Calculating the EMF of cell Find the EMF of the following cell :
  $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$
  Given:  $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$
  and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$


I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$
And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$
Then using the Nernst equation,
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$
And substituting,
$n=2$,
$[\text{Products}]=2.53\times10^{-7}$,
$[\text{Reactants}]=0.004$,
$T=\pu{298K}$,
$R=\pu{8.314JK^{-1}mol^{-1}}$, and $F=\pu{96500C}$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i  misunderstood something?
Use RT/F = 0.059 or 0.06
It will simplify your calculations.

Note:- I have converted ln into log.
So formula becomes:-
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{0.059}{n}\log\frac{[\text{Products}]}{[\text{Reactants}]}$$

Just try this in problems. Calculations would become easier.