Question

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

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

Answer 1

Answer: (D)

$\left(Fe\right)^{3 +}+3e^{-} \rightarrow Fe\ldots \left(\right.i\left.\right)$
$\left(Fe\right)^{2 +}+2e^{-} \rightarrow Fe...\left(\right.ii\left.\right)$
$\left(Fe\right)^{3 +}+2e^{-} \rightarrow \left(Fe\right)^{2 +}...\left(\right.iii\left.\right)$
$\left(\right.iii\left.\right)=\left(\right.i\left.\right)-\left(\right.ii\left.\right)$
$ΔG_{3}^\circ =ΔG_{1}^\circ -ΔG_{2}^\circ $
$-FE_{3}^\circ =-3FE_{1}^\circ +2FE_{2}^\circ $ or
$E_{3}^\circ =3E_{1}^\circ -2E_{2}^\circ $
$=3\left(- 0 . 036\right)+2\left(0 . 440\right)$
$=+0.772V$