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A Level Chemistry H3 Redox Electrochemistry Quiz

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A Level Chemistry H3 AI Generated Generated by DeepSeek V4 Flash Sample 04 Updated 2026-08-17

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Questions

A-Level Chemistry H3 Quiz - Redox Electrochemistry

Name: _________________________________________

Class: _________________________________________

Date: _________________________________________

Score: ______________ / 60

Duration: 1 hour 30 minutes

Total Marks: 60

Instructions:

Answer all questions.
Write your answers in the spaces provided.
You may use a scientific calculator.
The Data Booklet will be provided.
For calculation questions, show all working clearly. Marks are awarded for correct working and reasoning, not just the final answer.

Section A: Multiple-Choice Questions (20 marks)

Questions 1 to 10 carry 2 marks each. Choose the best option (A, B, C, or D).

1. Which statement about the standard hydrogen electrode (SHE) is correct?

A) It consists of a platinum electrode in contact with 1.0 mol dm⁻³ H⁺(aq) and H₂(g) at 1 atm. B) It consists of a zinc electrode in contact with 1.0 mol dm⁻³ H⁺(aq) and H₂(g) at 1 atm. C) It is assigned a standard electrode potential of +1.00 V. D) It can only be used as an anode.

Answer: ________

[2 marks]

2. What is the oxidation state of chromium in the dichromate(VI) ion, Cr₂O₇²⁻?

A) +3 B) +6 C) +7 D) +12

Answer: ________

[2 marks]

3. In the reaction below, which species is the oxidising agent?

$\text{2Fe}^{2+}(\text{aq}) + \text{Cl}_2(\text{aq}) \rightarrow \text{2Fe}^{3+}(\text{aq}) + \text{2Cl}^-(\text{aq})$

A) Fe²⁺(aq) B) Cl₂(aq) C) Fe³⁺(aq) D) Cl⁻(aq)

Answer: ________

[2 marks]

4. Which of the following is the correct cell notation for the Daniell cell?

A) Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) B) Cu(s) | Cu²⁺(aq) || Zn²⁺(aq) | Zn(s) C) Zn²⁺(aq) | Zn(s) || Cu(s) | Cu²⁺(aq) D) Zn(s) | Cu²⁺(aq) || Zn²⁺(aq) | Cu(s)

Answer: ________

[2 marks]

5. A galvanic cell has a standard cell potential, E°cell, of +1.10 V. Which statement about the cell reaction is correct?

A) The reaction is non-spontaneous. B) The reaction is spontaneous under standard conditions. C) The reaction is at equilibrium. D) The reaction requires an external power source.

Answer: ________

[2 marks]

6. What is the standard electrode potential, E°, of a half-cell where the oxidised species is a stronger oxidising agent than H⁺(aq)?

A) Positive B) Negative C) Zero D) Cannot be determined

Answer: ________

[2 marks]

7. Which quantity is related to the Gibbs free energy change, ΔG°, by the equation ΔG° = –nFE°?

A) The equilibrium constant B) The rate constant C) The activation energy D) The reaction quotient

Answer: ________

[2 marks]

8. In the electrolysis of molten sodium chloride, which product is formed at the anode?

A) Sodium metal B) Chlorine gas C) Hydrogen gas D) Oxygen gas

Answer: ________

[2 marks]

9. What is the quantity of charge, in coulombs, required to deposit 1 mole of silver atoms from a Ag⁺(aq) solution? (F = 96500 C mol⁻¹)

A) 48250 C B) 96500 C C) 193000 C D) 289500 C

Answer: ________

[2 marks]

10. Which statement about the Nernst equation is correct?

A) It is used to calculate the cell potential under non-standard conditions. B) It is used to calculate the standard cell potential. C) It is used to calculate the equilibrium constant only. D) It is used to calculate the rate of reaction.

Answer: ________

[2 marks]

Section B: Short-Answer Questions (20 marks)

Questions 11 to 15 carry 4 marks each. Answer all questions in the spaces provided.

11. (a) Define the term standard electrode potential. [2]

(b) State the conditions under which standard electrode potentials are measured. [2]

[4 marks]

12. A student sets up the following electrochemical cell:

$\text{Pt(s)} | \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) || \text{Ag}^+(\text{aq}) | \text{Ag(s)}$

Given: E°(Fe³⁺/Fe²⁺) = +0.77 V; E°(Ag⁺/Ag) = +0.80 V

(a) Identify the anode and the cathode of this cell. [2]

Anode: ____________________

Cathode: ____________________

(b) Calculate the standard cell potential, E°cell. [2]

[4 marks]

13. The following reaction occurs in an acidic solution:

$\text{MnO}_4^-(\text{aq}) + \text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{Fe}^{3+}(\text{aq})$

(a) Write the half-equation for the reduction of manganate(VII) ions in acidic solution. [2]

(b) Write the overall balanced ionic equation for this reaction. [2]

[4 marks]

14. A current of 2.50 A is passed through an electrolytic cell containing molten aluminium oxide for 30.0 minutes.

(a) Calculate the quantity of charge passed. [2]

(b) Calculate the mass of aluminium metal produced at the cathode. (F = 96500 C mol⁻¹; Ar(Al) = 27.0) [2]

[4 marks]

15. The standard electrode potential for the reduction of oxygen in acidic solution is:

$\text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^- \rightarrow 2\text{H}_2\text{O}(\text{l}) \qquad E^\circ = +1.23 \text{ V}$

(a) State what the positive value of E° indicates about the oxidising power of O₂(g). [1]

(b) Using the E° value, predict whether O₂(g) can oxidise Cl⁻(aq) to Cl₂(g) under standard conditions. E°(Cl₂/Cl⁻) = +1.36 V. Justify your answer. [3]

[4 marks]

Section C: Extended-Response Questions (20 marks)

Questions 16 to 20 carry 4 marks each. Answer all questions in the spaces provided.

16. A student constructs a cell using the following half-cells:

Half-cell 1: Cu²⁺(aq) / Cu(s), E° = +0.34 V
Half-cell 2: Zn²⁺(aq) / Zn(s), E° = –0.76 V

(a) Write the overall cell reaction. [1]

(b) Calculate the standard cell potential. [1]

[4 marks]

17. The Nernst equation is given as:

$E = E^\circ - \frac{0.0592}{n} \log_{10} Q$

For the cell: Zn(s) | Zn²⁺(aq, 0.010 mol dm⁻³) || Cu²⁺(aq, 1.00 mol dm⁻³) | Cu(s)

E°cell = +1.10 V

(a) Write the expression for the reaction quotient, Q, for this cell. [1]

(b) Calculate the cell potential, E, under these non-standard conditions. [3]

[4 marks]

18. (a) State Faraday's first law of electrolysis. [1]

(b) A solution of copper(II) sulfate is electrolysed using platinum electrodes. A current of 3.00 A is passed for 40.0 minutes.

(i) Write the half-equation for the reaction at the cathode. [1]

(ii) Calculate the mass of copper deposited. (F = 96500 C mol⁻¹; Ar(Cu) = 63.5) [2]

[4 marks]

19. The standard electrode potentials for two half-cells are:

$\text{Br}_2(\text{aq}) + 2e^- \rightarrow 2\text{Br}^-(\text{aq}) \qquad E^\circ = +1.07 \text{ V}$

$\text{I}_2(\text{aq}) + 2e^- \rightarrow 2\text{I}^-(\text{aq}) \qquad E^\circ = +0.54 \text{ V}$

(a) Deduce which halogen is a stronger oxidising agent. Explain your reasoning. [2]

(b) Predict whether bromine water will react with iodide ions. Write the balanced equation if a reaction occurs. [2]

[4 marks]

20. A fuel cell uses the reaction between hydrogen and oxygen:

$\text{2H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{2H}_2\text{O}(\text{l})$

(a) Write the half-equations for the reactions at the anode and cathode in an alkaline fuel cell. [2]

Anode: ______________________________________________________________________

Cathode: ______________________________________________________________________

(b) State one advantage of a fuel cell over a conventional battery. [1]

[4 marks]

END OF QUIZ

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Answers

A-Level Chemistry H3 Quiz - Redox Electrochemistry: Answer Key

Total Marks: 60

Section A: Multiple-Choice Questions (20 marks)

1. Answer: A

Marks: 2
Explanation: The standard hydrogen electrode (SHE) is the reference electrode. It consists of a platinum electrode (which is inert and provides a surface for the reaction) immersed in a solution of 1.0 mol dm⁻³ H⁺(aq), with hydrogen gas at 1 atm pressure bubbled over the electrode. The SHE is assigned a potential of exactly 0.00 V. Option B is incorrect because zinc is not used; it would be a different half-cell. Option C is incorrect because the assigned potential is 0.00 V, not +1.00 V. Option D is incorrect because the SHE can act as either the anode or cathode depending on the other half-cell it is connected to.

2. Answer: B

Marks: 2
Explanation: The oxidation state of oxygen is almost always –2. In Cr₂O₇²⁻, the total charge is –2. Let the oxidation state of Cr be x. The equation is: 2x + 7(–2) = –2. Solving: 2x – 14 = –2, so 2x = +12, and x = +6. Therefore, the oxidation state of chromium is +6. Option D (+12) is a common mistake where students forget to divide by the number of chromium atoms (2).

3. Answer: B

Marks: 2
Explanation: The oxidising agent is the species that is reduced (gains electrons). In this reaction, Cl₂ is reduced to Cl⁻ (oxidation state changes from 0 to –1). Fe²⁺ is oxidised to Fe³⁺ (oxidation state changes from +2 to +3), making it the reducing agent. Fe³⁺ and Cl⁻ are products, not reactants.

4. Answer: A

Marks: 2
Explanation: Cell notation is written with the anode (oxidation) on the left and the cathode (reduction) on the right. A single vertical line (|) represents a phase boundary, and a double vertical line (||) represents the salt bridge. In the Daniell cell, zinc is oxidised (Zn(s) → Zn²⁺(aq) + 2e⁻) and copper is reduced (Cu²⁺(aq) + 2e⁻ → Cu(s)). Therefore, the correct notation is Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

5. Answer: B

Marks: 2
Explanation: A positive standard cell potential (E°cell > 0) indicates that the cell reaction is spontaneous under standard conditions. This is because ΔG° = –nFE°cell. A positive E°cell gives a negative ΔG°, which is the condition for a spontaneous process. Option A is incorrect because a positive E°cell means the reaction is spontaneous. Option C is incorrect because a spontaneous reaction is not at equilibrium. Option D is incorrect because a galvanic cell produces electrical energy spontaneously; it does not require an external power source.

6. Answer: A

Marks: 2
Explanation: The standard electrode potential measures the tendency of a species to be reduced. A stronger oxidising agent has a greater tendency to gain electrons. By convention, the SHE is assigned 0.00 V. Species that are stronger oxidising agents than H⁺(aq) will have positive E° values (e.g., F₂, Cl₂, Ag⁺). Species that are weaker oxidising agents (stronger reducing agents) will have negative E° values (e.g., Zn²⁺, Na⁺).

7. Answer: A

Marks: 2
Explanation: The equation ΔG° = –nFE°cell connects the standard Gibbs free energy change to the standard cell potential. Since ΔG° = –RT ln K, we can also relate E°cell to the equilibrium constant K. A larger positive E°cell corresponds to a larger equilibrium constant, meaning the reaction proceeds further towards completion. The rate constant (Option B) is related to activation energy, not thermodynamics. The reaction quotient (Option D) is used in the Nernst equation for non-standard conditions.

8. Answer: B

Marks: 2
Explanation: In the electrolysis of molten sodium chloride (NaCl), the ions present are Na⁺ and Cl⁻. At the anode (positive electrode), oxidation occurs. Chloride ions are oxidised to chlorine gas: 2Cl⁻(l) → Cl₂(g) + 2e⁻. At the cathode (negative electrode), reduction occurs: Na⁺(l) + e⁻ → Na(l). Hydrogen gas and oxygen gas are not produced because there is no water present in molten NaCl.

9. Answer: B

Marks: 2
Explanation: The reduction of Ag⁺ to Ag is: Ag⁺(aq) + e⁻ → Ag(s). This shows that 1 mole of Ag requires 1 mole of electrons. The charge of 1 mole of electrons is 1 Faraday (F) = 96500 C. Therefore, the charge required is 96500 C. Option C (193000 C) would be for a 2+ ion like Cu²⁺.

10. Answer: A

Marks: 2
Explanation: The Nernst equation, E = E° – (0.0592/n) log₁₀ Q, is used to calculate the cell potential under non-standard conditions (when concentrations are not 1.0 mol dm⁻³ or pressures are not 1 atm). Under standard conditions, Q = 1, log₁₀(1) = 0, and E = E°. It is not used to calculate the standard cell potential directly (that is done from standard electrode potentials), nor is it used for rate calculations.

Section B: Short-Answer Questions (20 marks)

11. (a) Answer: The standard electrode potential is the electromotive force (emf) of a cell in which the electrode is combined with a standard hydrogen electrode, measured under standard conditions.

Marks: 2
Explanation: This is a definition question. Key terms to include are "standard hydrogen electrode" (the reference), "combined with" (to form a cell), and "standard conditions". A complete definition must mention the reference electrode.

(b) Answer: Standard conditions are: 1.0 mol dm⁻³ concentration for all aqueous solutions, 1 atm pressure for all gases, and a temperature of 298 K (25 °C).

Marks: 2
Explanation: These are the three standard conditions. Missing any one of them would lose a mark. Note that for solids and liquids, their standard state is the pure substance, and they are not included in the conditions.

12. (a) Answer:

Anode: Pt(s) | Fe²⁺(aq), Fe³⁺(aq) (or the Fe³⁺/Fe²⁺ half-cell)
Cathode: Ag⁺(aq) | Ag(s) (or the Ag⁺/Ag half-cell)
Marks: 2 (1 mark for each correct identification)
Explanation: The anode is where oxidation occurs. The Fe²⁺ is oxidised to Fe³⁺ (oxidation state increases from +2 to +3). The cathode is where reduction occurs. Ag⁺ is reduced to Ag (oxidation state decreases from +1 to 0). Since E°(Ag⁺/Ag) = +0.80 V is more positive than E°(Fe³⁺/Fe²⁺) = +0.77 V, Ag⁺ is the stronger oxidising agent and will be reduced at the cathode.

(b) Answer: E°cell = E°(cathode) – E°(anode) E°cell = (+0.80 V) – (+0.77 V) = +0.03 V

Marks: 2 (1 mark for correct formula/substitution, 1 mark for correct answer with sign and unit)
Explanation: The standard cell potential is calculated by subtracting the standard electrode potential of the anode from that of the cathode. A positive E°cell confirms the reaction is spontaneous.

13. (a) Answer: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

Marks: 2
Explanation: This is a standard half-equation that must be memorised. To balance it: (1) Balance the atoms other than O and H (Mn is balanced). (2) Balance O by adding H₂O: 4 H₂O on the right. (3) Balance H by adding H⁺: 8 H⁺ on the left. (4) Balance charge by adding electrons: Left side charge is –1 + 8 = +7; right side charge is +2. To balance, add 5e⁻ to the left: +7 + (–5) = +2.

(b) Answer: MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Marks: 2
Explanation: To combine the half-equations, the electrons must cancel. The reduction half-equation (from part a) involves 5 electrons. The oxidation half-equation is: Fe²⁺(aq) → Fe³⁺(aq) + e⁻. Multiply the oxidation half-equation by 5: 5Fe²⁺(aq) → 5Fe³⁺(aq) + 5e⁻. Add the two half-equations and cancel the electrons to get the overall equation.

14. (a) Answer: Q = I × t Q = 2.50 A × (30.0 × 60) s = 2.50 A × 1800 s = 4500 C

Marks: 2 (1 mark for correct formula, 1 mark for correct answer with unit)
Explanation: The quantity of charge is the product of current and time. Time must be converted from minutes to seconds. 1 A = 1 C s⁻¹.

(b) Answer: The half-equation for the reduction of Al³⁺ is: Al³⁺(l) + 3e⁻ → Al(l) Moles of electrons = Q / F = 4500 C / 96500 C mol⁻¹ = 0.0466 mol Moles of Al = 0.0466 mol / 3 = 0.0155 mol Mass of Al = moles × Ar = 0.0155 mol × 27.0 g mol⁻¹ = 0.419 g

Marks: 2 (1 mark for correct moles of electrons or correct stoichiometric ratio, 1 mark for correct final mass)
Explanation: First, calculate the moles of electrons using Faraday's constant. Then, use the stoichiometry of the half-equation: 3 moles of electrons produce 1 mole of Al. Finally, convert moles of Al to mass using the relative atomic mass.

15. (a) Answer: The positive E° value indicates that O₂(g) is a strong oxidising agent (it has a strong tendency to be reduced).

Marks: 1
Explanation: A more positive E° value means a greater tendency for the species on the left of the half-equation (the oxidised form) to gain electrons and be reduced. Therefore, O₂ is a good oxidising agent.

(b) Answer: E°cell for the reaction: O₂(g) + 4H⁺(aq) + 4Cl⁻(aq) → 2H₂O(l) + 2Cl₂(g) E°cell = E°(cathode) – E°(anode) = E°(O₂/H₂O) – E°(Cl₂/Cl⁻) = (+1.23 V) – (+1.36 V) = –0.13 V

Since E°cell is negative, the reaction is non-spontaneous under standard conditions. Therefore, O₂(g) cannot oxidise Cl⁻(aq) to Cl₂(g) under standard conditions.

Marks: 3 (1 mark for correct calculation of E°cell, 1 mark for correct sign, 1 mark for correct conclusion with justification)
Explanation: To determine if a reaction is spontaneous, calculate the E°cell. The more positive half-cell is the cathode (reduction), and the less positive is the anode (oxidation). Here, Cl₂/Cl⁻ has a higher E° than O₂/H₂O, so Cl₂ would be reduced, not Cl⁻ oxidised. The negative E°cell means ΔG° is positive, so the reaction is not spontaneous. This is consistent with the fact that chlorine is a stronger oxidising agent than oxygen.

Section C: Extended-Response Questions (20 marks)

16. (a) Answer: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Marks: 1
Explanation: Zinc is the anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻. Copper is the cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s). Adding the two half-equations gives the overall reaction.

(b) Answer: E°cell = E°(cathode) – E°(anode) = (+0.34 V) – (–0.76 V) = +1.10 V

Marks: 1
Explanation: The copper half-cell has the more positive E° and is the cathode. The zinc half-cell is the anode. Subtract the anode potential from the cathode potential.

(c) Answer: Electrons flow from the zinc electrode (anode) through the external circuit to the copper electrode (cathode). At the anode, Zn atoms lose electrons and go into solution as Zn²⁺ ions. At the cathode, Cu²⁺ ions in solution gain electrons and are deposited as Cu atoms on the electrode.

Marks: 2 (1 mark for direction of electron flow, 1 mark for explanation of processes at each electrode)
Explanation: Oxidation occurs at the anode, releasing electrons. These electrons travel through the wire to the cathode, where reduction occurs. The salt bridge maintains electrical neutrality by allowing ions to flow between the half-cells.

17. (a) Answer: Q = [Zn²⁺(aq)] / [Cu²⁺(aq)]

Marks: 1
Explanation: The reaction quotient Q is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. Solids (Zn(s) and Cu(s)) are not included in the expression. The overall reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

(b) Answer: E = E° – (0.0592 / n) log₁₀ Q n = 2 (two electrons transferred) Q = 0.010 / 1.00 = 0.010 E = 1.10 – (0.0592 / 2) log₁₀(0.010) E = 1.10 – (0.0296) × (–2.00) E = 1.10 + 0.0592 = +1.16 V

Marks: 3 (1 mark for correct n, 1 mark for correct substitution into Nernst equation, 1 mark for correct final answer with unit)
Explanation: The Nernst equation corrects the standard cell potential for non-standard concentrations. Here, the concentration of Zn²⁺ is lower than standard, which makes the reaction more favourable (Q < 1), so the cell potential increases. Remember that log₁₀(0.010) = –2.

18. (a) Answer: Faraday's first law states that the mass of a substance produced or consumed during electrolysis is directly proportional to the quantity of electricity passed.

Marks: 1
Explanation: This is a statement of the law. The key relationship is m ∝ Q.

(b) (i) Answer: Cu²⁺(aq) + 2e⁻ → Cu(s)

Marks: 1
Explanation: In an aqueous solution of copper(II) sulfate, Cu²⁺ ions are reduced at the cathode. Platinum electrodes are inert, so they do not participate in the reaction.

(ii) Answer: Q = I × t = 3.00 A × (40.0 × 60) s = 7200 C Moles of electrons = Q / F = 7200 C / 96500 C mol⁻¹ = 0.0746 mol Moles of Cu = 0.0746 mol / 2 = 0.0373 mol Mass of Cu = moles × Ar = 0.0373 mol × 63.5 g mol⁻¹ = 2.37 g

Marks: 2 (1 mark for correct moles of electrons, 1 mark for correct final mass)
Explanation: This is a standard electrolysis calculation. Remember to convert time to seconds. The stoichiometry of the half-equation shows that 2 moles of electrons are needed to deposit 1 mole of copper. The final mass is calculated using the relative atomic mass of copper.

19. (a) Answer: Br₂ is the stronger oxidising agent. This is because its standard electrode potential is more positive (+1.07 V) than that of I₂ (+0.54 V). A more positive E° value indicates a greater tendency to gain electrons and be reduced.

Marks: 2 (1 mark for correct identification, 1 mark for correct justification)
Explanation: The oxidising agent is the species that is reduced. The half-equation given is the reduction half-equation. The more positive the E° value, the more favourable the reduction, and hence the stronger the oxidising agent.

(b) Answer: Yes, bromine water will react with iodide ions. Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)

Marks: 2 (1 mark for correct prediction, 1 mark for correct balanced equation)
Explanation: Since Br₂ is a stronger oxidising agent than I₂, it will oxidise I⁻ to I₂ and be reduced to Br⁻. The E°cell for this reaction is: E°cell = E°(Br₂/Br⁻) – E°(I₂/I⁻) = +1.07 V – (+0.54 V) = +0.53 V. Since E°cell is positive, the reaction is spontaneous. The equation is balanced with 2 moles of I⁻ for every 1 mole of Br₂.

20. (a) Answer: Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ (or H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻) Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

Marks: 2 (1 mark for each correct half-equation)
Explanation: In an alkaline fuel cell, the electrolyte is a basic solution (e.g., KOH). At the anode, hydrogen is oxidised. It reacts with hydroxide ions to form water and release electrons. At the cathode, oxygen is reduced. It reacts with water and gains electrons to form hydroxide ions. The equations must be balanced for both atoms and charge.

(b) Answer: Fuel cells have a higher efficiency than conventional batteries. / Fuel cells produce electricity continuously as long as fuel is supplied, whereas batteries have a limited lifespan. / Fuel cells produce water as the only product, making them more environmentally friendly.

Marks: 1
Explanation: Any one valid advantage is acceptable. Common points include continuous power supply, higher energy density, or cleaner products.

Marks: 1
Explanation: The cathode is where reduction occurs (O₂ is reduced). The anode is where oxidation occurs (H₂ is oxidised). The standard cell potential is the difference between the two standard electrode potentials. The positive value confirms the reaction is spontaneous.