Electrochemistry: NEET previous year questions

Charge passed $Q = i\,t = 1.5 \times (10\times 60) = 900$ C. Deposition follows $\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}$, so mass $= \frac{E\,i\,t}{F} = \frac{63 \times 1.5 \times 600}{2 \times 96487} \approx 0.294$ g, i.e. 0.2938 g. The other values arise from using the wrong n-factor or the wrong time.

For $\mathrm{H}_{2} \rightarrow 2\mathrm{H}^{+} + 2\mathrm{e}^{-}$, the Nernst equation gives $E = E^{\circ} - \frac{0.059}{2}\log\frac{[H^+]^2}{p_{H_2}}$, with [H$^{+}$] = 0.02 M and p(H$_{2}$) = 2 atm. So $E = 0 - 0.0295\log\frac{(0.02)^2}{2} = -0.0295\log(2\times10^{-4}) = -0.0295(0.301 - 4) \approx +0.109$ V.

Electrode potentials are NOT additive when the electron counts differ; the additive quantity is $\Delta G^\circ = -nFE^\circ$. For $\mathrm{Fe}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Fe}$, $\Delta G_1^\circ = -3F(-0.04) = 0.12F$; for $\mathrm{Fe}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Fe}$, $\Delta G_2^\circ = -2F(-0.44) = 0.88F$. Subtracting, $\Delta G_3^\circ = 0.12F - 0.88F = -0.76F$, and since $n = 1$, $E^\circ = +0.76\ \text{V}$. NCERT Class 12, Electrochemistry (electrode potential and Gibbs energy).

For a strong electrolyte the Debye-Huckel-Onsager relation is $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$, where the plot's slope is $-A$. At $c = 0.01\ \text{M}$, $\sqrt{c} = 0.1$, so $\Lambda_m^0 = 145.0 + (90.0)(0.1) = 154.0$. Kohlrausch's law then gives $\lambda^0_{Y^-} = 154.0 - 74.0 = 80.0\ \text{S cm}^2\text{mol}^{-1}$. NCERT Class 12, Electrochemistry (variation of conductivity with concentration).

The reaction for the production of calcium is $\mathrm{Ca}^{2+} + 2\mathrm{e}^- → \mathrm{Ca}$. The number of moles of calcium produced is $\frac{20\ g}{40\ g mol^{-1}} = 0.5\ mol$. Since 2 moles of electrons are required to produce 1 mole of calcium, the number of Faradays required is $0.5 \times 2 = 1\ F$. Chapter: Electrochemistry.

On electrolysis of dilute sulfuric acid with platinum electrodes, oxygen gas is evolved at the anode due to the oxidation of water. The reaction is $2\mathrm{H}_{2}\mathrm{O} → \mathrm{O}_{2} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-}$. Chapter: Electrochemistry.