(a) (i) When the concentration of the electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity. It is represented by $\Lambda_m$. (ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc. Given,For $0.1 \mathrm{mol} \mathrm{L}^{-1} \mathrm{KCl}$ solutionResistance $(R) = 100 \Omega$Conductivity $(k) = 1.29 \times 10^{-2} \Omega^{-1} \mathrm{cm}^{-1}$Cell constant $(G^*) = k \times R$$= 1.29 \times 10^{-2} \Omega^{-1} \mathrm{cm}^{-1} \times 100 \Omega$$= 1.29 \mathrm{cm}^{-1}$For $0.02 \mathrm{mol} \mathrm{L}^{-1} \mathrm{KCl}$ solutionResistance $(R) = 520 \Omega$$\text{Conductivity} = \frac{\text{Cell constant } (k)}{\text{Concentration } (C)}$$K = \frac{1.29 \mathrm{cm}^{-1}}{520 \Omega}$$K = 2.48 \times 10^{-3} \Omega^{-1} \mathrm{cm}^{-1}$$(C) = 0.02 \mathrm{mol} \mathrm{L}^{-1}$$= 0.02 \times 10^{-3} \mathrm{mol} \mathrm{cm}^{-3}$Concentration $(C) = 0.02 \mathrm{mol} \mathrm{L}^{-1}$$= 0.02 \times 10^{-3} \mathrm{mol} \mathrm{cm}^{-3}$ Molar conductivity $(\Lambda_m)$ $= \frac{\text{Conductivity } (k)}{\text{Concentration } (C)}$$= \frac{2.48 \times 10^{-3} \Omega^{-1} \mathrm{cm}^{-1}}{0.02 \times 10^{-3} \mathrm{mol} \mathrm{cm}^{-3}}$$\Lambda_m = 124 \Omega^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1}$ OR (a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed." $m = Z \times Q$where, $m =$ mass of a substance deposited or liberated at an electrode.$Q =$ amount of charge passed through it and$Z =$ electrochemical equivalentThe reduction of one $\mathrm{mol}$ of $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}$ can be represented as:$\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$Since, $2 \mathrm{mol}$ of electrons are involved in the reduction, so the amount of charge required is $2 \mathrm{F}$. (b) The given cell reaction can be represented as $\mathrm{Mg}(\mathrm{s}) + \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq})$ $+ \mathrm{Cu}(\mathrm{s})$$E_{\text{cell}} = E^{\circ} - \frac{2.303 RT}{n F} \log \frac{\mathrm{Mg}^{2+}}{\mathrm{Cu}^{2+}}$$E_{\text{cell}} = 2.71 - \frac{2.303 \times 0.0831 \times 298}{2 \times 96500}$ $\log \frac{0.1}{0.01}$$E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log 10$$E_{\text{cell}} = 2.68 \mathrm{V}$