Question

Given $: E^{\circ}_{ Fe ^{3+} / Fe }=-0.036\, V , E^{\circ}_{ Fe ^{2+} / Fe }=-0.439 \, V$. The value of standard electrode potential for the change, $Fe _{(a q)}^{3+}+e^{-} \rightarrow Fe ^{2+}{ }_{(a q)}$ will be

• A. $-0.072\, V$
• B. $0.385 \,V$
• C. $0.770 \,V$
• D. $-0.270 \,V$

Answer 1

Answer: (C)

Given, $Fe ^{3+}+3 e^{-} \longrightarrow Fe ; E^{\circ}{ }_{1}=-0.036 \,V$

$Fe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}{ }_{2}=-0.439\, V$

Required equation is

$Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+} ; E_{3}^{\circ}=?$

Applying, $\Delta G^{\circ}=-n F E^{\circ} \,\,\,\, \therefore \Delta G_{3}^{\circ}=\Delta G_{1}^{\circ}-\Delta G_{2}^{\circ}$

$\left(-n_{3} F E^{\circ}{ }_{3}\right)=\left(-n_{1} F E^{\circ}{ }_{1}\right)-\left(-n_{2} F E_{2}^{\circ}\right)$

$E_{3}^{\circ}=3 E_{1}^{\circ}-2 E_{2}^{\circ}=3 \times(-0.036)-2 \times(-0.439)$

$E_{3}^{\circ}=-0.108+0.878=0.77\, V$