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.

  • 2
    $\begingroup$ Have a look at the Nernst equation $\endgroup$
    – porphyrin
    Sep 21, 2016 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, 2019 at 16:56

1 Answer 1


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


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