JEE Main 2025 — Electrochemistry Question with Solution

The cell reaction is:

${\mathrm{QH}}_2+2{\mathrm{Ag}}^{+}\to 2\mathrm{Ag}+\mathrm{Q}+2{\mathrm{H}}^{+}$

The Nernst equation for the reaction can be expressed as:

$\mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.06}{2}\log {\left[{\mathrm{H}}^{+}\right]}^2$

Simplifying, we obtain:

$\mathrm{E}={\mathrm{E}}^{\circ }-0.06\log \left[{\mathrm{H}}^{+}\right]$

Given data includes:

${\mathrm{E}}_{\text{cell}}=+0.4\mathrm{V}$

Standard potential for ${\mathrm{Ag}}^{+}/\mathrm{Ag}$, ${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^0=+0.8\mathrm{V}$

Using the cell potential and the standard potential, we calculate:

$\mathrm{pH}=-\log \left({\mathrm{H}}^{+}\right)=\frac{\mathrm{E}-{\mathrm{E}}^{\circ }}{0.06}=\frac{0.4-0.1}{0.06}$

$\mathrm{pH}=\frac{0.3}{0.06}=5$

This established pH must now relate to the buffer equation involving ${\mathrm{NH}}_4\mathrm{X}$, at

$\mathrm{pH}+{\mathrm{NH}}_4\mathrm{X}=7-\frac{1}{2}{\mathrm{pK}}_{\mathrm{b}}-\frac{1}{2}\log \mathrm{C}$

Given the concentration $\mathrm{C}=0.01\mathrm{M}=10^{-2}$, we substitute into the equation:

$5=7-\frac{1}{2}\times {\mathrm{pK}}_{\mathrm{b}}-\frac{1}{2}\log (10^{-2})$

Simplifying further, the equation resolves to:

${\mathrm{pK}}_{\mathrm{b}}=6$

Thus, the calculated ${\mathrm{pK}}_{\mathrm{b}}$ value of the ammonium halide salt ${\mathrm{NH}}_4\mathrm{X}$ is 6.