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# The voltage of cell given is -0.46V $Pt/H_2$(atm)/$NaHSO_3(0.4M),Na_2SO_3(6.44\times 10^{-3}M)\parallel Zn^{+2}(0.3M)/Zn(g)$.Also $Zn^{+2}+2e^-\rightarrow Zn(s) \leftrightharpoons E^0=-0.763$V.Then the value of $K_2=\large\frac{[H^+][SO_3^{-2}]}{[HSO_3^-]}$

$\begin{array}{1 1}(a)\;7.2\times 10^{-8}\\(b)\;1.2\times 10^{-5}\\(c)\;6.4\times 10^{-6}\\(d)\;3.6\times 10^{-6}\end{array}$

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A)
The Nernst expression for this cell is
$E_{cell}=E^0_{Zn^{+2}/Zn}-E^0_{H^+/H_2}-\large\frac{0.59}{2}$$\log \large\frac{[H^+]^2}{P_{H_2}[Zn^{+2}]} -0.46=0.763-\large\frac{0.59}{2}$$\log \large\frac{[H^+]^2}{1\times 0.3}$
$[H^+]=4\times 10^{-6}$
This $[H^+]$ is an odd compartment comes from $HSO_3^-$.Dissociation of $HSO_3^-$ is suppressed in presence of $SO_3^{-2}$ due to common ion effect.
Let degree of dissociation of $HSO_3^-$ in presence of $SO_3^{-2}$ ion is $\alpha$
$HSO_3^-\leftrightharpoons H^++SO_3^{-2}$
$Na_2SO_3\rightarrow 2Na^++SO_3^{--}(6.44\times 10^{-3})$
As it can be seen that $[H^+]=C\alpha=4\times 10^{-6}$,these $C\alpha$ is ignorable with respect to $6.44\times 10^{-3}$
$\therefore [HSO_3^-]=0.4-4\times 10^{-6}\approx 0.4,[H^+]=4\times 10^{-6}$
$[SO_3^{--}]=4\times 10^{-6}+6.4\times 10^{-3}=6.4\times 10^{-3}$
$K_2=\large\frac{[H^+][SO_3^{-2}]}{[HSO_3^-]}=\large\frac{4\times 10^{-6}\times 6.44\times 10^{-3}}{0.4}$
$\Rightarrow 6.4\times 10^{-6}$
Hence (c) is the correct answer.