) and (e)(ii) were pleasing and about half of all students gained the mark in each case.

 

E26.          Redox questions involving electrochemical cells are usually found to be very difficult. It was therefore very pleasing to find that this question was answered well in all parts except for parts (a)(iv) and (b)(i). In part (a)(iv) candidates were expected to recall from Unit 1 that lithium reacts with water but only about 20% were able to do this and these were not necessarily the best candidates. As expected, part (b)(i) proved to be discriminating. Although about one third of candidates scored both marks, others failed to include platinum or put the various ions in the wrong order.

E27.As usual, this question that tested candidates’ understanding of redox and cells proved to be difficult but it was pleasing to note that even the weaker candidates were able to gain a significant number of marks. Parts (a) and (b) were answered well. Many candidates were able to score two out of three marks for part (c). The mark lost was usually due to incorrect use of data from the table. Many candidates stated that the electrode potential values showed that fluorine is a better oxidising agent than water. In this case, water is the reducing agent and candidates should have stated that the data shows that fluorine is a better oxidising agent than oxygen. In part (d)(i), a majority of candidates scored one mark out of the two available; only the best candidates showed metal species on the outside of the cell and all the phase boundaries, correctly indicated. Parts (d)(ii) and (e)(i) were answered well but only about a third of candidates gave a correct answer to part (d)(iii). In part (e)(ii) marks were lost because candidates did not simplify the overall equation after combining the half equations or the equation was for discharge rather than for recharging. As expected, part (f) proved to be challenging, particularly part (f)(iii). Fuel cells were included in the Specification as an example of ‘How Science Works’. Their value is that they can produce efficiently a steady current to drive electric motors because the concentrations of the reactants remain constant when the fuel and the oxidant are supplied continuously.

E28.Part (a) was answered well. In part (b), answers were disappointing. Only about 20% of students used data from the table correctly. An effective way of using the data would have been to state that the electrode potential for the chlorine/chloride half equation is more positive than for the oxygen/water half equation. However, in many cases the species involved in the half equation were not mentioned or were in the wrong order. Answers to part (c) were often correct. Answers to part (d) were often disappointing and many students suggested that the oxidation state of hydrogen rather than oxygen was changing. Part (e) required similar understanding and use of electrode potentials as in part (b) and responses were similarly disappointing.

 

E29.The correct answer was given by the majority of the students in part (a)(i). Part (a)(ii) was well done although a few students could not state the correct units of concentration. In part (b) the two half-cells were generally identified but the cell was often reversed as was the equation given. In part (c) the correct cobalt species was generally identified as the oxidising agent but the equation was often not balanced and the oxidation product given as Co2+ or H+. Students found part (d) challenging. Very few gave E⊖ with the correct terminology; very often this was given the wrong way round e.g. E⊖ Au/Au+. Most students thought, incorrectly, that the reaction was between gold and water. The E⊖ − values should have indicated to them that a redox reaction between gold and water is not feasible.

E30.Only the more able students scored well in part (a)(i). Many students failed to give the correct equations at the electrodes and the overall equation was often not cancelled down to the simplest form. Most students scored the first mark in part (a)(ii) but only a very few students referred to an equation to score M2. Most students could draw the correct line on the graph in part (b) although several assorted incorrect curves were seen. The effect of changing the surface area was well known in part (c)(i). Some students failed to gain credit in part (d) because their answers were vague and did not identify pollutants. In part (e) a statement about carbon dioxide being released in the generation of hydrogen was required to score the mark; some students did not identify hydrogen as the fuel in the fuel cell and described general ways that carbon dioxide would be released when the fuel-cell was made e.g. transportation of hydrogen etc.