1. Identify the electrodes

The cell is $\mathrm{Au}(s)|\mathrm{QH_2},\mathrm{Q}|\mathrm{NH_4X}(0.01\,M)||\mathrm{Ag^+}(1\,M)|\mathrm{Ag}(s)$ with $E_{\text{cell}}=+0.4\,V$

Right electrode is silver: $\mathrm{Ag^+ + e^- \rightarrow Ag}$ Since $[\mathrm{Ag^+}]=1\,M$, $E_{\mathrm{Ag^+/Ag}}=E^\circ_{\mathrm{Ag^+/Ag}}=0.8\,V$

2. Find the potential of the quinhydrone electrode

Using $E_{\text{cell}}=E_{\text{right}}-E_{\text{left}}$ we get $0.4=0.8-E_{\text{QH}_2/\text{Q}}$ So, $E_{\text{QH}_2/\text{Q}}=0.4\,V$

3. Use the quinhydrone electrode equation

For quinhydrone electrode, $\mathrm{Q + 2H^+ + 2e^- \rightarrow QH_2}$ Its Nernst equation is $E=E^\circ-\frac{0.06}{2}\log\frac{[\mathrm{QH_2}]}{[\mathrm{Q}][\mathrm{H^+}]^2}$

Since quinhydrone contains equimolar quinone and hydroquinone, $\frac{[\mathrm{QH_2}]}{[\mathrm{Q}]}=1$ Hence, $E=E^\circ-\frac{0.06}{2}\log\frac{1}{[\mathrm{H^+}]^2}$ $E=E^\circ-0.06\,\text{pH}$

For quinhydrone electrode, $E^\circ=0.7\,V$ Thus, $0.4=0.7-0.06\,\text{pH}$ $0.06\,\text{pH}=0.3$ $\text{pH}=5$

4. Relate pH to hydrolysis of $\mathrm{NH_4X}$

$\mathrm{NH_4X}$ is a salt of weak base $\mathrm{NH_4OH}$ and strong acid, so its solution is acidic.

For a salt of weak base and strong acid, $\text{pH}=\frac{1}{2}(pK_w-pK_b-\log C)$ Here,

• $C=0.01=10^{-2}$
• $pK_w=14$
• $\text{pH}=5$

So, $5=\frac{1}{2}(14-pK_b-\log 10^{-2})$ Since $\log 10^{-2}=-2$ therefore, $5=\frac{1}{2}(14-pK_b+2)$ $10=16-pK_b$ $pK_b=6$

5. Final answer

$\boxed{6}$