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 $K_2=\large\frac{[H^+][SO_3^{-2}]}{[HSO_3^-]}$

Question

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