3
$\begingroup$

For the galvanic cell: $\ce{Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag}$, calculate the electromotive force (EMF) generated.
$K_\mathrm{sp}(\ce{AgCl}) = 2.8\times10^{-10}$; $K_\mathrm{sp}(\ce{AgBr}) = 3.3\times10^{-13}$

I have tried this forming the cell reaction which is coming out to be $$\ce{AgCl + Br- <=> AgBr + Cl-}$$ then using the $K_\mathrm{sp}$ of both salts I found the equilibrium constant which should be $$K_\mathrm{eq} = \sqrt{\frac{K_\mathrm{sp}(\ce{AgCl})}{K_\mathrm{sp}(\ce{AgBr})}}$$ then I put it in the formula $$E_\mathrm{cell} = -0.059 \log(K_\mathrm{eq})$$ but my answer is coming wrong.

|improve this question
$\endgroup$
  • 2
    $\begingroup$ Have a look at the Nernst equation $\endgroup$ – porphyrin Sep 21 '16 at 8:00
  • $\begingroup$ Consider them as 2 separate Ag/AgX/X- half cells and do no forget to involve X- concentrations. $\endgroup$ – Poutnik Apr 25 at 16:56
1
$\begingroup$

$$\begin{align} E_{\ce{Ag/AgBr}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgBr}}{[\ce{Br-}]} \\ E_{\ce{Ag/AgCl}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgCl}}{[\ce{Cl-}]} \\ EMF&=0.059 \cdot \left| \log \left( {\frac { K_{\rm sp, AgCl} \cdot [\ce{Br-}]} { K_{\rm sp, AgBr} \cdot [\ce{Cl-}]} } \right) \right| \\ EMF&=0.059 \cdot \left| \log \left( {\frac { 2.8\times10^{-10} \cdot 0.001} { 3.3\times10^{-13} \cdot 0.2} } \right) \right| \\ EMF&=0.059 \cdot \left| \log \left( {\frac { 2.8}{ 0.66} } \right) \right| \\ EMF&=0.037 \rm \ V\\ \end{align}$$

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.