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Solve the following.
Calculate emf of the following cell at : [Given , ] State Faraday’s first law of electrolysis…
(i)[3.0]
Calculate emf of the following cell at $298\text{ K}$:
$\text{Mg}(s) \mid \text{Mg}^{2+}(0\cdot1\text{ M}) \parallel \text{Cu}^{2+}(0\cdot01\text{ M}) \mid \text{Cu}(s)$
[Given $E^\circ_\text{cell} = +2\cdot71\text{ V}$, $1\text{ Faraday} = 96500\text{ C mol}^{-1}$]
(ii)[2.0]
State Faraday’s first law of electrolysis. Calculate the charge required in terms of Faraday for the reduction of $1\text{ mole}$ of $\text{Cu}^{2+}$ to $\text{Cu}$.
Answer
Answer (i)
AIFormula used: Nernst equation at 298 K: E = E0 - 0.0591/n*log10(Q) with Q = 10, n = 2, E0 = 2.71 V, so E = 2.68 V
Cell reaction: $\mathrm{Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu}$, $n = 2$
$E = E^\circ - \dfrac{0\cdot0591}{n}\log\dfrac{[\mathrm{Mg^{2+}}]}{[\mathrm{Cu^{2+}}]} = 2\cdot71 - \dfrac{0\cdot0591}{2}\log\dfrac{0\cdot1}{0\cdot01}$
$E = 2\cdot71 - 0\cdot02955 \times 1 = 2\cdot68\ \mathrm{V}$
Final answer: 2.68 V
Answer (ii)
AIFaraday's first law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of charge passed through the electrolyte.
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$: 2 mol of electrons are needed for 1 mol of Cu.
Final answer: 2 F
From ISC 2018 Specimen Chemistry Paper 1, question 16(b).
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