A Level Chemistry H3 Redox Electrochemistry Quiz

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A-Level Chemistry H3 Quiz - Redox Electrochemistry

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Duration: 60 minutes
Total Marks: 40

Instructions:
This quiz covers Redox Electrochemistry as assumed H2 prerequisite knowledge for H3 Chemistry. Answer all 20 questions. Section A is short structured items (1 mark each unless stated). Section B requires calculations and explanations. Section C involves data interpretation and extended reasoning. Show all working for calculation questions. Use the Data Booklet if needed.

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Section A: Fundamentals (Questions 1–8)

1. Define oxidation in terms of electron transfer. [1]

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2. Write the half-equation for the reduction of $MnO_4^-$ to $Mn^{2+}$ in acidic solution. [1]

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3. State the Nernst equation for a half-cell. [1]

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4. Calculate the oxidation state of Cr in $K_2Cr_2O_7$. [1]

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5. Identify the species oxidised in: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$. [1]

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6. What is the standard electrode potential $E^\circ$ of the SHE? [1]

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7. State two conditions for a galvanic cell to operate. [1]

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8. Write the cell notation for a Daniell cell. [1]

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Section B: Calculations and Applications (Questions 9–14)

9. For the cell $Zn|Zn^{2+}(1.0\,M)||Cu^{2+}(1.0\,M)|Cu$, given $E^\circ_{Zn^{2+}/Zn} = -0.76\,V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\,V$, calculate $E^\circ_{cell}$. [2]

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10. Using the Nernst equation at 298 K, calculate the cell potential for the above cell if $[Zn^{2+}] = 0.10\,M$ and $[Cu^{2+}] = 1.0\,M$. ($RT/F = 0.0257\,V$) [3]

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11. A solution contains $Fe^{2+}$ and $Fe^{3+}$. The $E^\circ_{Fe^{3+}/Fe^{2+}} = +0.77\,V$. If $[Fe^{3+}] = 0.50\,M$ and $[Fe^{2+}] = 0.10\,M$, calculate the electrode potential at 298 K. [3]

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12. In the electrolysis of aqueous $NaCl$ with inert electrodes, write the half-equations at anode and cathode, and state the overall cell reaction. [3]

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13. Calculate the charge in coulombs required to deposit 2.00 g of Ag from $Ag^+$ solution. ($A_r$ of Ag = 108, $F = 96500\,C\,mol^{-1}$) [3]

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14. For the reaction $2Ag^+ + Cu \rightarrow 2Ag + Cu^{2+}$, given $E^\circ_{Ag^+/Ag} = +0.80\,V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\,V$, calculate $\Delta G^\circ$ (in kJ mol$^{-1}$) for the reaction as written. ($n=2$, $F=96500\,C\,mol^{-1}$) [3]

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Section C: Data Interpretation and Extended Reasoning (Questions 15–20)

15. A student measures the following standard reduction potentials:

• $I_2 + 2e^- \rightarrow 2I^-$: $+0.54\,V$
• $Br_2 + 2e^- \rightarrow 2Br^-$: $+1.07\,V$
• $Cl_2 + 2e^- \rightarrow 2Cl^-$: $+1.36\,V$

(a) Arrange $Cl_2, Br_2, I_2$ in order of decreasing oxidizing strength. [1]
(b) Predict whether $Cl_2$ will oxidize $Br^-$ to $Br_2$. Explain. [2]

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16.

Image pending generation: graph for Q16.

Using the plotted graph, explain how the cell potential varies with concentration ratio in a concentration cell at 298 K. [3]

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17. Compare the operation of a galvanic cell and an electrolytic cell under the headings: (i) energy change, (ii) sign of $E_{cell}$, (iii) electrode labels. [3]

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18. A fuel cell uses $H_2$ and $O_2$ to produce water. Write the half-equations and overall equation. Given $E^\circ_{O_2/H_2O} = +1.23\,V$ and $E^\circ_{H^+/H_2} = 0.00\,V$, calculate $E^\circ_{cell}$. [4]

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19. Discuss how overpotential affects the products of electrolysis of aqueous $Na_2SO_4$ with inert electrodes. [3]

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20. The table shows $E^\circ$ values for some metals:

Metal	$E^\circ_{M^{2+}/M}$ / V
Mg	-2.37
Fe	-0.44
Ni	-0.25
Au	+1.50

(a) Which metal is the strongest reducing agent? [1]
(b) Calculate $E^\circ_{cell}$ for a cell with Mg anode and Ni cathode. [2]
(c) State one application of such a cell. [1]

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

Total Marks: 40
Topic: Redox Electrochemistry (H2 assumed knowledge for H3)

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Section A: Fundamentals

1. [1 mark] Oxidation is loss of electrons (OIL).
Teaching note: In redox, oxidation = electrons lost; reduction = electrons gained. A common mistake is defining by O₂, which is only one case.

2. [1 mark] $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Teaching note: Balance O with H₂O, H with H⁺, charge with e⁻. In acidic medium, H⁺ is used.

3. [1 mark] $E = E^\circ - \frac{RT}{nF}\ln Q$ (or $E = E^\circ - \frac{0.0591}{n}\log Q$ at 298 K)
Teaching note: Q is reaction quotient. Nernst adjusts potential for non-standard conditions.

4. [1 mark] +6
Working: K = +1 (×2 = +2), O = -2 (×7 = -14); total 0 → 2 + 2x -14 = 0 → x = +6.

5. [1 mark] Zn (or Zn(s))
Teaching note: Zn goes from 0 to +2, loses e⁻ → oxidized.

6. [1 mark] 0.00 V by definition.
Teaching note: Standard Hydrogen Electrode is reference, assigned 0 V.

7. [1 mark] Any two: two different half-cells; salt bridge/ion pathway; external circuit; redox couples at non-equal potentials.
Marking: 1 mark total for two correct conditions.

8. [1 mark] $Zn|Zn^{2+}(1\,M)||Cu^{2+}(1\,M)|Cu$
Teaching note: Anode left, cathode right, || salt bridge.

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Section B: Calculations

9. [2 marks]
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = +1.10\,V$
Marking: 1 for identifying cathode Cu, anode Zn; 1 for correct value.

10. [3 marks]
Reaction: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$, n = 2.
$Q = [Zn^{2+}]/[Cu^{2+}] = 0.10/1.0 = 0.10$
$E = 1.10 - \frac{0.0257}{2}\ln(0.10)$
$\ln(0.10) = -2.303$
$E = 1.10 - 0.01285 \times (-2.303) = 1.10 + 0.0296 = 1.13\,V$
Marking: 1 Nernst substitution, 1 ln/Q, 1 final.

11. [3 marks]
$E = E^\circ - \frac{0.0257}{1}\ln\frac{[Fe^{2+}]}{[Fe^{3+}]}$ (n=1)
$= 0.77 - 0.0257 \ln(0.10/0.50) = 0.77 - 0.0257 \ln(0.20)$
$\ln(0.20) = -1.609$
$E = 0.77 - 0.0257(-1.609) = 0.77 + 0.0414 = 0.811\,V$
Marking: 1 equation, 1 calc, 1 answer.

12. [3 marks]
Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (or $2H^+ + 2e^- \rightarrow H_2$ if acidic)
Anode: $2Cl^- \rightarrow Cl_2 + 2e^-$ (overpotential makes Cl⁻ oxidize instead of O₂)
Overall: $2NaCl + 2H_2O \rightarrow H_2 + Cl_2 + 2NaOH$
Marking: 1 each.

13. [3 marks]
$m = 2.00\,g$, $M = 108\,g\,mol^{-1}$
$n(Ag) = 2.00/108 = 0.01852\,mol$
$Ag^+ + e^- \rightarrow Ag$, so mol e⁻ = 0.01852
$Q = n \times F = 0.01852 \times 96500 = 1787\,C \approx 1.79 \times 10^3\,C$
Marking: 1 moles, 1 Faraday, 1 answer.

14. [3 marks]
$E^\circ_{cell} = 0.80 - 0.34 = 0.46\,V$
$\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 0.46 = -88780\,J\,mol^{-1} = -88.8\,kJ\,mol^{-1}$
Marking: 1 Ecell, 1 formula, 1 value.

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Section C

15. [3 marks]
(a) [1] $Cl_2 > Br_2 > I_2$ (higher E° = stronger oxidant)
(b) [2] Yes. $E^\circ_{cell} = 1.36 - 1.07 = +0.29\,V > 0$, spontaneous. Cl₂ is stronger oxidant than Br₂.

16. [3 marks]
Graph shows $E = 0.0296 \log([cathode]/[anode])$ for n=1 at 298 K. As ratio increases, E increases linearly. Slope matches $\frac{0.0591}{n}$ with n=2? Actually slope 0.0296 = 0.0591/2 so n=2.
Marking: 1 linear increase, 1 slope meaning, 1 reference to Nernst.

17. [3 marks]
(i) Galvanic: chemical→electrical (spontaneous); Electrolytic: electrical→chemical (non-spontaneous).
(ii) Galvanic E>0; Electrolytic E<0 (needs applied).
(iii) Galvanic: anode -, cathode +; Electrolytic: anode +, cathode -.
Marking: 1 each.

18. [4 marks]
Anode: $2H_2 \rightarrow 4H^+ + 4e^-$
Cathode: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$
Overall: $2H_2 + O_2 \rightarrow 2H_2O$
$E^\circ_{cell} = 1.23 - 0.00 = 1.23\,V$
Marking: 1+1 half, 1 overall, 1 Ecell.

19. [3 marks]
At cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (not Na due to reactivity). At anode: O₂ evolution expected from water ($E^\circ=+1.23$) but overpotential for O₂ on inert electrodes raises required V, yet still > S₂O₈²⁻ formation; Na₂SO₄ inert, so H₂ and O₂ produced. Overpotential prevents SO₄²⁻ oxidation.
Marking: 1 cathode, 1 anode, 1 overpotential note.

20. [4 marks]
(a) [1] Mg (most negative E° = best reducer)
(b) [2] $E^\circ = -0.25 - (-2.37) = +2.12\,V$
(c) [1] Battery / corrosion protection / reference.

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Caveat: No past-year H3 papers exist (first exam 2026). This is syllabus-first practice generated from H2 redox electrochemistry assumed knowledge.