Vanadium can adopt four oxidation states in aqueous solution. Half-reaction E° at 298 K, V VO2+(aq) — Electrochemistry Chemistry Question

VO²⁺(aq) + 2 H⁺(aq) + e– → VO²⁺(aq) + H₂O(l) ∆G° = –(1)(F)(1.00 V)
VO²⁺(aq) + 2 H⁺(aq) + e– → V³⁺(aq) + H₂O(l) ∆G° = –(1)(F)(0.34 V)
V³⁺(aq) + e– → V²⁺(aq) ∆G° = –(1)(F)(–0.26 V)
VO²⁺(aq) + 4 H⁺(aq) + 3 e– → V²⁺(aq) + 2 H₂O(l) ∆G° = –(3)(F)(E°)
–(3)(F)(E°) = –(F)(1.00 V + 0.34 V – 0.26 V)
E° = +0.36 V

∆E° = 1.00 V – (–0.26 V) = 1.26 V

∆G° = –nF∆E° = ∆H° – T∆S°, so ∆E° = (–∆H°/nF) + T∆S°/nF
If ∆E° increases by 1.76 × 10-4 V K^-1, then ∆S°/nF = 1.76 × 10-4 V K^-1.
∆S° = (1)(96500 J V^-1 mol^-1)(1.76 × 10-4 V K^-1) = 17.0 J mol^-1 K^-1
At 298 K, ∆G° = –nF∆E° = –(1)(96.5 kJ V^-1 mol^-1)(1.26 V) = –121.6 kJ mol^-1
∆G° = ∆H° – T∆S°
–121.6 kJ mol^-1 = ∆H° – (298 K)(0.0170 kJ mol^-1 K^-1)
∆H° = –117 kJ mol^-1

Using the Nernst equation (with base 10 logarithm, at 298 K):
∆E = ∆E° – (0.0591/n)log([VO²⁺][V³⁺]/([VO²⁺][V²⁺][H⁺]^2))
1.14 V = 1.26 V – 2(0.0591)(pH) – 0.0591log([VO²⁺][V³⁺]/([VO²⁺][V²⁺]))
1.14 V = 1.14 V – 0.0591log([VO²⁺][V³⁺]/([VO²⁺][V²⁺]))
log([VO²⁺][V³⁺]/([VO²⁺][V²⁺])) = 0
So the numerator and denominator must be equal, which means (given the initial concentrations of 0.50 M for both vanadium reactants and 0.10 M for both vanadium products) that all vanadium species must have equal concentrations. So [V³⁺] = 0.30 M.

Since [V³⁺] increases by 0.20 M in a 0.100 L solution, this corresponds to the passage of (1)(96500 C mol^-1)(0.20 mol L^-1)(0.100 L) = 1930 C. At a current of 10.0 A,
(10.0 A)t = 1930 C
t = 193 s