JEE Main 2025 — Electrochemistry Question with Solution

$$\underset{(Anode)}{Mg\left|M{g}^{2+}(aq)\right|}\underset{(Cathode)}{\left|A{g}^{+}(aq)\right|Ag}$$

Nernst equation (general)

$$E_{Cell}=E_{Cell}^0-\frac{RT}{nF}\ln Q$$

For the cell, the chemical equation can be written as

$${\mathrm{Mg}}_{(\mathrm{s})}+2{\mathrm{Ag}}_{(\text{aq })}^{+}⟶{\mathrm{Mg}}_{(\text{aq })}^{2+}+2\mathrm{Ag}(\mathrm{s})$$

$n$ (number of electrons transferred $)=2$

$$E_{cell}=E_{cell}^o-\frac{RT}{2F}\ln \frac{\left[M{g}^{2+}\right]{\left[Ag\right]}^2}{\left[Mg\right]{\left[A{g}^{+}\right]}^2}$$

$$E_{cell}=E_{cell}^o-\frac{RT}{2F}\ln \frac{\left[M{g}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}$$

$$E_{cell}=E_{cell}^o+\frac{RT}{2F}\ln \frac{{\left[A{g}^{+}\right]}^2}{\left[M{g}^{2+}\right]}$$

Concentration of solid species

$$[\mathrm{Mg}]=[\mathrm{Ag}]=1$$

Correct answer :

Option D) $$E_{cell}=E_{cell}^o+\frac{RT}{2F}\ln \frac{{\left[A{g}^{+}\right]}^2}{\left[M{g}^{2+}\right]}$$