Question

The equation$\Delta G=-nF{E}^o$ also can be applied to half-reactions. Use standard reduction potentials to estimate$\Delta {G^o}_i$ for Fe²⁺ (aq) and Fe³⁺ (aq). ($\Delta {G^o}_i$ for e^-=0.)

Short answer

$\Delta {G^0}_f$for $F{e}^{2+}\left(aq\right)$is $-84.907 kJ$and for $F{e}^{3+}\left(aq\right)$is $-10.42 kJ$

Step-by-step solution

Reduction reaction and its calculation

The reduction reaction of $F{e}^{2+}\left(aq\right)$are as follows:

$$\begin{gathered} F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right) E^o=-0.44 V---\left(1\right) \\ \Delta {G^o}_f\left(Fe\right(s\left)\right)=0 \\ \Delta {G^o}_f\left(e^{-}\right)=0 \end{gathered}$$

Then, applying the values,

$$\begin{gathered} \Delta G^0=-nF{E}^0 \\ =-2\times 96485 mo{l}^{-1}{e}^{-1}\times (-0.44V) \\ =84906.8 J \end{gathered}$$

Calculation by applying the equation

In equation (1),

$$\begin{gathered} \Delta G^o=\Delta {G^0}_f\left(Fe\right(s\left)\right)-\left(\Delta {G^0}_f\right(F{e}^{2+}\left(aq\right))+\Delta {G^0}_f(e^{-}\left)\right) \\ 84.907kJ=0-\left(\Delta {G^0}_f\right(F{e}^{2+}\left(aq\right))+0) \\ {G^0}_f\left(F{e}^{2+}\right(aq)=-84.907 kJ \end{gathered}$$

Calculation for Fe3+ (aq)

On calculating $\Delta G^o,$

$$\begin{gathered} \Delta G^o=-nF{E}^o \\ =-3\times 96845 mo{l}^{-1}{e}^{-}\times (-0.036V) \\ =10420 J \end{gathered}$$

On using equation (2),

$$\begin{gathered} \Delta G^o=\Delta {G^o}_f\left(Fe\right(s\left)\right)-\left(\Delta {G^o}_f\right(F{e}^{3+}\left(aq\right))+\Delta {G^o}_f(e^{-}\left)\right) \\ 10.42 kJ=0-\left(\Delta {G^o}_f\right(F{e}^{3+}\left(aq\right))+0) \\ \Delta {G^o}_f\left(F{e}^{3+}\right(aq)=-10.42 kJ \end{gathered}$$