Question

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

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

Answer 1

Answer: (C)

Cell reaction :
$2 Fe ^{3+}+2 I ^{-} \rightarrow 2 Fe ^{2+}+ I _{2}$
$Fe ^{3+}$ 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,
$E_{\text {cathode }}^{\circ}\left( Fe ^{3+} / Fe ^{2+}\right)=0.771\, V$
and $ E_{\text {anode }}^{\circ}\left(I / I_{2}\right)=0.536\, V$
$\therefore E_{\text {cell }}^{\circ}=0.771-0.536 $
$=0.235\, V$