Question

$Cr_2O_7^{2-} I^- \to I_2 + Cr^{3+}$ $ E^{\circ}_{\text{cell}} = 0.79\,V$ $E^{\circ}_{Cr_2O_7^{2-}} = 1.33\,V , E^{\circ}_{I_2}$ 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
$E_{c e l l}^{o}=E_{C}^{o}-E_{A}^{o}$
$ C r_{2} O_{7}^{2-}+I^{-} \rightarrow I_{2}+C r^{3+}$
In this reaction $I^{-}$is oxidised to $I_{2}$,
$\therefore I_{2}$ is anode.
$Cr _{2} O _{7}{ }^{2-}$ is reduced to $C r^{3+}$
$ \therefore C r$ forms cathode.
Given, $E_{\text {cell }}^{o}=0.79 \,V $
$E_{\text {cathode }}^{o}=1.33 \,V$
$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
$E^{o}{ }_{\text {anode }}=1.33-0.79=0.54 \,V$