Calculate the equilibrium constant at 25 ∘C for the reaction Fe (s) + 2Ag + (aq) →Fe2 + (aq) + 2Ag (s)

Standard Reduction Potentials at 25 ∘C Fe2 + (aq) + 2e-→Fe (s) E∘ = - 0.45 V Ag + (aq) + e-→Ag (s) E∘ = 0.80 V

1.7 * 10 ^42

Explanation:

Using Nernst equation

E°cell = RT/nF Inq

at equilibrium

Q=K

E°cell = 0.0257 / n Ink = 0.0592/n log K

Fe2 + (aq) + 2e-→Fe (s) E∘ = - 0.45 V

Ag+aq) + e-→Ag (s) E∘ = 0.80 V

Fe (s) + 2Ag + (aq) →Fe2 + (aq) + 2Ag (s)

balance the reaction

Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v

2 Ag⁺ + 2e⁻ → 2Ag

n = 2 moles and K = equilibrium constant

E° cell = 0.80 + 0.45 = 1.25 V

E° cell = (0.0592 / n) log K

substitute the value into the equations and solve for K

(1.25 * 2) / 0.0592 = log K

42.23 = log K

k = 10^ 42.23

K = 1.7 * 10 ^42