${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{I}}^{+}\to {\mathrm{I}}_2+{\mathrm{Cr}}^{3+} E_{cell}^0=0.79\mathrm{V}$; ${\mathrm{E}}_{{\mathrm{cr}}_2{\mathrm{O}}_7^{2-}}^0=1.33\mathrm{V}. \mathrm{Then} {\mathrm{E}}_{{\mathrm{I}}_2}^0 =?$

 

${\mathrm{Cr}}_{2 }{\mathrm{O}}_7^{2-}+{\mathrm{I}}^{+}\to {\mathrm{I}}_2+{\mathrm{Cr}}^{3+}$ in this reaction,

${\mathrm{I}}^{-}\to {\mathrm{I}}^2$ oxidation takes place.

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\to {\mathrm{Cr}}^{2+}$ reduction takes place.

${\mathrm{I}}^{-}$ get oxidized to ${\mathrm{I}}_2$ hence will form anode and get reduced to ${\mathrm{Cr}}^{3+}$ hence will form cathode.

Given: ${\mathrm{E}}_{\mathrm{cell}}^0=0.79\mathrm{V} , {\mathrm{E}}_{{\mathrm{Cr}}_2 {\mathrm{O}}_7^{2-}}^0=1.33\mathrm{V}$

By putting above values in following equation

$$\begin{gathered} E_{cell}^0 =E_{cathode}^0 - E_{anode}^0 \\ {E}_{cell }^0=E_{c{r}_2 o_7^{2-}}^0 E_{I_2}^0 \\ 0.79 = 1.33- E_{I_2}^0 \\ {E}_{I_{2 }}^0= 1.33 - 0.79 \\ {E}_{I_2}^0 = 0.54 V \end{gathered}$$

Hence option A is correct.