Question

From the given standard electrode potentials of two half cells

$Fe^{+ +}+2e^{-} \rightarrow Fe;E^{o}=-0.440V$

$Fe^{+ + +}+3e^{-} \rightarrow Fe;E^{o}=-0.036V$

Calculate the standard electrode potential $\left(\right.E^{o}\left.\right)$ for

$Fe^{+ + +}+e^{-} \rightarrow Fe^{+ +}$ is

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

Answer 1

Answer: (A)

$\Delta G^{o}=-nE^{o}F$

$Fe^{2 +}+2e^{-} \rightarrow Fe$ …..(i)

$\Delta G^{o}=-2\times F\times \left(\right.-0.440V\left.\right)=0.880F$

$Fe^{3 +}+3e^{-} \rightarrow Fe$ …..(ii)

$\Delta G^{o}=-3\times F\times \left(\right.-0.036\left.\right)=0.108F$

On subtracting equation (i) from (ii)

$Fe^{3 +}+e^{-} \rightarrow Fe^{2 +}$

$\Delta G^{o}=0.108F-0.880F=-0.772F$

$E^{o}$ for the reaction $=-\frac{\Delta G^{o}}{n F}=-\frac{\left(\right. - 0.772 F \left.\right)}{1 \times F}=+0.772V$