Question

If $E_{F{e}^{-2}\mid Fe}^{\circ }$ is $x_1,E_{F{e}^{+3}\mid Fe}^{\circ }$ is $x_2$; then $E_{F{e}^{+3}\mid F{e}^{+2}}^{\circ }$ will be

• A. $3x_2-2x_1$
• B. $x_2-x_1$
• C. $x_2+x_1$
• D. $2x_1+3x_2$

Answer 1

Answer: (A)

(i) $F{e}^{+2}+2e^{-}⟶Fe$
$E^{\circ }=x_0,\Delta G_1=-2F{x}_1$
(ii) $F{e}^{+3}+3e^{-}⟶Fe$
$E^{\circ }=x_2,\Delta G_2=-3F{x}_2$
$F{e}^{+3}+e^{-}⟶F{e}^{+2}$
This reaction is obtained by subtraction of (ii) - (i)

$-1\times F{E}_3^{\circ }\Rightarrow -3F{x}_2-(-2F)x_1$
$E_3^{\circ }\Rightarrow 3x_2-2x_1$