Question

The voltage of the cell given below is $-0.46V$.
$Pt\left|\begin{array}{cc}NaHS{O}_3& N{a}_{2}S{O}_3\\ 0.4M& 6.44\times {10}^{-3}M\end{array}\right|\left|\begin{array}{c}Z{n}^{2+}\\ 0.3M\end{array}\right|Zn\left(s\right)$
Also, $Z{n}^{2+}+2e\to Zn\left(s\right);E^o=-0.763V$
The value of $K_2$ where $K_2=\frac{\left[H^{+}\right]\left[S{O}_3^{2-}\right]}{\left[HS{O}_3^{-}\right]}$ is:

• A. $6.44\times {10}^{-8}$
• B. $5.43\times {10}^{-8}$
• C. $7.34\times {10}^{-8}$
• D. None of these

Answer 1

The correct option is A $6.44\times {10}^{-8}$

$E_{cell}=E_{O{P}_H}^o+E_{R{P}_{Z{n}^{2+}/Zn}}^o+\frac{0.059}{2}log\frac{\left[Z{n}^{2+}\right]}{[H^{+}{]}^2}$
$-0.46=0-0.763+\frac{0.059}{2}log\frac{\left[0.03\right]}{[H^{+}{]}^2}$
$\therefore \left[H^{+}\right]=4.0\times {10}^{-6}$
$\because HS{O}_3^{-}\rightleftharpoons H^{+}+S{O}_3^{2-}$
Now, $K_2=\frac{\left[H^{+}\right]\left[S{O}_3^{2-}\right]}{\left[HS{O}_3^{-}\right]}$
The dissociation of $HS{O}_3^{-}$ is suppressed in presence of $S{O}_3^{2-}$ due to common ion effect. Thus $\left[S{O}_3^{2-}\right]=6.44\times {10}^{-3}M$ and $\left[HS{O}_3^{-}\right]=0.4M$
$K_2=\frac{4\times {10}^{-6}\times 6.44\times {10}^{-3}}{0.4}=6.44\times {10}^{-8}$