Question
Chemistry Question on Gibbs Free Energy
Given standard electrode potentials Fe2:+2e−⟶FeE∘=−0⋅440V Fe3++3e−⟶FeE∘=−0⋅036V The standard electrode potential (E∘) for Fe3++e−⟶Fe2+ is :
A
+ 0.772 V
B
- 0.772 V
C
+ 0.417 V
D
-0.414 V
Answer
+ 0.772 V
Explanation
Solution
We know, △G∘=−nFE∘
Fe2++2e−→Fe
ΔG∘=−2×F×(−0.440V)=0.880F... (1)
Fe3++3e−→Fe
ΔG∘=−3×F×(−0.036)=0.108F... -(2)
On subtracting Eqs. (1) from (2),
Fe3++e−→Fe2+
ΔG∘=0.108F−0.880F=−0.772F
ΔG∘=−nFE∘
E∘=−nFΔG∘=−1×F−0.772F=+0.772V