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Question: Given : \(E_{Fe^{3 +}/Fe = - 0.036}^{0}V,E_{Fe^{2 +}/Fe}^{0} = - 0.439V\)The value of standard elec...

Given : EFe3+/Fe=−0.0360V,EFe2+/Fe0=−0.439VE_{Fe^{3 +}/Fe = - 0.036}^{0}V,E_{Fe^{2 +}/Fe}^{0} = - 0.439VThe value of

standard electrode potential for the change, +e−→+ e^{-}\overset{\quad\quad}{\rightarrow} will be :

A

0.385V

B

0.770V

C

– 0.270V

D

– 0.072V

Answer

0.770V

Explanation

Solution

Fe3+ + 3e−→Fe ΔG1 = -3 × F ×EFe3+/Fe0\text{F}\text{e}^{\text{3+}}\text{ + 3}\text{e}^{-}\overset{\quad\quad}{\rightarrow}\text{Fe } \Delta G_{1}\text{ = -3 }\text{×}\text{ F }\text{×}\text{E}_{\text{F}\text{e}^{3 +}\text{/Fe}}^{0}

Fe2+→ Fe  ΔG2 = -2 × F × EFe2+/Fe0Fe3+ + e−→ Fe2+ Δ G = ΔG1 - ΔG2 \text{F}\text{e}^{\text{2+}}\overset{\quad\quad}{\rightarrow}\text{ Fe } \text{ }\Delta G_{2}\text{ = -2 × F × }\text{E}_{\text{F}\text{e}^{2 +}\text{/Fe}}^{0}\text{F}\text{e}^{\text{3+}}\text{ + }\text{e}^{-}\overset{\quad\quad}{\rightarrow}\text{ F}\text{e}^{\text{2+}}\ \Delta\text{ G = }\Delta G_{1}\text{ - }\Delta G_{2}\ ΔG=3×0.036 F−2×0.439×F=−1×E0(Fe3+/Fe+2)×F\Delta \mathrm { G } = 3 \times 0.036 \mathrm {~F} - 2 \times 0.439 \times \mathrm { F } = - 1 \times \mathrm { E } ^ { 0 } \left( \mathrm { Fe } ^ { 3 + } / \mathrm { Fe } ^ { + 2 } \right) \times \mathrm { F } E0 (Fe3+/Fe+2)= 2 × 0.439 - 3 × 0.036E^{0}\ (\text{F}\text{e}^{\text{3+}}\text{/F}\text{e}^{\text{+2}}) \text{= 2 }\text{×}\text{ 0.439 - 3 }\text{×}\text{ 0.036}

}\text{= 0.770 V}$$