Question

If it is desired to construct the following voltaic cell to have $E_{cell}=0.0860V$, what $\left[C{l}^{-}\right]$ must be present in the cathodic half cell to achieve the desired e.m.f.? Given $K_{sp}$ of $AgCl$ and $AgI$ are $1.8\times {10}^{-10}$ and $8.5\times {10}^{-17}$ respectively.
$Ag\left(s\right)|A{g}^{+}[Sat.AgI(aq\left)\right]||A{g}^{+}(Sat.AgCl,MC{l}^{-})|Ag\left(s\right)$

• A. $8.8\times {10}^{-4}M$
• B. $6.8\times {10}^{-4}M$
• C. $4.8\times {10}^{-4}M$
• D. $5.8\times {10}^{-4}M$

Answer 1

The correct option is B $6.8\times {10}^{-4}M$

$E_{cell}=E_{R{P}_{A{g}^{+}/Ag}}^o+\frac{0.059}{1}log\frac{[A{g}^{+}{]}_{R.H.S.}}{[A{g}^{+}{]}_{L.H.S.}}$

or $0.0860=\frac{0.059}{1}log\frac{[A{g}^{+}{]}_{R.H.S.}}{[A{g}^{+}{]}_{L.H.S.}}$

Also $[A{g}^{+}{]}_{L.H.S.}$ can be derived as $\left[A{g}^{+}\right]=\sqrt{K_{s{p}_{Agl}}}=\sqrt{(8.5\times {10}^{-17})}$ $=9.22\times {10}^{-9}M$

$\therefore 0.0860=\frac{0.059}{1}log\frac{[A{g}^{+}{]}_{R.H.S.}}{9.22\times {10}^{-9}}$

or $\frac{[A{g}^{+}{]}_{R.H.S.}}{9.22\times {10}^{-9}}=28.68$

$\therefore [A{g}^{+}{]}_{R.H.S.}=18.68\times 9.22\times {10}^{-9}M$

Also for $R.H.S.,\left[A{g}^{+}\right]\left[C{l}^{-}\right]=K_{s{p}_{AgCl}}$

$\therefore \left[C{l}^{-}\right]=\frac{K_{s{p}_{AgCl}}}{\left[A{g}^{+}\right]}=\frac{1.8\times {10}^{-10}}{28.68\times 9.22\times {10}^{-9}}$

or $\left[C{l}^{-}\right]=6.8\times {10}^{-4}M$