Question

Given standard electrode potential;
$F{e}^{3+}+3e^{-}\to Fe;E^{\circ }=-0.036volt$
$F{e}^{2+}+2e^{-}\to Fe;E^{\circ }=-0.440volt$
The standard electrode potential $E^{\circ }$ for
$F{e}^{+3}+e^{-}\to F{e}^{2+}$ is

• A. $-0.476$ volt
• B. $-0.404$ volt
• C. $0.440$ volt
• D. $+0.772$ volt

Answer 1

Answer: (D)

${\left(Fe\right)}^{3+}+3e^{-}\to Fe\dots (i)$
${\left(Fe\right)}^{2+}+2e^{-}\to Fe...(ii)$
${\left(Fe\right)}^{3+}+2e^{-}\to {\left(Fe\right)}^{2+}...(iii)$
$(iii)=(i)-(ii)$
$\Delta G_3^{\circ }=\Delta G_1^{\circ }-\Delta G_2^{\circ }$
$-F{E}_3^{\circ }=-3F{E}_1^{\circ }+2F{E}_2^{\circ }$ or
$E_3^{\circ }=3E_1^{\circ }-2E_2^{\circ }$
$=3\left(-0.036\right)+2\left(0.440\right)$
$=+0.772V$