Question

$C{r}_2{O}_7^{2-}{I}^{-}\to I_2+C{r}^{3+}$ $E_{\text{cell}}^{\circ }=0.79V$ $E_{C{r}_2{O}_7^{2-}}^{\circ }=1.33V,E_{I_2}^{\circ }$ is

• A. - 0.10 V
• B. + 0.18 V
• C. - 0.54 V
• D. 0.54 V

Answer 1

Answer: (D)

(i) Decide cathode and anode with the help of given cell reaction.
(ii) Find electrode potential by using formula
$E_{cell}^o=E_C^o-E_A^o$
$C{r}_2{O}_7^{2-}+I^{-}\to I_2+C{r}^{3+}$
In this reaction $I^{-}$is oxidised to $I_2$,
$\therefore I_2$ is anode.
$C{r}_2{O}_7{}^{2-}$ is reduced to $C{r}^{3+}$
$\therefore Cr$ forms cathode.
Given, $E_{\text{cell }}^o=0.79V$
$E_{\text{cathode }}^o=1.33V$
$E_{\text{cell }}^o=E_{\text{cathode }}^o-E_{\text{anode }\left(I_2\right)}^o$
$0.79=1.33-E^o{}_{\text{anode }\left(I_2\right)}$ or
$E^o{}_{\text{anode }}=1.33-0.79=0.54V$