Calculate the standard cell potential (in V)of the cell in which following reaction  takes place  ${\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)⟶{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+\mathrm{Ag}\left(\mathrm{s}\right)$  Given that,

${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{o}}=\mathrm{xV}$          ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{O}}=\mathrm{yV}$       ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}^{\mathrm{o}}=\mathrm{ZV}$

$F{e}^{2+}\left(aq\right)+A{g}^{+}\left(aq\right)⟶F{e}^{3+}\left(aq\right)Ag\left(s\right)$

$\therefore E_{\text{cell}}^o=E_{A^{+}/Ag}^o-E_{F{e}^{3+}/F{e}^{2+}}^0=xV-E_{F_e^{3+}/F{e}^{2+}}^o$   ..........(i)

Now, for two half‐cells

(i)  $F{e}^{2+}+2e^{-}⟶Fe; E_{F^{2+}/Fe}^o=E_1^o=yV \mathrm{\Delta }{G}_2^o=-2F{E}_1^o$

(ii) $F{e}^{3+}+3e^{-}⟶Fe; E_{F{e}^{3+/Fe}}^o=E_2^o=zV \mathrm{\Delta }{G}_2^o=-3F{E}_2^o$

So,  $F{e}^{3+}+e^{-}⟶F{e}^{2+}; E_{F{e}^{3+}/F{e}^{2+}}^o=E_3^o=?$

$; \mathrm{\Delta }{G}_3^o=-1\times F{E}_3^o$

Again,  $\mathrm{\Delta }{G}_3^o=\mathrm{\Delta }{G}_2^o-\mathrm{\Delta }{G}_1^o\Rightarrow -F{E}_3^o=-3F{E}_2^o-(-2F{E}_1^o)$

$\Rightarrow -E_3^o=2E_1^o-3E_2^o\Rightarrow 3E_3^o=3E_2^o-2E_1^o\Rightarrow E_{F^{3+}/F_e^{2+}}^o=(3z-2y)V$

So, from equation (i)

$E_{\text{cell}}^o=xV-(3z-2y)V=(x-3z+2y)V$