Question

At what $\frac{\left[B{r}^{-1}\right]}{\sqrt{\left[C{O}_3^2{}^{-}\right]}}$ does the following cell have its reaction at equilibrium?

$Ag\left(s\right)|A{g}_{2}C{O}_3\left(s\right)|N{a}_{2}C{O}_3\left(aq\right)||KBr\left(aq\right)|AgBr\left(s\right)|Ag\left(s\right)$

$K_{sp}$ = $8\times {10}^{-12}forA{g}_{2}C{O}_{3}and{K}_{sp}=4\times {10}^{-13}forAgBr$

• A. $\sqrt{1}\times {10}^{-7}$
• B. $\sqrt{2}\times {10}^{-7}$
• C. $\sqrt{3}\times {10}^{-7}$
• D. $\sqrt{4}\times {10}^{-7}$

Answer 1

The correct option is B

$\sqrt{2}\times {10}^{-7}$

$AnodeAg\left(s\right)\to A{g}^{+}\left(aq\right)+e^{-}$

$\underset{–}{CathodeA{g}^{+}\left(aq\right)+e^{-}\to Ag}$

Net : $A{g}^{+}\left(AgBr\right)\to A{g}^{+}\left(A{g}_{2}C{O}_3\right)$

$E=E^{\circ }-\frac{0.059}{1}logK$

This is a concentration cell at equilibrium, both $E$ and $E^{\circ }$ will be zero here!

$K=\frac{A{g}^{+}\left(A{g}_{2}C{O}_3\right)}{A{g}^{+}\left(AgBr\right)}$

Let's write this in terms of $K_{sp}$ values since that's what is given in the question.

$K_{sp}\left(A{g}_{2}C{O}_3\right)=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}],K_{sp}(AgBr)=[A{g}^{+}\left]\right[B{r}^{-}]$.

$K=\frac{A{g}^{+}\left(A{g}_{2}C{O}_3\right)}{A{g}^{+}\left(AgBr\right)}$

$A{g}^{+}\left(A{g}_{2}C{O}_3\right)=\sqrt{\frac{K_{sp}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}}$

$A{g}^{+}\left(AgBr\right)=\frac{K_{sp}\left(AgBr\right)}{\left[B{r}^{-}\right]}$

$0=0+\frac{0.059}{1}log\frac{\frac{K_{sp}AgBr}{\left[B{r}^{-}\right]}}{\sqrt{\frac{K_{sp}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}}}\Rightarrow \frac{K_{sp}AgBr}{\left[B{r}^{-}\right]}=\sqrt{\frac{K_{sp}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}}$

$\Rightarrow \frac{4\times {10}^{-13}}{\sqrt{8\times {10}^{-12}}}=\frac{B{r}^{-}}{\sqrt{\left[C{O}_3^2{}^{-}\right]}}=\frac{B{r}^{-}}{\sqrt{\left[C{O}_3^2{}^{-}\right]}}=\sqrt{2}\times {10}^{-7}$