Level 3 — ProductionRedox & Electrochemistry (Intro)

Redox & Electrochemistry (Intro)

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 (Production — derivations, reasoning from memory, explain-out-loud) Time limit: 45 minutes Total marks: 60

Use $F = 96485\ \text{C mol}^{-1}$ , $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , $T = 298\ \text{K}$ unless told otherwise. Show all reasoning.

Question 1 — Derive the Nernst equation from first principles (10 marks)

Starting from memory with the thermodynamic relation $\Delta G = \Delta G^\circ + RT\ln Q$ and the electrochemical link between free energy and cell potential:

(a) State the two founding equations and explain physically why the sign is negative in $\Delta G = -nFE$ . (3)

(b) Derive the Nernst equation $E = E^\circ - \dfrac{RT}{nF}\ln Q$ . (4)

Question 2 — Build a galvanic cell and compute its EMF (12 marks)

You are given the half-cells: $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}, \quad E^\circ = -0.76\ \text{V}$ $\text{Ag}^+ + e^- \rightarrow \text{Ag}, \quad E^\circ = +0.80\ \text{V}$

(a) Identify anode and cathode and justify using the electrochemical series. (2)

(b) Write the balanced cell reaction and the standard cell notation (line diagram). (3)

(d) Compute $E_{\text{cell}}$ when $[\text{Zn}^{2+}] = 0.10\ \text{M}$ and $[\text{Ag}^+] = 0.010\ \text{M}$ . (3)

(e) Compute $\Delta G^\circ$ for the cell reaction (kJ). (2)

Question 3 — Equilibrium constant from $E^\circ$ (8 marks)

For the reaction in Question 2:

(a) Derive the relation $\ln K = \dfrac{nFE^\circ}{RT}$ from the two founding equations. (3)

(b) Compute $K$ for the Zn/Ag cell (state $n$ used). (3)

Question 4 — Concentration cell (10 marks)

A concentration cell is built from two copper electrodes in $\text{Cu}^{2+}$ solutions of $0.001\ \text{M}$ and $1.0\ \text{M}$ .

(a) Explain out loud (in words) why this cell produces a voltage despite identical electrodes; identify which electrode is the cathode. (3)

(b) Write the Nernst expression for $E_{\text{cell}}$ and derive the simplified form for a concentration cell (where $E^\circ_{\text{cell}} = 0$ ). (3)

Question 5 — Faraday's laws & industrial electrolysis (12 marks)

(a) State Faraday's first law and derive $m = \dfrac{MIt}{nF}$ from the definition of moles of electrons. (3)

(b) Molten $\text{Al}_2\text{O}_3$ is electrolysed at $40{,}000\ \text{A}$ . Calculate the mass of aluminium deposited in 1.0 hour. Take $M(\text{Al}) = 27\ \text{g mol}^{-1}$ . (4)

(c) For the electrolysis of brine (aqueous NaCl), write the electrode reactions and name the three commercial products. (3)

(d) Explain why aqueous $\text{Na}^+$ is not reduced at the cathode in brine electrolysis, referencing electrode potentials. (2)

Question 6 — Corrosion, protection & fuel cells (8 marks)

(a) Describe the electrochemical mechanism of iron rusting: identify anodic and cathodic regions and write both half-reactions. (3)

(b) Explain why galvanization (zinc coating) protects iron even if the coating is scratched — refer to the electrochemical series. (2)

(c) For a H₂/O₂ fuel cell (alkaline), write both electrode reactions and give one reason it is preferred on spacecraft over a battery. (3)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Founding equations: $\Delta G = -nFE$ and $\Delta G = \Delta G^\circ + RT\ln Q$ . (1 each) Sign is negative because a spontaneous cell ( $E>0$ ) does electrical work on the surroundings, releasing free energy ( $\Delta G<0$ ); $nF$ is positive charge transferred so $-nFE$ enforces $\Delta G<0$ when $E>0$ . (1)

(b) Substitute both: $-nFE = -nFE^\circ + RT\ln Q$ . (2) Divide by $-nF$ : $E = E^\circ - \frac{RT}{nF}\ln Q$ (2)

(c) Convert: $\ln Q = 2.303\log_{10}Q$ , so prefactor $=\dfrac{2.303RT}{nF}$ . (1) Numerically: $\dfrac{2.303\times 8.314\times 298}{96485} = 0.05916\ \text{V}$ . (2) $E = E^\circ - \frac{0.0592}{n}\log_{10}Q$

Question 2 (12)

(a) Zn has more negative $E^\circ$ → more readily oxidised → anode. Ag → cathode. (2)

(b) Reaction: $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$ (2) (balance electrons: Zn gives 2e⁻, need 2 Ag⁺). Cell: $\text{Zn}\,|\,\text{Zn}^{2+}\,\|\,\text{Ag}^+\,|\,\text{Ag}$ (1)

(c) $E^\circ_{\text{cell}} = E^\circ_{\text{cat}} - E^\circ_{\text{an}} = 0.80 - (-0.76) = 1.56\ \text{V}$ (2)

(d) $n=2$ , $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} = \dfrac{0.10}{(0.010)^2} = 1000$ . (1) $E = 1.56 - \dfrac{0.0592}{2}\log_{10}(1000) = 1.56 - 0.0296\times 3 = 1.56 - 0.0888 = 1.471\ \text{V}$ (2)

(e) $\Delta G^\circ = -nFE^\circ = -2\times 96485\times 1.56 = -301{,}033\ \text{J} \approx -301\ \text{kJ}$ (2)

Question 3 (8)

(a) At equilibrium $\Delta G=0$ , $E=0$ , $Q=K$ . From $\Delta G^\circ = -nFE^\circ$ and $\Delta G^\circ = -RT\ln K$ : (2) $-RT\ln K = -nFE^\circ \Rightarrow \ln K = \frac{nFE^\circ}{RT}$ (1)

(b) $n=2$ : $\ln K = \dfrac{2\times 96485\times 1.56}{8.314\times 298} = \dfrac{301{,}033}{2477.6} = 121.5$ . (2) $K = e^{121.5} \approx 5\times 10^{52}$ . (1)

(c) $K$ is astronomically large → reaction goes essentially to completion. (2)

Question 4 (10)

(a) Both electrodes/reactions identical, but potential depends on ion concentration (Nernst). The dilute side has a lower Cu²⁺ potential and undergoes oxidation (anode); the concentrated side is reduced (cathode = 1.0 M side). The system drives toward equalising concentrations. (3)

(b) $E = E^\circ - \dfrac{0.0592}{n}\log\dfrac{[\text{Cu}^{2+}]_{\text{anode/dilute}}}{[\text{Cu}^{2+}]_{\text{cathode/conc}}}$ ; with $E^\circ=0$ : $E = -\frac{0.0592}{2}\log\frac{[\text{dilute}]}{[\text{conc}]} = \frac{0.0592}{2}\log\frac{[\text{conc}]}{[\text{dilute}]}$ (3)

(c) $n=2$ : $E = \dfrac{0.0592}{2}\log\dfrac{1.0}{0.001} = 0.0296\times 3 = 0.0888\ \text{V}$ (4)

Question 5 (12)

(a) First law: mass deposited ∝ charge passed. Moles of electrons $= \dfrac{It}{F}$ ; moles of metal $= \dfrac{It}{nF}$ ; mass $= \dfrac{MIt}{nF}$ . (3)

(b) $\text{Al}^{3+}+3e^-\rightarrow\text{Al}$ , $n=3$ . $Q = It = 40000\times 3600 = 1.44\times 10^8\ \text{C}$ . (1) $m = \dfrac{27\times 1.44\times 10^8}{3\times 96485} = \dfrac{3.888\times 10^9}{289455} = 13{,}432\ \text{g} \approx 13.4\ \text{kg}$ (3)

(c) Cathode: $2\text{H}_2\text{O}+2e^-\rightarrow \text{H}_2 + 2\text{OH}^-$ ; Anode: $2\text{Cl}^-\rightarrow \text{Cl}_2 + 2e^-$ . (2) Products: H₂, Cl₂, NaOH. (1)

(d) $E^\circ(\text{Na}^+/\text{Na}) = -2.71\ \text{V}$ is far more negative than water reduction (~ $-0.83$ V at pH 7); water is preferentially reduced, so H₂ evolves not Na. (2)

Question 6 (8)

(a) Anodic region: $\text{Fe}\rightarrow \text{Fe}^{2+}+2e^-$ . Cathodic region (in presence of O₂/water): $\text{O}_2 + 2\text{H}_2\text{O}+4e^-\rightarrow 4\text{OH}^-$ . Fe²⁺ further oxidised → hydrated Fe₂O₃ (rust). (3)

(b) Zn is more electronegative (more negative $E^\circ = -0.76$ V vs Fe $-0.44$ V), so Zn acts as sacrificial anode and is oxidised preferentially, protecting iron even when scratched (cathodic protection). (2)

(c) Anode: $\text{H}_2 + 2\text{OH}^-\rightarrow 2\text{H}_2\text{O}+2e^-$ ; Cathode: $\text{O}_2 + 2\text{H}_2\text{O}+4e^-\rightarrow 4\text{OH}^-$ . (2) Preferred on spacecraft: continuous power as long as fuel supplied (no recharge needed), and by-product water is drinkable. (1)

[
  {"claim":"Q2c E_cell standard = 1.56 V","code":"Ec=0.80-(-0.76); result=abs(Ec-1.56)<1e-9"},
  {"claim":"Q2d E_cell at given conc = 1.471 V","code":"import math; E=1.56-(0.0592/2)*math.log10(0.10/(0.010**2)); result=abs(E-1.471)<1e-3"},
  {"claim":"Q2e dG standard approx -301 kJ","code":"dG=-2*96485*1.56; result=abs(dG/1000+301)<1"},
  {"claim":"Q3b lnK approx 121.5","code":"import math; lnK=2*96485*1.56/(8.314*298); result=abs(lnK-121.5)<0.5"},
  {"claim":"Q4c concentration cell E = 0.0888 V","code":"import math; E=(0.0592/2)*math.log10(1.0/0.001); result=abs(E-0.0888)<1e-3"},
  {"claim":"Q5b Al mass approx 13.4 kg","code":"m=27*40000*3600/(3*96485); result=abs(m/1000-13.4)<0.1"}
]