The cell reaction is:

$ \mathrm{QH}_2 + 2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag} + \mathrm{Q} + 2 \mathrm{H}^{+} $

The Nernst equation for the reaction can be expressed as:

$ \mathrm{E} = \mathrm{E}^{\circ} - \frac{0.06}{2} \log \left[\mathrm{H}^{+}\right]^2 $

Simplifying, we obtain:

$ \mathrm{E} = \mathrm{E}^{\circ} - 0.06 \log \left[\mathrm{H}^{+}\right] $

Given data includes:

$\mathrm{E}_{\text{cell}} = +0.4 \, \mathrm{V}$

Standard potential for $\mathrm{Ag}^{+}/\mathrm{Ag}$, $\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0 = +0.8 \, \mathrm{V}$

Using the cell potential and the standard potential, we calculate:

$ \mathrm{pH} = -\log \left(\mathrm{H}^{+}\right) = \frac{\mathrm{E} - \mathrm{E}^{\circ}}{0.06} = \frac{0.4 - 0.1}{0.06} $

$ \mathrm{pH} = \frac{0.3}{0.06} = 5 $

This established pH must now relate to the buffer equation involving $\mathrm{NH}_4 \mathrm{X}$, at

$ \mathrm{pH} + \mathrm{NH}_4 \mathrm{X} = 7 - \frac{1}{2} \mathrm{pK}_{\mathrm{b}} - \frac{1}{2} \log \mathrm{C} $

Given the concentration $\mathrm{C} = 0.01 \, \mathrm{M} = 10^{-2}$, we substitute into the equation:

$ 5 = 7 - \frac{1}{2} \times \mathrm{pK}_{\mathrm{b}} - \frac{1}{2} \log (10^{-2}) $

Simplifying further, the equation resolves to:

$ \mathrm{pK}_{\mathrm{b}} = 6 $

Thus, the calculated $\mathrm{pK}_{\mathrm{b}}$ value of the ammonium halide salt $\mathrm{NH}_4 \mathrm{X}$ is 6.