Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction
Fe³⁺(aq) + V²⁺ (aq) → Fe²⁺(aq) + V³⁺ (aq)
At a temperature of 298 K.

Question

Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction Fe³⁺(aq) + V²⁺ (aq) → Fe²⁺(aq) + V³⁺ (aq) At a temperature of 298 K.

Quizplus · Accepted Answer

We can use the Nernst equation to calculate the standard cell potential and then use it to calculate the equilibrium constant.

The half-reactions are: 
FeS1U1P13 + 2e- → FeS1U1P12 (E° = -0.535 V) 
VS1U1P13+S1S1P0 + e- → VS1U1P12+S1S1P0 (E° = 0.34 V)

Adding the two gives: 
FeS1U1P13+ S1S1P0(aq) + VS1U1P12+S1S1P0 (aq) → FeS1U1P12+ S1S1P0(aq) + VS1U1P13+S1S1P0 (aq)
E°cell = -0.535 + 0.34 = -0.195 V

At 298 K, the Nernst equation is: 
Ecell = E°cell - (RT/nF)ln(Q)
where Q is the reaction quotient and n is the number of electrons transferred (in this case, 3).

The equilibrium constant is related to the reaction quotient by: 
K = exp(-nF∆G°/RT)
where ∆G° is the standard free energy change, which is related to E°cell by:
∆G° = -nFE°cell

Substituting these equations and simplifying, we get: 
K = exp((nF/RT)(E°cell))
K = exp((3 * 96485 / (8.314 * 298))(-0.195))
K = 2.6 × 10^17

Therefore, the best choice is C.

Using the Standard Reduction Potentials in Table 17