JEE Main 2025 Apr 2 Shift 1, Chemistry Q22: Electrochemical Cell And Nernst Equation

JEE Main2025Apr 2, Shift 1Chemistry

Consider the following electrochemical cell at standard condition.

$A u (s) ∣ Q H_{2}, Q ∣ N H_{4} X (0.01 M) ∣∣ A g^{+} (1 M) ∣ A g (s)$

$E_{ce l l} = + 0.4 V$

The couple $Q H_{2} / Q$ represents quinhydrone electrode, the half cell reaction is given below

$[Given : E_{A g^{+} / A g}^{\circ} = + 0.8 V and \frac{2.303 R T}{F} = 0.06 V]$

The $p K_{b}$ value of the ammonium halide salt $(N H_{4} X)$ used here is ______. (nearest integer)

Your answer

Solution

Cell reaction: oxidation at the quinhydrone electrode, reduction at the silver electrode.

$Q H_{2} + 2 A g^{+} \to Q + 2 A g + 2 H^{+}$

Standard cell EMF:

$E_{ce l l}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{Q / Q H_{2}}^{\circ} = 0.8 - 0.7 = 0.1 V$

Apply the Nernst equation (with $[A g^{+}] = 1 M$ , so it drops out):

$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.06}{2} lo g [H^{+}]^{2} = E_{ce l l}^{\circ} - 0.06 lo g [H^{+}]$

$0.4 = 0.1 + 0.06 \cdot p H \Rightarrow p H = 5$

Relating pH to $p K_{b}$ : $N H_{4} X$ is the salt of a weak base $(N H_{3})$ and the conjugate acid $N H_{4}^{+}$ hydrolyses, giving an acidic solution. For such a salt:

$p H = 7 - \frac{1}{2} p K_{b} - \frac{1}{2} lo g C$

With $C = 0.01 M$ , so $lo g C = - 2$ :

$5 = 7 - \frac{1}{2} p K_{b} - \frac{1}{2} (- 2) = 8 - \frac{1}{2} p K_{b}$

$\frac{1}{2} p K_{b} = 3 \Rightarrow p K_{b} = 6$

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JEE Main 2025 Apr 2 Shift 1, Chemistry Q22: Electrochemical Cell And Nernst Equation

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