Solve the following.
(i)[2.5]
Calculate the mass of silver deposited at cathode when a current of $2\text{ amperes}$ is passed through a solution of $\text{AgNO}_3$ for $15\text{ minutes}$.
(at. wt. of $\text{Ag} = 108, 1\text{ F} = 96,500\text{ C}$)
(ii)[2.5]
Calculate the emf and $\Delta G$ for the cell reaction 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 $E^\circ_\text{cell} = 2\cdot71\text{ V}$, $1\text{F} = 96,500\text{ C}$
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Answer (i)
AIFormula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, I = 2 A, M = 108 g mol^-1, n = 1, t = 900 s, so w = 2.015 g
Formula: $w = \dfrac{M \times I \times t}{n \times F}$
$t = 15 \times 60 = 900\text{ s}$, $Q = It = 2\text{ A} \times 900\text{ s} = 1800\text{ C}$
$w = \dfrac{108\text{ g mol}^{-1} \times 1800\text{ C}}{1 \times 96500\text{ C mol}^{-1}} = 2\cdot015\text{ g}$
Final answer: 2.015 g
Answer (ii)
AIFormulas used: Nernst equation at 298 K: E = E0 - 0.0591/n*log10(Q) with Q = 10, n = 2, E0 = 2.71 V, so E = 2.68045 V Gibbs energy from emf: dG = -n*F*E with E = 2.68045 V, F = 96500 C mol^-1, n = 2, so dG = -517327 J mol^-1
Cell reaction: $\mathrm{Mg} + \mathrm{Cu^{2+}} \rightarrow \mathrm{Mg^{2+}} + \mathrm{Cu}$, $n = 2$
$Q = \dfrac{[\mathrm{Mg^{2+}}]}{[\mathrm{Cu^{2+}}]} = \dfrac{0\cdot1}{0\cdot01} = 10$
Nernst equation: $E = E^\circ - \dfrac{0\cdot0591}{n}\log Q = 2\cdot71 - \dfrac{0\cdot0591}{2}\log 10 = 2\cdot68\text{ V}$
$\Delta G = -nFE = -2 \times 96500\text{ C mol}^{-1} \times 2\cdot68045\text{ V} = -5\cdot17\times10^{5}\text{ J mol}^{-1}$
Final answer: $E = 2.68\text{ V},\ \Delta G = -517\text{ kJ mol}^{-1}$