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

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Redox Reactions and Electrochemistry

Consider the following electrochemical cell at standard condition.

Au (s) ∣ QH_{2}, Q ∣ NH_{4} X (0.01 M) ∣∣ Ag^{+} (1 M) ∣ Ag (s) E_{cell} = + 0.4 V

The couple

QH_{2} / Q

represents quinhydrone electrode, the half cell reaction is given below

The

pK_{b}

value of the ammonium halide salt (

NH_{4} X

) used here is ______. (nearest integer)

Answer: 6

The quinhydrone couple is the oxidation site,

QH X_{2} Q + 2 H X^{+} + 2 e X^{-}

, while silver ion is reduced,

Ag X^{+} + e X^{-} Ag

, giving the overall change

QH X_{2} + 2 Ag X^{+} Q + 2 Ag + 2 H X^{+}

. The standard potential is

E^{\circ} = E_{Ag X^{+} / Ag}^{\circ} - E_{Q / Q H_{2}}^{\circ} = 0.8 - 0.7 = 0.1 V

. With

[Ag X^{+}] = 1

and equimolar Q and

QH X_{2}

, the Nernst equation reads

E = E^{\circ} - \frac{0.06}{2} lo g [H X^{+}]^{2} = 0.1 + 0.06 pH

, so

E = 0.4 V

fixes

pH = 5

. The salt

NH X_{4} X

is that of a weak base and a strong acid, for which

pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)

; putting

C = 0.01 M

gives

5 = \frac{1}{2} (14 - p K_{b} + 2)

, so the value of

p K_{b}

6

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Integer answer