Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

The relationship between conductivity $S$, area $A$, concentration $c$, and length $l$ is expressed by the formula $S = \kappa \frac{A \cdot c}{l}$, where $\kappa$ denotes the constant of proportionality. Solving for $\kappa$ yields the equation $\kappa = \frac{S \cdot l}{A \cdot c}$. By substituting the corresponding SI units-Siemens for $S$, meters for $l$, square meters for $A$, and $mol \cdot m^{-3}$ for $c$-the calculation results in $\kappa = \frac{S \cdot m}{m^2 \cdot mol \cdot m^{-3}}$. This simplifies to the final dimensional unit of $\kappa = S \cdot m^2 \cdot mol^{-1}$.

In electrochemical cell notation, it is standard practice to place the anode, where oxidation takes place, on the left, while the cathode, where reduction occurs, is placed on the right. The electromotive force of the cell is calculated by finding the potential difference between these two electrodes, which is done by subtracting the reduction potential of the left electrode from the reduction potential of the right electrode. This calculation is represented by the formula $E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}$.

Anode reactions are defined by the process of oxidation, which involves the loss of electrons and a corresponding increase in the oxidation state of the chemical species. In the reaction $2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cr}_2\mathrm{O}_7^{2-} + 14 \mathrm{H}^{+}$, the oxidation state of chromium increases from $+3$ to $+6$, which confirms this as an oxidation process suitable for the anode. Since the other provided reactions represent reduction where oxidation states decrease, they are characteristic of the cathode rather than the anode.

The principle of electrolytic refining relies on the fact that the pure metal consistently deposits at the cathode while the impure metal serves as the anode. To purify copper contaminated with zinc, the impure sample must function as the anode to release ions into the solution, while a strip of pure copper acts as the cathode to facilitate the reduction process $Cu^{2+} + 2e^{-} \rightarrow Cu(s)$.

This concentration cell generates potential by maintaining hydrogen gas at different pressures across two identical electrodes. Since the oxidation of hydrogen at pressure $P_1$ and the reduction at pressure $P_2$ result in a net transfer of gas with $n = 2$ electrons, the standard cell potential is zero. Using the Nernst equation, we express the cell potential as $E_{\text{cell}} = 0 - \frac{RT}{2F} \log_e \frac{P_2}{P_1}$. Applying properties of logarithms to rearrange this expression leads to the final value of

$\frac{RT}{2F}\log_e\frac{P_1}{P_2}$