Question

Consider the cell $Ag\left(s\right)|AgBr\left(x\right)|B{r}^{-}\left(aq\right)||C{l}^{-}\left(aq\right)|AgCl\left(s\right)|Ag\left(s\right)$ at 25${}^o$C. The solubility product constants of $AgBr$ and $AgCl$ are $5\times {10}^{-13}$ and $1\times {10}^{-10}$ respectively. For what ratio of the concentrations of Br${}^{-}$ and Cl${}^{-}$ ions, would the emf of the cell be zero?

• A. $1:200$
• B. $1:100$
• C. $1:500$
• D. $200:1$

Answer 1

Answer: (A)

$1:200$

$\left[A{g}^{+}\right]LHS=K_{sp}\frac{AgBr}{B{r}^{-}}=\frac{5\times {10}^{-13}}{a}$

$\left[A{g}^{+}\right]RHS=K_{sp}\frac{AgCl}{C{l}^{-}}=\frac{1\times {10}^{-10}}{b}$

$E_{B{r}^{-}/AgBr/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{0.059}{1}log{K}_{sp}\left(AgBr\right)=E_{A{g}^{+}/Ag}^0-0.7257$ and

$E_{C{l}^{-}/\left(AgCl\right)/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{0.059}{1}log{K}_{sp}\left(AgCl\right)=E_{A{g}^{+}/Ag}^0-0.59$

Now cell reaction is

$Ag+B{r}^{-}\to AgBr+1e^{-}$
$AgCl+1e^{-}\to Ag+C{l}^{-}$
----------------------------------------------------------
$B{r}^{-}+AgCl\stackrel{1e^{-}}{\to }C{l}^{-}+AgBr$

If emf of the cell is zero, then

$0=(0.7257-0.59)+\frac{0.059}{1}log\frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}$

$\Rightarrow \frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}=0.005=1/200i.e.1:200$

Hence, the correct option is $A$