How to calculate the electromotive force of a silver chloride/ silver bromide cell? 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 
$$\ce{AgCl + Br- &lt;=&gt; AgBr + Cl-}$$
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.
$$\begin{align}
E_{\ce{Ag/AgBr}}&amp;=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgBr}}{[\ce{Br-}]} \\
E_{\ce{Ag/AgCl}}&amp;=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgCl}}{[\ce{Cl-}]} \\
EMF&amp;=0.059 \cdot \left| \log \left( {\frac  
{ K_{\rm sp, AgCl} \cdot [\ce{Br-}]}
{ K_{\rm sp, AgBr} \cdot [\ce{Cl-}]}
} \right) \right| \\
EMF&amp;=0.059 \cdot \left| \log \left( {\frac  
{ 2.8\times10^{-10} \cdot 0.001}
{ 3.3\times10^{-13} \cdot 0.2}
} \right) \right| \\
EMF&amp;=0.059 \cdot \left| \log \left( {\frac { 2.8}{ 0.66} } \right) \right| \\
EMF&amp;=0.037 \rm \ V\\
\end{align}$$