Question

The solubility product of $AgCl$ and $AgBr$ are $2\times 10^{-10}$ and $1\times 10^{-12}$ respectively at $25^{\circ }C$. Calculate the ratio of concentrations of $C{l}^{-}$and $B{r}^{-}$ions at which the cell reaction will be in equilibrium $25^{\circ }C$.
$A{g}_{(s)},AgB{r}_{(s)},B{r}^{-}\Vert C{l}^{-},AgC{l}_{(s)},A{g}_{(s)}$

Answer 1

Answer: 200

In order that cell reaction may be in equilibrium, $E_{\text{cell }}$ must be zero
The half-cell reactions and cell reactions for one Faraday of electricity are as given below

$=E_{AgC{l}_{(s)},A{g}_{(s)},C{l}^{-}}^{\circ }-E_{AgB{r}_{(s)},A{g}_{(s)},B{r}^{-}}^{\circ }-0.0591\log \frac{\left[C{l}^{-}\right]}{\left[B{r}^{-}\right]}$ ...(i)
For the half cell $A{g}_{(s)},Ag{X}_{(s)},X^{-}$the half cell reaction (reduction) is

$0.0591\log K_{sp}=E_{AgX,Ag,X^{-}}^{\circ }-E_{A{g}^{+},Ag}^{\circ }$
$\therefore E_{AgX,Ag,X^{-}}^0=0.0591\log K_{sp}+E_{A{g}^{+},Ag}^0$
So, equation (i) may be put as
$E_{\text{cell }}=0.0591\log K_{sp\left(AgCl\right)}+E_{A{g}^{+},Ag}^{\circ }-0.0591\log K_{sp\left(AgBr\right)}-E_{A{g}^{+},Ag}^{\circ }-0.0591\log \frac{\left[C{l}^{-}\right]}{\left[B{r}^{-}\right]}$
$=0.0591\left[\log \left(2\times 10^{-10}\right)-\log \left(1\times 10^{-12}\right)-\log \frac{\left[C{l}^{-}\right]}{\left[B{r}^{-}\right]}\right]$
$=0.0591\left(\log \frac{2\times 10^{-10}}{1\times 10^{-12}}-\log \frac{\left[C{l}^{-}\right]}{\left[B{r}^{-}\right]}\right)$
At equilibrium
$E_{\text{cell }}=0$
so $\frac{\left[C{l}^{-}\right]}{\left[B{r}^{-}\right]}=\frac{2\times 10^{-10}}{1\times 10^{-12}}$
$=200$