# 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?

• Sorry for being rusty, but can you clarify the setup more? What reactions are happening, are the cells compartmentalized and what exactly are the values given in brackets in the question? May 10 '18 at 11:50

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