A question in my book asks us to find the EMF of the following cell : $$Pb(s),PbSO_4|SO_4^{-2}(0.100M)||Pb^{+2}(0.004M)|Pb(s)$$ Given: $$E^0_{PbSO_4|Pb,SO_4^{-2}}=-0.359V$$ $$E^0_{Pb^{+2}|Pb}=-0.126V$$

I first found the $Pb^{+2}$ 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 $E^0_{cell}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{cell}=E^0_{cell}-\frac{RT}{nF}\ln\frac{[Products]}{[Reactants]}$$ And substituting, $$n=2$$ $$[Products]=2.53\times10^{-7}$$ $$[Reactants]=0.004$$ $$T=298K$$ $$R=8.314$$ $$F=96500$$ I get $E_{cell}=0.357V$

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

So, is my answer correct or have i misunderstood something?