JEE Main 2024 — 5 April, Shift 2 — Question 73

Given:

$$E^\circ(\mathrm{M^{2+}/M}) = -0.46\,\text{V}$$ $$E^\circ(\mathrm{X/X^{2-}}) = +0.34\,\text{V}$$

Step 1: Identification of cathode and anode

Since, $E^\circ(\mathrm{X/X^{2-}}) > E^\circ(\mathrm{M^{2+}/M})$ Thererfore, X undergoes reduction and M undergoes oxidation.

Step 2: Half-cell reactions

Oxidation (anode): $\mathrm{M \rightarrow M^{2+} + 2e^-}$ Reduction (cathode): $\mathrm{X + 2e^- \rightarrow X^{2-}}$ Step 3: Calculation of cell potential

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 0.34 - (-0.46) = +0.80\,\text{V}$$

Since, $E^\circ_{\text{cell}} > 0$, the reaction is spontaneous.

Overall spontaneous reaction: $\mathrm{M^{2+} + X^{2-} \rightarrow M + X}$