Electrode potentials and electrochemical cellsAQA A-Level Chemistry: Topic test

20 questions, 54 marks

AQA A-Level Chemistry

Electrode potentials and electrochemical cells topic test

Total 54 marks

Name

Class

Date

1
A student sets up a cell from an iron half-cell and a tin half-cell connected by a salt bridge. Standard electrode potentials: Fe²⁺(aq) + 2e⁻ ⇌ Fe(s), E° = −0.44 V; Sn²⁺(aq) + 2e⁻ ⇌ Sn(s), E° = −0.14 V.
(a)
What is the EMF of this cell under standard conditions?
[1 mark]
A+0.58 V
B+0.30 V
C−0.30 V
D−0.58 V
(b)
Which set of conditions applies to a standard electrode potential?
[1 mark]
A273 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions
B298 K, 100 kPa and 0.100 mol dm⁻³ solutions of ions
C298 K, 1000 kPa and 1.00 mol dm⁻³ solutions of ions
D298 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions
(c)
Write the conventional representation of this cell, with the negative electrode on the left.
[2 marks]
Total for question 1: 4 marks
2
A car battery is a lead–acid cell with sulfuric acid as the electrolyte. The electrode reactions, written as reductions, are: PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l), E° = +1.69 V; PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq), E° = −0.36 V.
(a)
Which equation represents the reaction at the negative electrode when the cell is discharging?
[1 mark]
APbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq)
BPbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
CPb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
DPb(s) → Pb²⁺(aq) + 2e⁻
(b)
What happens to the concentration of the sulfuric acid as the cell discharges?
[1 mark]
AIt decreases, because sulfate and hydrogen ions are used up at the electrodes
BIt stays the same, because the electrolyte is not a reactant
CIt increases, because sulfate ions are released at the negative electrode
DIt falls to zero immediately
(c)
State what is meant by a rechargeable cell and explain why the lead–acid cell can be recharged.
[2 marks]
Total for question 2: 4 marks
3
Vanadium forms ions in several oxidation states. Standard electrode potentials: VO₂⁺(aq) + 2H⁺(aq) + e⁻ ⇌ VO²⁺(aq) + H₂O(l), E° = +1.00 V; VO²⁺(aq) + 2H⁺(aq) + e⁻ ⇌ V³⁺(aq) + H₂O(l), E° = +0.34 V; V³⁺(aq) + e⁻ ⇌ V²⁺(aq), E° = −0.26 V; Sn⁴⁺(aq) + 2e⁻ ⇌ Sn²⁺(aq), E° = +0.15 V; Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), E° = −0.76 V.
(a)
Use the data to explain why zinc in acidic solution can reduce VO₂⁺ all the way to V²⁺.
[3 marks]
(b)
Tin(II) ions are used instead of zinc to reduce VO₂⁺ in acidic solution. Use the data to deduce the lowest oxidation state of vanadium formed and write the equation for the first reduction step.
[4 marks]
Total for question 3: 7 marks
4
A student sets up a cell from an aluminium half-cell and a nickel half-cell, joined by a salt bridge of filter paper soaked in aqueous potassium nitrate. Each metal is in contact with a 1.00 mol dm⁻³ solution of its ions at 298 K. Standard electrode potentials: Al³⁺(aq) + 3e⁻ ⇌ Al(s), E° = −1.66 V; Ni²⁺(aq) + 2e⁻ ⇌ Ni(s), E° = −0.25 V.
(a)
Write the half-equation at each electrode when the cell delivers a current, write the overall equation, and calculate the EMF of the cell. State the direction of electron flow in the external circuit and the purpose of the salt bridge.
[6 marks]
(b)
The student repeats the experiment with 0.0100 mol dm⁻³ Ni²⁺(aq) in the nickel half-cell and everything else unchanged. Predict how the EMF compares with 1.41 V and explain your answer using the equilibrium at the nickel electrode. State why the voltmeter must have a very high resistance, and state the conditions under which standard electrode potentials are measured.
[6 marks]
Total for question 4: 12 marks
5
A zinc–air button cell is used in hearing aids. The electrolyte is aqueous potassium hydroxide and air enters the cell through a small hole. The electrode reactions, written as reductions, are: ZnO(s) + H₂O(l) + 2e⁻ ⇌ Zn(s) + 2OH⁻(aq), E° = −1.26 V; O₂(g) + 2H₂O(l) + 4e⁻ ⇌ 4OH⁻(aq), E° = +0.40 V.
(a)
Which equation represents the reaction at the negative electrode when the cell is in use?
[1 mark]
AO₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
BZnO(s) + H₂O(l) + 2e⁻ → Zn(s) + 2OH⁻(aq)
CZn(s) → Zn²⁺(aq) + 2e⁻
DZn(s) + 2OH⁻(aq) → ZnO(s) + H₂O(l) + 2e⁻
(b)
What is the EMF of the cell?
[1 mark]
A+0.86 V
B+1.66 V
C−1.66 V
D+0.40 V
(c)
Write the overall equation for the reaction in the cell. Explain why a hearing aid powered by this cell stops working after a time.
[2 marks]
Total for question 5: 4 marks
6
A remote weather station is powered by an alkaline hydrogen–oxygen fuel cell. The electrode reactions are: H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻ and O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq). Hydrogen is supplied from a gas cylinder.
(a)
At which electrode does the hydrogen react and what happens to it?
[1 mark]
AAt the negative electrode, where it is oxidised
BAt the negative electrode, where it is reduced
CAt the positive electrode, where it is oxidised
DAt the positive electrode, where it is reduced
(b)
Which statement about this fuel cell is correct?
[1 mark]
AIt must be electrically recharged when the hydrogen cylinder is empty
BThe electrodes are used up as the cell operates
CIt produces a current for as long as hydrogen and oxygen are supplied, so it does not need to be electrically recharged
DHydroxide ions are used up overall, so the electrolyte must be replaced
(c)
Give one benefit and one risk to society of using hydrogen fuel cells.
[2 marks]
Total for question 6: 4 marks
7
Hydrogen peroxide solution is stored in a sealed bottle in a laboratory. Standard electrode potentials: H₂O₂(aq) + 2H⁺(aq) + 2e⁻ ⇌ 2H₂O(l), E° = +1.77 V; O₂(g) + 2H⁺(aq) + 2e⁻ ⇌ H₂O₂(aq), E° = +0.68 V.
(a)
Use the data to show that hydrogen peroxide can decompose into water and oxygen. Write an equation for the reaction.
[3 marks]
(b)
A student says, "The EMF is +1.09 V, so the hydrogen peroxide in the bottle must be decomposing rapidly." Evaluate this statement.
[4 marks]
Total for question 7: 7 marks
8
A manufacturer is developing a rechargeable nickel–metal hydride (NiMH) cell for hybrid vehicles. The electrode reactions, written as reductions, are: NiO(OH)(s) + H₂O(l) + e⁻ → Ni(OH)₂(s) + OH⁻(aq), E° = +0.52 V; M(s) + H₂O(l) + e⁻ → MH(s) + OH⁻(aq), E° = −0.83 V, where M is a hydrogen-absorbing metal alloy. The electrolyte is aqueous potassium hydroxide.
(a)
Deduce the equation for the reaction at each electrode when the cell discharges, the overall equation and the EMF. Explain how the cell is recharged.
[6 marks]
(b)
Evaluate the benefits and risks to society of using rechargeable cells such as NiMH cells, compared with non-rechargeable cells, in hybrid vehicles. Reach a justified conclusion.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).