Question

$E^0_{Fe^{2+}/Fe}=-0.441 \, V$ and $E^0_{Fe^{3+}/Fe^{2+}} = 0.771\, V$, the standard emf of the reaction $Fe+2Fe^{3+} \rightarrow 3Fe^{2+}$ will be :

• A. 0.111 V
• B. 0.330 V
• C. 1.653 V
• D. 1.212 V

Answer 1

Answer: (D)

$Fe \rightarrow Fe ^{2+}+2 e ^{-}$..........(i)
$E _{ FeF e ^{2+}}^{0}=- E _{ Fe ^{2+} / Fe }^{\circ}=+0.441$
$Fe ^{3+}+ e ^{-} \rightarrow Fe ^{2+}$
$2 Fe ^{3+}+2 e ^{-} \rightarrow 2 Fe ^{2+}$............(ii)
Adding (i) and (ii) we get,
$Fe +2 Fe ^{3+} \rightarrow 3 Fe ^{2+}$
For this reaction, no. of electron used in oxidation is equal to no. of electron used for reduction. So,
$E _{\text {cell }}^{0}= E _{ OPFeF e ^{2+}}^{0}+ E _{ ROF e ^{3+} / Fe ^{2+}}^{0}$
$E ^{0}=0.441+0.771=1.212 \,V$