Consider the following electrochemical cell: Pt(s) | MnO4- (0.00100 mol dm-3), Mn2+ (0.00200 mol dm- — Analytical Chemistry Chemistry Question
Basic Electrochemistry
Consider the following electrochemical cell:
Pt(s) | MnO4- (0.00100 mol dm-3), Mn2+ (0.00200 mol dm-3), pH = 3.00 || Ce4+ (0.0100 mol dm-3), Ce3+ (0.0100 mol dm-3) | Pt(s)
The relevant reduction half-reactions are:
Ce4+ + e- ⟶ Ce3+ E° = 1.70 V
MnO4- + 8 H+ + 5e- ⟶ Mn2+ + 4 H2O E° = 1.507 V
Write a balanced net reaction for this cell and determine the values of E o cell and K for the net reaction.
Model Answer
Anode: Mn2+ + 4 H2O ⟶ MnO4- + 8 H+ + 5 e-
Cathode: 5 Ce4+ + 5 e- ⟶ 5 Ce3+
Net: Mn2+ + 4 H2O + 5 Ce4+ ⟶ MnO4- + 8 H+ + 5 Ce3+
E°cell = E°cathode - E°anode = 1.70 V - 1.507 V = 0.193 V
K = 2.05 × 10^16 (calculated from E°cell = (0.05916 / n) * log K)
How many Coulombs of electron charge are transferred when 5.0 mg of Ce 4+ are consumed in the reaction from question 15.1?
Model Answer
5 mg Ce4+ * (1 g / 1000 mg) * (1 mol Ce4+ / 140.12 g Ce4+) * (5 mol e- / 5 mol Ce4+) * (96,485 C / 1 mol e-) = 3.44 C
Determine the cell potential for the electrochemical cell shown in the cell diagram.
Model Answer
Using Nernst equation:
Ecell = E°cell - (0.05916 / n) * log Q
Ecell = E°cell - (0.05916 / n) * log ([MnO4-][H+]^8 [Ce3+]^5 / [Mn2+][Ce4+]^5)
Substituting the values:
[MnO4-] = 0.00100 mol dm-3
[Mn2+] = 0.00200 mol dm-3
pH = 3.00 ⇒ [H+] = 1.00 × 10^-3 mol dm-3
[Ce4+] = 0.0100 mol dm-3
[Ce3+] = 0.0100 mol dm-3
Ecell = 0.193 - (0.05916 / 5) * log ( (0.001 * (10^-3)^8 * (0.01)^5) / (0.002 * (0.01)^5) )
Ecell = 0.193 - (0.05916 / 5) * log ( 0.5 * 10^-24 )
Ecell = 0.193 - (0.011832) * (-24.301)
Ecell = 0.193 + 0.2875 = 0.481 V