Consider the following electrochemical cell at standard condition. $ \begin{aligned} & \mathrm{Au}(\mathrm{s})\left|\mathrm{QH}_2, \mathrm{Q} ight| \mathrm{NH}_{4} \mathrm{X}(0.01 \mathrm{M})| | \mathrm{Ag}^{+}(1 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s}) \ & \mathrm{E}_{\mathrm{cell}}=+0.4…

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Consider the following electrochemical cell at standard condition. $ \begin{aligned} & \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}_{\mathrm{cell}}=+0.4 \mathrm{V} \end{aligned}$ 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 $\left(\mathrm{NH}_4 \mathrm{X}\right)$ used here is _________. (nearest integer)

Aicharya · Accepted Answer

$\begin{aligned} & QH_2+2 Ag^{+} ightarrow 2 Ag+Q+2 H^{+} \ & E=E^{\circ}-\frac{0.06}{2} \log \left[H^{+}ight]^2 \ & E=E^{\circ}-0.06 	imes \log \left[H^{+}ight] \ & pH=-\log \left(H^{+}ight)=\frac{E-E^{\circ}}{0.06}=\frac{0.4-0.1}{0.06} \ & =\frac{0.3}{0.06}=5 \end{aligned}$ $\begin{aligned} & pH+NH_4 X=7-\frac{1}{2} pK_{b}-\frac{1}{2} \log C \ & 5=7-\frac{1}{2} 	imes pK_{b}-\frac{1}{2} \log \left(10^{-2}ight) \ & pK_{b}=6 \end{aligned}$

Consider the following electrochemical cell at standard condition. $ \begin{aligned} &…

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