Electrochemistry: A Level Chemistry questions with answers
Free Electrochemistry questions for A Level Chemistry (9701), part of Physical chemistry. Each question has the answer and the marking points an examiner would look for.
A cell is made from the Fe2+/Fe and Ag+/Ag half-cells. E⦵(Fe2+/Fe) = −0.44 V; E⦵(Ag+/Ag) = 0.8 V.
(a)
Calculate E⦵cell, identify the positive electrode and write the overall equation.
[3]
(b)
The concentration of ions in the Ag+/Ag half-cell is reduced to 0.01 mol dm−3. Use the Nernst equation to calculate the new electrode potential of this half-cell and the new Ecell.
[3]
(c)
Copper is deposited by electrolysis of CuSO4(aq) using a current of 2 A for 20 minutes. Calculate the mass of copper deposited. (F = 96 500 C mol−1)
[3]
(d)
Explain why a reaction with a positive E⦵cell may not be observed in practice.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
+1.24 V; Ag positive
1.24 V B1 Ag electrode positive B1 balanced overall equation B1
3
(b)
0.68 V; 1.12 V
E = E⦵ + (0.059/z) log [ox] M1 correct z A1 new EcellA1
3
(c)
0.79 g
Q = It in seconds C1 ÷ 2F C1 0.79 A1
3
(d)
high activation energy / slow rate; non-standard conditions
kinetics: high activation energy, too slow B1 non-standard conditions change E B1
2
Question 2Structured · 13 marks
A cell is set up from a Zn²⁺/Zn half-cell and a Ni²⁺/Ni half-cell under standard conditions. E⦵(Zn²⁺/Zn) = −0.76 V; E⦵(Ni²⁺/Ni) = −0.25 V.
(a)
State the standard conditions for an electrode potential measurement.
[2]
(b)
Calculate the standard cell potential.
[1]
(c)
Write the equation for the overall cell reaction.
[2]
(d)
Use your answer to (b) to calculate ΔG⦵ for the reaction. (F = 96 500 C mol−1)
[2]
(e)
The concentration of Ni²⁺ in its half-cell is reduced to 0.010 mol dm⁻³. State and explain the effect on the cell potential.
[2]
(f)
Describe the standard hydrogen electrode.
[3]
(g)
Suggest why a salt bridge made from filter paper soaked in potassium nitrate is used rather than a wire.
correct species B1 correct direction (metal of the more negative half-cell and the ion of the more positive one as reactants) and balanced, no electrons shown B1
2
(d)
−98.4 kJ mol−1
ΔG = −nFE with n = 2 C1 −98.4 kJ mol−1A1
2
(e)
E decreases
the electrode potential of the Ni²⁺/Ni half-cell becomes less positive (equilibrium shifts to the left) B1 so Ecell decreases B1
2
(f)
Pt electrode; H₂ at 100 kPa; 1 mol dm⁻³ H⁺; 298 K
Any 3 of: platinum electrode (platinised); hydrogen gas at 100 kPa (allow 101 kPa); 1.00 mol dm−3 H⁺ (e.g. HCl); at 298 K B3
3
(g)
allows ions to move, completes circuit without reacting
it allows ions to flow between the half-cells to complete the circuit without metal electrode reactions / KNO₃ does not react or precipitate with the ions B1
1
Question 3Structured · 10 marks
A student electrolyses aqueous copper(II) sulfate with copper electrodes. A current of 3.0 A is passed for 60 minutes and the cathode gains 3.482 g.
(a)
Write the half-equation at the cathode and at the anode.
[2]
(b)
Calculate the charge passed.
[1]
(c)
Use your answer to (b) and the mass gained to calculate a value for the Avogadro constant. (charge on electron 1.60 × 10−19 C; Aᵣ Cu 63.5)
[4]
(d)
Explain why the colour of the solution does not change during this electrolysis.
[2]
(e)
Suggest why the experimental value of L may differ from 6.02 × 1023 mol−1.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
Cu²⁺ + 2e⁻ → Cu; Cu → Cu²⁺ + 2e⁻
cathode: Cu²⁺ + 2e⁻ → Cu B1 anode: Cu → Cu²⁺ + 2e⁻ B1
2
(b)
10800 C
Q = It = 10800 C B1
1
(c)
6.15 × 1023 mol−1
electrons = Q/e = 6.75 × 1022M1 moles Cu = 0.0548 M1 moles electrons = 2 × moles Cu M1 L = 6.15 × 1023 mol−1A1
4
(d)
Cu²⁺ removed at cathode replaced at anode at same rate
copper dissolves from the anode as Cu²⁺ at the same rate as Cu²⁺ is deposited at the cathode B1 so [Cu²⁺] stays constant B1
2
(e)
cathode not completely dry / current fluctuates / copper lost on washing
e.g. cathode not fully dried, current not constant, or copper flaking off B1
1
Question 4Structured · 12 marks
This question is about electrochemistry.
(a)
Define standard electrode potential.
[3]
(b)
A cell is made from Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V). Calculate E°cell and write the overall equation.
[3]
(c)
Predict, using E° values, whether acidified MnO₄⁻ (E° = +1.51 V) can oxidise Cl⁻ to Cl₂ (E° = +1.36 V). Explain.
[2]
(d)
Explain why a reaction predicted feasible by E° values may not be observed.
[1]
(e)
A current of 1.00 A passes through aqueous copper(II) sulfate for 30 minutes. Calculate the mass of copper deposited. (F = 96 500 C mol−1)
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
e.m.f. of half-cell relative to standard hydrogen electrode; 298 K, 1 mol dm⁻³, 101 kPa
the e.m.f. of a half-cell compared with a standard hydrogen electrode B1 under standard conditions: 298 K, 1.00 mol dm−3 solutions, 101 kPa B2
3
(b)
+1.10 V; Zn + Cu²⁺ → Zn²⁺ + Cu
E°cell = 0.34 − (−0.76) M1 +1.10 V A1 Zn + Cu²⁺ → Zn²⁺ + Cu A1
3
(c)
yes; E°cell = +0.15 V positive
yes B1 E°cell = 1.51 − 1.36 = +0.15 V, which is positive so the reaction is feasible B1
2
(d)
high activation energy (slow) / non-standard conditions
the rate may be very slow (high activation energy) / conditions are not standard B1
1
(e)
0.592 g
Q = It = 1800 C M1 moles Cu = Q/2F M1 0.592 g A1
3
Question 5Structured · 3 marks
A current of 1.6 A is passed through aqueous copper(II) sulfate for 39 minutes. Calculate the mass of copper deposited. (F = 96 500 C mol−1, Ar Cu = 63.5)
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
1.23 g
Q = It C1 mol e− ÷ 2 C1 1.23 A1
3
Question 6Structured · 11 marks
An electrochemical cell is made from the Ag⁺/Ag half-cell (E° = +0.80 V) and the I₂/I⁻ half-cell (E° = +0.54 V).
(a)
Draw a labelled diagram of the standard hydrogen electrode.
[3]
(b)
Calculate the standard cell potential and identify the negative electrode.
[2]
(c)
State the purpose of the salt bridge and suggest a suitable salt.
[2]
(d)
Predict, with a reason, the effect on Ecell of diluting the solution in the positive half-cell.
[2]
(e)
Explain why fuel cells are more efficient than combustion engines.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
H₂ at 101 kPa, 1 mol dm⁻³ H⁺, platinum (black) electrode, 298 K
H₂ gas at 101 kPa bubbling over a platinum electrode B1 1.00 mol dm−3 H⁺ (aq) B1 298 K and salt bridge / voltmeter B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in