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Q.
If $E_{ Fe ^{-2} \mid Fe }^{\circ}$ is $x _{1}, E _{ Fe ^{+3} \mid Fe }^{\circ}$ is $x _{2}$; then $E _{ Fe ^{+3} \mid Fe ^{+2}}^{\circ}$ will be
Electrochemistry
Solution:
(i) $Fe ^{+2}+2 e ^{-} \longrightarrow Fe $
$E ^{\circ}= x _{0}, \Delta G _{1}=-2 Fx _{1}$
(ii) $Fe ^{+3}+3 e ^{-} \longrightarrow Fe$
$E^{\circ}=x_{2}, \Delta G_{2}=-3 F x_{2}$
$Fe ^{+3}+ e ^{-} \longrightarrow Fe ^{+2}$
This reaction is obtained by subtraction of (ii) - (i)
$-1 \times FE _{3}^{\circ} \Rightarrow -3 F x _{2}-(-2 F ) x _{1}$
$E _{3}^{\circ} \Rightarrow 3 x _{2}-2 x _{1}$
