Question

Calculate the equilibrium constant for each of the reactions at 25 ∘C .

A.) 2Fe³⁺(aq)+3Sn(s)→2Fe(s)+3Sn²⁺(aq)

B.) O₂(g)+2H₂O(l)+2Cu(s)→4OH⁻(aq)+2Cu²⁺(aq)

C.) Br₂(l)+2I⁻(aq)→2Br⁻(aq)+I₂(s)

Answer 1

Calculate the equilibrium constant for each of the reactions at 25 ∘C .

We can always relate the equilibrium constant to cell potentials via Free Energy, dG

the formulas we have:

dG = -n*F*E°cell;

where n = number of electrons being transferred, F = 96500 C/mol, farady consant, E°cell, can be calcualted via E°cell = E°reduction - Eoxidation

Also,

dG = -RT*ln(K);

R = ideal gas constant, 8.314 J/molK, T = temperature, typically 298K, and K the equilibrium constant

so

dG = -RT*ln(K)

dG = -n*F*E°cell

dG1 = dG2

-RT*ln(K) =  -n*F*E°cell

solving for K

K = exp(n*F*E°cell/(RT))

K = exp(n*(96500)*E°cell/(8.314*T))

Now, get E°cell, T in abs, and n, number of e-

A.) 2Fe³⁺(aq)+3Sn(s)→2Fe(s)+3Sn²⁺(aq)

Fe3+ + 3 e− ⇌ Fe(s) −0.04

Sn2+ + 2 e− ⇌ Sn(s) −0.13

E°Cell = Ereduction - Eoxidation = -0.13 - -0.04 = -0.09 V

Now, substitute

K = exp(n*(96500)*E°cell/(8.314*T))

K = exp(2*(96500)*(-0.09 )/(8.314*298))

K = 0.0009019

B.) O₂(g)+2H₂O(l)+2Cu(s)→4OH⁻(aq)+2Cu²⁺(aq)

O2(g) + 2 H2O + 4 e− ⇌ 4 OH−(aq) +0.401

Cu2+ + 2 e− ⇌ Cu(s) +0.337

n =4

E°cell = Ecathode - EAnode = 0.401 - 0.337 = 0.064 V

K = exp(n*(96500)*E°cell/(8.314*T))

K = exp(4*(96500)*(0.064 )/(8.314*298))

K = 21397.9

C.) Br₂(l)+2I⁻(aq)→2Br⁻(aq)+I₂(s)

Br2(l) + 2 e− ⇌ 2 Br− +1.066

I2(s) + 2 e− ⇌ 2 I− +0.54

E°cell = 1.066-0.54 = 0.526 V

n = 2

K = exp(n*(96500)*E°cell/(8.314*T))

K = exp(2*(96500)*(0.526 )/(8.314*298))

K = 6.2391*10^17