I. Reduction-oxidation reactions have played an important role in chemistry due to their potential t — Physical Chemistry — Electrochemistry Chemistry Question
Applied electrochemistry
I. Reduction-oxidation reactions have played an important role in chemistry due to their potential to be valuable sources of energy for technology and life.
Write down chemical equation for the oxidation of glucose (C6H12O6) with KMnO4 solution in the presence of sulfuric acid to form gaseous CO2.
Model Answer
5 C6H12O6 + 24 KMnO4 + 36 H2SO4 → 12 K2SO4 + 24 MnSO4 + 30 CO2 + 66 H2O
Write down chemical equation for the oxidation of FeSO4 with KMnO4 in an acidic medium (sulfuric acid) to form Fe2(SO4)3.
Model Answer
2 KMnO4 + 10 FeSO4 + 8 H2SO4 → 2 MnSO4 + 5 Fe2(SO4)3 + 8 H2O
Based on the reaction in 10.2 determine the anodic reaction and cathodic reaction and the relevant cell diagram.
Model Answer
Anode: 2 Fe2+ →← 2 Fe3+ + 2 e–
Cathode: MnO4- + 8 H+ + 5 e– →← Mn2+ + 4 H2O
The cell diagram:
Pt | Fe3+, Fe2+ || MnO4-, Mn2+, H+ | Pt
Derive the expression for electromotive force E of the cell.
Model Answer
Electromotive force E of the cell can be calculated as follows:
E = E0 – 0.059/10 log ([Mn2+]^2[Fe3+]^10 / [MnO4-]^2[Fe2+]^10[H+]^16)
II. In the thermodynamics point of view, Gibbs free energy ∆G at constant p, T condition is closely related to electromotive force E of a redox reaction according to below expression:
∆G = – n F E → E = – ∆G / n F
where: n – number of electrons transferred,
F – Faraday constant.
The correlation of the standard reduction potential between Mn ions in acidic medium is:
Determine the standard reduction potential of the pair MnO4 2- / MnO2
Model Answer
In order to determine the reduction potential of the pair MnO4 2- / MnO2 we need to use the below diagram:
According to Hess’ Law:
∆G0 2 = ∆G0 1 + ∆G0 3
∆G0 3 = ∆G0 2 - ∆G0 1
We have ∆G0 = - n F E0 → E0 3 = 2.27V
Determine the standard reduction potential of the pair MnO2/ Mn3+
Model Answer
Similarly, we have:
∆G0 4 = ∆G0 5 + ∆G0 6
We have ∆G0 = - n F E0
E0 4 = 0.95 V
III. A process is spontaneous if Gibbs free energy is negative. Based on the thermodynamic data:
Determine Gibbs free energy of the following reaction:
3 MnO4 2- + 4 H+ 2 MnO4 - + MnO2 + 2 H2O
Model Answer
According to the standard reduction potential diagram, we have:
MnO4 2- + 4 H+ + 2 e MnO2 + 2 H2O (3) ∆G0 3 (E0 3 = 2.27 V)
2 MnO4 - + 2 e 2 MnO4 2- (1) ∆G0 1 (E0 1 = 0.56 V)
3 MnO4 2- + 4 H+ 2 MnO4 - + MnO2 + 2 H2O ∆rG0
Is the reaction spontaneous?
Model Answer
In order to know if the reaction is spontaneous, ∆G must be considered.
The reaction that is considered can be obtained by subtracting (1) from (3):
∆rG 0 = ∆G0 3 - ∆G0 1. We have ∆G0 = – n F E0 where ∆E0 reaction = 1.71 V, or ∆G3 < 0
and the reaction is spontaneous.
Calculate Kc for the reaction.
Model Answer
The equilibrium constant can also be calculated:
log K = n E0 / 0.059 = 2×1.71 / 0.059 = 57.96 → K = 9.25×10^57
The large value of K confirms the reaction to be spontaneous.