Question

Find the potential of a half - cell that has the reaction, $A{g}_{2}S+2e^{-}\to 2Ag+S^{2-}$ in a solution buffered at pH = 3 and which is also saturated with 0.1 M $H_{2}S$.
For $H_{2}S:K_1={10}^{-8}$ and $K_2=2\times {10}^{-13},K_{sp}\left(A{g}_{2}S\right)=2\times {10}^{-48},E_{A{g}^{+}|Ag}^0=0.8V$

• A. $0.432V$
• B. $1.658V$
• C. $-0.245V$
• D. $-0.144V$

Answer 1

The correct option is D $-0.144V$

Since the solution is buffered at pH = 3, so
$H_{2}S\stackrel{Ka=K_1\times K_2}{\rightleftharpoons }2H^{+}+S^{2-}$
$\therefore K_a={10}^{-8}\times 2\times {10}^{-13}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$
$\Rightarrow 2\times {10}^{-21}=\frac{\left({10}^{-3}{)}^2\right[S^{2-}]}{0.1}$ $\Rightarrow \left[S^{2-}\right]=2\times {10}^{-16}M$
Again
$A{g}_{2}S\left(s\right)\stackrel{K_{sp}}{\rightleftharpoons }2A{g}^{+}+S^{2-}$
$\therefore K_{sp}=\left[A{g}^{+}{]}^2\right[S^{2-}]$ $\Rightarrow 2\times {10}^{-48}=[A{g}^{+}{]}^2\times 2\times {10}^{-16}$ $\Rightarrow \left[A{g}^{+}\right]={10}^{-16}M$
So, for cell $A{g}_{2}S+2e^{-}\to 2Ag+S^{2-}when\left[S^{2-}\right]=2\times {10}^{-16}M$
Which is equivalent to emf of the cell $A{g}^{+}+e^{-}\to Ag$ $when\left[A{g}^{+}\right]={10}^{-16}M$
Hence $E=E^0-\frac{0.059}{1}log\frac{1}{\left[A{g}^{+}\right]}\Rightarrow E=0.8-\frac{0.059}{1}log\frac{1}{{10}^{-16}}=0.8-0.944=-0.144V$