Given  ; $2\text{Cr}\left(\text{s}\right)+3{\text{Cd}}^{2+}\left(\text{aq}\right)⟶2{\text{Cr}}^{3+}\left(\text{aq}\right)+3\text{Cd}$

${\text{E}}_{{\text{Cr}}^{3+}/\text{Cr}}^{\text{o}}=-0\text{.}7\mathrm{4 }\text{V; }{\text{E}}_{\left({\text{Cd}}^{2+}/\text{Cd}\right)}^{\text{o}}=-0\text{.}40\text{ V}$
(a)  Calculation of standard cell potential (E^o)
                         $$\begin{gathered} {\text{E}}_{\text{cell}}^{\text{o}}={\text{E}}_{\text{cathode}}^{\text{o}}-{\text{E}}_{\text{anode}}^{\text{o}} \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{ = (-0.40)-(-0.74) V} \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{ = -0.40 + 0.74V = 0.34V} \end{gathered}$$
${\text{E}}_{\text{cell}}^{\text{o}}=0\text{.}34\text{ V}$
(b)  Calculation of ${\Delta }_{\text{r}}{\text{G}}^{\text{o}}$. 

We know that , 
 ${\Delta }_{\text{r}}{\text{G}}^{\text{o}}=-\text{nF}{\text{E}}_{\text{cell}}^{\text{o}}$

Here the value of n = 6 

thus ,$$\begin{gathered} {\Delta }_{\text{r}}{\text{G}}^{\text{o}}=- 6 \times 96500 \times 0.34 J mo{l}^{-1} \\ {\Delta }_{\text{r}}{\text{G}}^{\text{o}}{\text{ = -196860 J mol}}^{-1} \\ {\Delta }_{\text{r}}{\text{G}}^{\text{o}}{\text{ = -196.86 kJmol}}^{-1} \end{gathered}$$  
(c)  Calculation of equilibrium constant (K_c)
 ${\Delta }_{\text{r}}{\text{G}}^{\text{o}}=-2.303 RT \log K_c$ 
  $\text{log }{\text{K}}_{\text{c}}=\frac{-{\Delta }_{\text{r}}{\text{G}}^{\text{o}}}{2\text{.}303\text{ RT}}$
 $$\begin{gathered} {\text{log K}}_c=\frac{-(-196860)}{2·303\times 8·314\times 298} \\ {\text{log K}}_c\text{ = 34.50} \\ {K}_c= Anti\log (34.501) =3.17 \times {10}^{34} \end{gathered}$$