Calculate the equilibrium constant for an oxidation-reduction reaction involving the transfer of 1 mole of electrons with ΔE0'= 0.25 V.(R = 8.315 JK-1mol-1,ℱ= 96.48 kJV-1mol-1,T = 298k)
A)1
B)6 × 10-⁵
C)1.7 × 10⁴
D)0.49

Question

Quizplus · Accepted Answer

The equilibrium constant for an oxidation-reduction reaction can be calculated using the following equation:

K = e^(-ΔG°/RT)

where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and e is the base of the natural logarithm.

We can use the following equation to relate ΔE°, ΔG°, and K:

ΔG° = -nℱΔE°

where n is the number of electrons transferred and ℱ is the Faraday constant.

Plugging in the given values, we have:

ΔG° = -1 mol × 96.48 kJ/mol × 0.25 V = -24.12 kJ/mol

Now we can calculate the equilibrium constant K:

K = e^(-ΔG°/RT) = e^(-(-24.12 kJ/mol) / (8.315 J/K/mol × 298 K)) = 1.7 × 10^4

Therefore, the correct answer is C.

Calculate the Equilibrium Constant for an Oxidation-Reduction Reaction Involving the Transfer