Question

Given the electrode potentials
$F{e}^{3+}+e^{-}⟶F{e}^{2+},E^{\circ }=0.771$ volt
$I_2+2e^{-}⟶2I^{-},E^{\circ }=0.536$ volt
$E_{\text{cell }}^{\circ }$ for the cell reaction
$2F{e}^{3+}+2I^{-}⟶2F{e}^{2+}+I_2$, is :

• A. 1.006 V
• B. 0.503 V
• C. 0.235 V
• D. -0.235 V

Answer 1

Answer: (C)

Cell reaction :
$2F{e}^{3+}+2I^{-}\to 2F{e}^{2+}+I_2$
$F{e}^{3+}$ shows reduction while I ${}^{-}$shows oxidation (reduction occurs at cathode while oxidation takes place at anode)
$E_{\text{cell }}^{\circ }=E_{\text{cathode }}^{\circ }-E_{\text{anode }}^{\circ }$
Given,
$E_{\text{cathode }}^{\circ }\left(F{e}^{3+}/F{e}^{2+}\right)=0.771V$
and $E_{\text{anode }}^{\circ }\left(I/I_2\right)=0.536V$
$\therefore E_{\text{cell }}^{\circ }=0.771-0.536$
$=0.235V$