JEE Main 2025 — 2 April, Morning Shift — Question 21

Consider the following electrochemical cell at standard condition.

$\mathrm{Au}(\mathrm{s})\left|\mathrm{QH}_{2}, \mathrm{Q}\right| \mathrm{NH}_{4} \mathrm{X}(0.01 \mathrm{M})| | \mathrm{Ag}^{+}(1 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s})$

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

The couple $\mathrm{QH}_{2} / \mathrm{Q}$ represents quinhydrone electrode, the half cell reaction is given below :

$\left[\right.$ Given : $\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.8 \mathrm{~V}$ and $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right]$

The $\mathrm{pK}_{b}$ value of the ammonium halide salt $\left(\mathrm{NH}_{4} \mathrm{X}\right)$ used here is $\_\_\_\_\_$ . (nearest integer)