Consider the following electrochemical cell at standard condition. The couple represents quinhydrone electrode, the half cell reaction is given below. [IMAGE 158] The value of the ammonium halide salt used here is _________. (nearest inte

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Solution

Q.
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.

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

The $p K_{b}$ value of the ammonium halide salt $(N H_{4} X)$ used here is _________. (nearest integer). [2025]

Ans.
(6)

Anode half reaction:

$Q H_{2} \to Q + 2 e^{-} + 2 H^{+}, E_{Q / Q H_{2}}^{\circ} = 0.7 V$

Cathode half reaction:

$A g^{+} + e^{-} \to A g, E_{A g^{+} / A g}^{\circ} = 0.8 V$

Cell reaction:

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

By Nernst equation:

$E_{cell} = E_{cell}^{\circ} - \frac{0.06}{2} \log \frac{{[H^{+}]}^{2} [Q]}{[Q H_{2}] {[A g^{+}]}^{2}}$

In quinone-hydroquinone electrode, the concentration of Q and ${QH}_{2}$ is maintained equal.

So, $E_{cell} = E_{cell}^{\circ} - \frac{0.06}{2} \log \frac{{[H^{+}]}^{2}}{{[A g^{+}]}^{2}}$

$0.4 = 0.1 - \frac{0.06}{2} \log \frac{{[H^{+}]}^{2}}{1^{2}}$

$0.4 = 0.1 - 0.06 \log [H^{+}]$

$pH = 5$

$H^{+}$ in the anode electrolyte comes from $N H_{4} X$ , which is a salt of weak base and strong acid.

$pH for such a salt = \frac{1}{2} (p K_{w} - p K_{b} - \log [N H_{4} X])$

$5 = \frac{1}{2} (14 - p K_{b} - \log 0.01)$

$10 = 14 - p K_{b} + 2$

$p K_{b} = 6$

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