Redox & Electrochemistry (Intro)

Q1 (10)

(a) Ni has lower (more negative) $E^\circ$ → oxidised (anode); Ag reduced (cathode).

• Anode: $\text{Ni} \to \text{Ni}^{2+} + 2e^-$ (1)
• Cathode: $\text{Ag}^+ + e^- \to \text{Ag}$ (1)
• Balance electrons (×2 on Ag): overall $\text{Ni} + 2\text{Ag}^+ \to \text{Ni}^{2+} + 2\text{Ag}$ (1) correct balancing; (1) correct anode/cathode ID.

(b) $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - (-0.25) = +1.05\ \text{V}$. (2)

(c) $n = 2$. $\Delta G^\circ = -nFE^\circ = -(2)(96485)(1.05) = -2.026\times10^5\ \text{J} = -202.6\ \text{kJ}$. (2) (–ve → spontaneous.)

(d) $\ln K = \dfrac{nFE^\circ}{RT} = \dfrac{2\times96485\times1.05}{8.314\times298} = 81.79$ $K = e^{81.79} \approx 2.5\times10^{35}$. (2) (very large → reaction essentially complete.)

Q2 (10)

(a) $E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \text{V}$. (2)

(b) Reaction: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$, $n=2$. $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.010}{0.50} = 0.020$. (1) $E = E^\circ - \dfrac{0.0592}{2}\log Q = 1.10 - 0.0296\log(0.020)$ (2) $\log(0.020) = -1.699$; $E = 1.10 - 0.0296(-1.699) = 1.10 + 0.0503 = 1.150\ \text{V}$. (1)

(c) At equilibrium $E_{\text{cell}} = 0$. (1) Equilibrium = the cell is "dead"; forward and reverse rates equal, $Q=K$, no net current/no more usable work. (1)

(d) Lowering $[\text{Cu}^{2+}]$ raises $Q$, so $-\frac{0.0592}{2}\log Q$ becomes more negative → $E_{\text{cell}}$ decreases. (2)

Q3 (10)

(a) Both electrodes are the same metal/couple, so $E^\circ_{\text{cathode}} = E^\circ_{\text{anode}}$; their difference is $0$. EMF arises solely from concentration difference. (2)

(b) $E = -\dfrac{0.0592}{2}\log\dfrac{[\text{dilute}]}{[\text{conc}]}$ where the dilute side is the anode. Cell drives ions from low→high implicitly; use $E = \dfrac{0.0592}{2}\log\dfrac{[\text{Cu}^{2+}]_{\text{cathode}}}{[\text{Cu}^{2+}]_{\text{anode}}}$. (1) $0.0296 = \dfrac{0.0592}{2}\log\dfrac{0.100}{x}$ (1) $0.0296 = 0.0296\log\dfrac{0.100}{x} \Rightarrow \log\dfrac{0.100}{x} = 1$ (2) $\dfrac{0.100}{x} = 10 \Rightarrow x = 0.010\ \text{M}$. (1) The dilute half-cell ($x = 0.010\ \text{M}$) is the anode (oxidation raises its ion concentration). (1)

(c) e.g. ion-selective electrodes / pH meters use concentration-cell EMF to measure unknown ion concentrations. (2)

Q4 (12)

(a) $Q = It = 5.0 \times (2.0\times3600) = 36000\ \text{C}$. (1) $\text{Cu}^{2+} + 2e^- \to \text{Cu}$, so $n=2$. $m = \dfrac{Q\,M}{nF} = \dfrac{36000\times63.5}{2\times96485} = 11.85\ \text{g}$. (3)

(b) $\text{Al}^{3+} + 3e^- \to \text{Al}$, $n=3$. $m = \dfrac{36000\times27.0}{3\times96485} = 3.36\ \text{g}$. (3)

(c) By Faraday's laws mass $\propto$ equivalent weight $M/n$. Same charge → same number of moles of electrons → chemically equivalent (mole-electrons) amounts, but different $n$ and $M$ give different masses. (2)

(d) Anode reaction: $\text{C} + \text{O}^{2-} \to \text{CO}_2/\text{CO} + \text{electrons}$, e.g. $\text{C} + 2\text{O}^{2-} \to \text{CO}_2 + 4e^-$ — oxygen released oxidises the carbon anode, consuming it. (2) Al cannot come from aqueous solution because water/$\text{H}^+$ is reduced in preference to $\text{Al}^{3+}$ (Al is far more negative than the H reduction), so hydrogen forms instead of aluminium. (1)

Q5 (8)

(a) Anode (oxidation): $\text{Fe} \to \text{Fe}^{2+} + 2e^-$ (1) Cathode (reduction): $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^-$ (1) The region of the iron that is oxidised (often less oxygenated area) is the anode. (1)

(b) Zn has more negative $E^\circ$ (−0.76 V) than Fe (−0.44 V), so Zn is oxidised preferentially — it is the sacrificial anode and Fe becomes the cathode (protected). This holds even if scratched, so the whole surface is protected, not just coverage. (3)

(c) Sacrificial anode / impressed-current cathodic protection: connect a more reactive metal (Mg, Zn) to the pipe so it corrodes instead; or apply an external DC to force the pipe to be cathode. (2)

[
 {"claim":"Q1b E_cell = 1.05 V","code":"E=0.80-(-0.25); result = abs(E-1.05)<1e-9"},
 {"claim":"Q1c dG = -202.6 kJ","code":"dG=-2*96485*1.05/1000; result = abs(dG-(-202.6))<0.5"},
 {"claim":"Q1d ln K approx 81.79","code":"lnK=2*96485*1.05/(8.314*298); result = abs(lnK-81.79)<0.1"},
 {"claim":"Q2b E = 1.150 V","code":"E=1.10-(0.0592/2)*log(0.010/0.50,10); result = abs(E-1.150)<0.005"},
 {"claim":"Q3b x = 0.010 M","code":"from sympy import symbols,solve,log,Eq;x=symbols('x',positive=True);sol=solve(Eq(0.0296,(0.0592/2)*log(0.100/x,10)),x);result = abs(float(sol[0])-0.010)<1e-4"},
 {"claim":"Q4a Cu mass 11.85 g","code":"m=36000*63.5/(2*96485); result = abs(m-11.85)<0.05"},
 {"claim":"Q4b Al mass 3.36 g","code":"m=36000*27.0/(3*96485); result = abs(m-3.36)<0.05"}
]