Galvanic Cells, theoritical voltage [closed]

This reaction proceeds in the galvanic cell:

$$\ce{Zn + Pb^{2+} -> Zn^{2+} + Pb}$$

I need the equation for the half-reactions of this cell and the theoretical voltage produced by this cell if concentration of $\ce{Zn^{2+}}$ is $\mathrm{0.1\ mol/L}$ and concentration of $\ce{Zn^{2+}}$ is $\mathrm{10^{-4}\ mol/L}$.

closed as off-topic by Klaus-Dieter Warzecha, Geoff Hutchison, Philipp, ron, jerepierreDec 16 '14 at 16:02

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$$Pb^{2+} + 2e^{-}->Pb E_{0}(Pb^{2+}) = -0.1263$$ $$Zn->Zn^{2+} +2e^{-} E_{0}(Zn^{2+}) = -0.76$$
$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$
$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$ $$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76)$$
$$E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$