Question

Part A

Calculate the equilibrium constant at 25 ∘C for the reaction
Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if
Ni2+(aq)+2e−→Ni(s), E∘ = -0.26 V,
Al+(aq)+e−→Al(s), E∘ = 0.80 V

Express your answer using one significant figure.

Part B

Calculate the equilibrium constant at 25 ∘C for the reaction
Hg2+2(aq)→Hg(l)+Hg2+(aq)
See Appendix D for standard reduction potentials.

Express your answer using one significant figure.

Answer 1

Part A

Calculate the equilibrium constant at 25 ∘C for the reaction
Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if

assume Al --> Ag ( error )
Ni2+(aq)+2e−→Ni(s), E∘ = -0.26 V,
Al+(aq)+e−→Al(s), E∘ = 0.80 V

Express your answer using one significant figure.

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-

E°cell = Ecathode - Eanode

E°cell = 0.80 - -0.26 = 1.06 V

K = exp(2*(96500)*1.06 /(8.314*298))

K = 7.2594*10^35

K = 7*10^35