Chemistry - Electrochemistry Question with Solution | TestHub

$$\begin{array}{rl}& \because {\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe};{\mathrm{E}}^{\circ }=-0.036\mathrm{V}\\ & \therefore \Delta {\mathrm{G}}_1^{\circ }=-{\mathrm{nFE}}^{\circ }=-3\mathrm{F}(-0.036)\\ & =+0.108\mathrm{F}\\ & \text{ Also }{\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Fe};{\mathrm{E}}^{\circ }=-0.439\mathrm{V}\\ & \therefore \Delta {\mathrm{G}}_2^{\circ }=-{\mathrm{nFE}}^{\circ }\\ & =-2\mathrm{F}(-0.439)\\ & =0.878\mathrm{F}\\ & \text{ To find }{\mathrm{E}}^{\circ }\text{ for }{\mathrm{Fe}}_{(\mathrm{aq})}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}(\mathrm{aq})\\ & \Delta {\mathrm{G}}^{\circ }=-\mathrm{nFE}{}^{\circ }=-1\mathrm{FE}\circ \\ & \because {\mathrm{G}}^{\circ }={\mathrm{G}}_1^{\circ }-{\mathrm{G}}_2^{\circ }\\ & \therefore {\mathrm{G}}^{\circ }=0.108\mathrm{F}-0.878\mathrm{F}\\ & \therefore -{\mathrm{FE}}^{\circ }=+0.108\mathrm{F}-0.878\mathrm{F}\\ & \therefore {\mathrm{E}}^{\circ }=0.878-0.108\\ & =0.77\mathrm{V}\end{array}$$