Question:
Given, standard electrode potentials of following reactions,
$Fe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}=-0.440\, V $
$Fe ^{3+}+3 e^{-} \longrightarrow Fe ; E^{\circ}=-0.036\, V$
The standard electrode potential $\left(E^{\circ}\right)$ for
$Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$, is
Given, standard electrode potentials of following reactions,
$Fe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}=-0.440\, V $
$Fe ^{3+}+3 e^{-} \longrightarrow Fe ; E^{\circ}=-0.036\, V$
The standard electrode potential $\left(E^{\circ}\right)$ for
$Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$, is
Updated On: Apr 1, 2024
- -0.476 V
- -0.404 V
- + 0.404 V
- + 0.772 V
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The Correct Option is D
Solution and Explanation
$\Delta G^{\circ}=-n F E^{\circ}$
$ Fe ^{2+}+2 e^{-} \longrightarrow Fe$ ... (i)
$ \Delta G^{\circ} =-2 \times F \times(-0.440 \,V ) $
$=0.880\, F $
$ Fe ^{3+}+3 e^{-} \longrightarrow Fe $ ... (ii)
$ \Delta G^{\circ}=-3 \times F \times(-0.036)=0.108\, F $
On subtracting E (i) from E (ii)
$ Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$
$\Delta G^{\circ}=0.108\, F -0.880 \,F =-0.772 \,F$
$E_{\text {cell }}^{\circ}=\frac{-\Delta G^{\circ}}{n F}$
$=-\frac{(-0.772 F)}{1 \times F}=+0.772 \,V$
$ Fe ^{2+}+2 e^{-} \longrightarrow Fe$ ... (i)
$ \Delta G^{\circ} =-2 \times F \times(-0.440 \,V ) $
$=0.880\, F $
$ Fe ^{3+}+3 e^{-} \longrightarrow Fe $ ... (ii)
$ \Delta G^{\circ}=-3 \times F \times(-0.036)=0.108\, F $
On subtracting E (i) from E (ii)
$ Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$
$\Delta G^{\circ}=0.108\, F -0.880 \,F =-0.772 \,F$
$E_{\text {cell }}^{\circ}=\frac{-\Delta G^{\circ}}{n F}$
$=-\frac{(-0.772 F)}{1 \times F}=+0.772 \,V$
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Concepts Used:
Nernst Equation
This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.
