Question
Chemistry Question on Nernst Equation
Given, standard electrode potentials of following reactions, Fe2++2e−⟶Fe;E∘=−0.440V Fe3++3e−⟶Fe;E∘=−0.036V The standard electrode potential (E∘) for Fe3++e−⟶Fe2+, is
A
-0.476 V
B
-0.404 V
C
+ 0.404 V
D
+ 0.772 V
Answer
+ 0.772 V
Explanation
Solution
ΔG∘=−nFE∘
Fe2++2e−⟶Fe ... (i)
ΔG∘=−2×F×(−0.440V)
=0.880F
Fe3++3e−⟶Fe ... (ii)
ΔG∘=−3×F×(−0.036)=0.108F
On subtracting E (i) from E (ii)
Fe3++e−⟶Fe2+
ΔG∘=0.108F−0.880F=−0.772F
Ecell ∘=nF−ΔG∘
=−1×F(−0.772F)=+0.772V