Question

From the given standard electrode potentials of two half cells

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

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

Calculate the standard electrode potential $(E^o)$ for

$F{e}^{+++}+e^{-}\to F{e}^{++}$ is

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

Answer 1

Answer: (A)

$\Delta G^o=-n{E}^{o}F$

$F{e}^{2+}+2e^{-}\to Fe$ …..(i)

$\Delta G^o=-2\times F\times (-0.440V)=0.880F$

$F{e}^{3+}+3e^{-}\to Fe$ …..(ii)

$\Delta G^o=-3\times F\times (-0.036)=0.108F$

On subtracting equation (i) from (ii)

$F{e}^{3+}+e^{-}\to F{e}^{2+}$

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

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