tutorial questions on electrochemistry

7/29/2019 Tutorial Questions on Electrochemistry

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CHM1312/ Tutorial on Electrochemistry (A2 syllabus)

The diagram below shows the apparatus used for determining the standard electrode potential of a

solution of an element in different oxidation states.

1 a Name the pieces of apparatus and chemicals labelled A to E. [5]

b i What are the two functions of part E? [2]

ii Explain how part E is prepared. [3]

2 Give the contents of the left-hand beaker required to determine the electrode potentials for the

following half-cell reactions:

a Br2(aq) + 2e 2Br

(aq) [2]

b MnO4(aq) + 8H+(aq) + 5eMn

2+(aq) + 4H2O(l) [3]

c Fe3+

(aq) + e

Fe2+

(aq) [2]

d Cr3+(aq) + e Cr2+(aq) [2]

3 Using the data sheet on standard electrode potentials, calculate the standard cell potentials

of the following cells:

a Cl2/Cl

and Fe2+/Fe [1]

b Pb2+/Pb and Ag

+/Ag [1]

c Zn2+/Zn and Cu

2+/Cu [1]

d Fe3+/Fe

2+and Cl2/Cl

[1]

e Sn4+/Sn

2+and MnO4

/Mn2+

[1]

4 For each cell in question 3, give the negative and positive terminals. [5]

5 For the cells featured in question 3, give the equation for the cell reaction taking place. [10]

6 Using standard electrode potentials, explain whether or not the following reactions can

take place under standard conditions:

a 2Fe3+

+ Zn 2Fe2+

+ Zn2+

[3]

b H2 + Zn2+

2H+

+ Zn [3]

c 2H+

+ Pb H2 + Pb2+

[3]

d I2 + 2Fe2+

2I

+ 2Fe3+

[3]

e Ni + 2Ag+ Ni

2++ 2Ag [3]

7 The diagram below shows an electrochemical cell involving two metal/metal-ion systems.


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a Using the data sheet, calculate the cell voltage. Show your working. [2]

b Give the balanced ionic equation for the overall cell reaction. [2]

c In this reaction:

i Which substance is oxidised? Explain your answer. [1]ii Which substance is reduced? Explain your answer. [1]

iii In which direction do electrons flow? [1]

8 The diagram below shows an electrochemical cell.

a What is the pressure of the chlorine gas under standard conditions? [1]

b Name the apparatus labelled A and give its function in the cell. [3]

c i Give the two ionic equations for the reactions happening in the two half-cells. [2]

ii Give the equation for the overall cell reaction occurring between chlorine and

Fe2+

ions. [1]

d Using the values for the standard electrode potentials for these two half-reactions fromthe data sheet, calculate the voltage of the cell. Show your working. [2]

9 When a solution of copper(I) sulfate is left to stand, a red solid appears and the solution

changes from colourless to blue. Explain these observations.

Your should:

identify the products use the electrode potentials shown on the data sheet to explain the reaction taking place name the type of reaction taking place give a balanced ionic equation for the reaction. [6]


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CHM1312/MS Tutorial on Electrochemistry (A2 syllabus)

1 a A = hydrogen gas at 1 atm pressure (101 kPa) [1]

B = 1.00 mol dm3 hydrochloric acid [1]

C = platinum electrode [1]

D = platinum electrode [1]

E = salt bridge [1]

b i to complete the circuit [1]

to maintain an ionic balance [1]

ii soak filter paper [1]

in saturated [1]

potassium nitrate solution [1]

2 a 1.00 mol dm3 Br2(aq) [1]

1.00 mol dm3

Br

(aq), e.g. KBr(aq) [1]

b 1.00 mol dm3 KMnO4(aq) [1]

1.00 mol dm3

Mn2+

(aq), e.g. MnSO4(aq) [1]0.5 mol dm3 H2SO4or1.00 mol dm

3 HNO3 (not HCl) [1]

c 1.00 mol dm3

FeCl3(aq) orFe(NO3)3(aq) [1]

1.00 mol dm3 Fe2+(aq), e.g. FeSO4(aq) [1]

d 1.00 mol dm3

Cr3+

(aq), e.g. CrCl3 [1]

1.00 mol dm3

Cr2+

(aq), e.g. CrCl2 [1]

3 a Eo

cell = +1.36 (0.44) = +1.80 V [1]

b Eo

cell = +0.80 (0.13) = +0.93 V [1]

c Eo

cell = +0.34 (0.76) = +1.10 V [1]

d Eo

cell = +1.36 (+0.77) = +0.59 V [1]

e Eo

cell = +1.52 (+0.15) = +1.37 V [1]

4 a The Cl2/Cl

(platinum electrode) is the positive; Fe2+/Fe is negative. [1]

b The Ag+/Ag is the positive; Pb2+/Pb is negative. [1]

c The Cu2+/Cu is the positive; Zn2+/Zn is negative. [1]

d The Cl2/Cl

(platinum electrode) is the positive; Fe3+/Fe

2+is negative. [1]

e The MnO4/Mn2+ (platinum electrode) is the positive; Sn4+/Sn2+ is negative. [1]

5 a Cl2 + Fe Fe2+

+ 2Cl

correct reactants and products [1]

correctly balanced [1]

b Pb + 2Ag+ Pb2+ + 2Ag

correct reactants and products [1]correctly balanced [1]

c Zn + Cu2+ Zn2+ + Cu



d Cl2 + 2Fe2+

2Fe3+

+ 2Cl



e 2MnO4 + 16H+ + 5Sn2+ 2Mn

2+ + 8H2O + 5Sn4+




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6 a It will take place because the redox equilibrium 2e

+ Zn2+

Zn has a more negative

standard electrode potential than e

+ Fe3+

Fe2+

[1]

This means that the first equilibrium has a greater tendency to lose electrons [1]

therefore it can proceed to the left and the second one can proceed to right. [1]

b It will not take place because the redox equilibrium 2e

+ 2H+ H2 has a more positive standard

electrode potential than 2e + Zn2+ Zn [1]

This means that the first equilibrium has a greater tendency to accept electrons. [1]

Therefore it cannot proceed to left and hence the second one cannot proceed to right. [1]

c It will take place because the redox equilibrium 2e + Pb2+ Pb has a more negative

standard electrode potential than 2e

+ 2H+ H2 [1]


therefore it can proceed to left and the second one can proceed to right. [1]

d It will not take place because the redox equilibrium e + Fe3+ Fe2+ has a more positive standard

electrode potential than 2e + I2 2I [1]

This means that the first equilibrium has a greater tendency to accept electrons. [1]Therefore it cannot proceed to left and hence the second one cannot proceed to right. [1]

e It will take place because the redox equilibrium 2e + Ni2+ Ni has a more negative

standard electrode potential than e

+ Ag+ Ag [1]


therefore it can proceed to left and the Ag+/Ag equilibrium can proceed to right. [1]

9 The two relevant half-cell reactions and standard electrode potentials are:

Cu2+

+ e

Cu+ Eo = +0.15 V

Cu+ + e Cu Eo = +0.52 V [1]both are needed for the 1 mark


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The Cu2+/Cu+ half-equation has the more negativeEo value; [1]

therefore Cu+

(first half-equation) releases electrons to reduce Cu+

(second half-equation) or

the first half-equation proceeds to the left, and the second half-equation proceeds to the right. [1]

The Cu+

disproportionates oris both oxidised and reduced [1]

to give Cu (copper metal, the red solid) and Cu2+

ions (the blue solution). [1]The overall reaction is 2Cu

+(aq) Cu(s) + Cu

2+(aq). [1]

tutorial questions on electrochemistry

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