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?

  • $\begingroup$ 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? $\endgroup$ – stochastic13 May 10 '18 at 11:50

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.

| improve this answer | |
  • $\begingroup$ But the answer i am getting by doing these calculations is 0.357 whereas the answer given in the book is 0.133. So my question is where exactly am i making a mistake.. $\endgroup$ – Rutwik May 4 '18 at 16:57

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