Calculate the emf of  $Mg\left(s\right)\left|M{g}^{2+}\right(0.01M)\parallel l^{-}(0.01M\left)\right|{\mathrm{Cl}}_2\left(1\mathrm{atm}\right)$ 

$(E_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^0=-2.37\mathrm{V} ;$

$E_{Pt,{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^0=+1.36\mathrm{V})$

$Mg\left(s\right)\to M^{2+}\left(aq\right)+2e; E_{\mathrm{Mg}/M{g}^{2+}}^0=+2.37\mathrm{V}$

$\underline{{\mathrm{Cl}}_2\left(\mathrm{g}\right)+2\mathrm{e}\to 2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right);{\mathrm{E}}_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^0=+1.36\mathrm{V}}$

$Q_C=\frac{\left[{\mathrm{Mg}}^{2+}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2}{PC{l}_2}$

$E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{2}\log \frac{\left[{\mathrm{Mg}}^{2+}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2}{P_{{\mathrm{Cl}}_2}}$

$=3.73-\frac{0.0591}{2}\log {10}^{-2}\times {\left({10}^{-2}\right)}^2$

$=3.73+0.0591\times 4=3.90$