Question

Compute the equilibrium constant for the reaction between Cd^(2+)(aq) and Ni(s).

Express your answer using two significant figures.

Answer 1

The first thing to do is to write the complete reaction

Cd⁺²(aq) + Ni(s) ===== Ni⁺²(aq) + Cd(s)

write the half reactions

Cd⁺² + 2e ==== Cd(s) E = -0.40 V

Ni(s) ===== Ni⁺²(aq) + 2e E = 0.23 V

Now add these 2 values = 0.23 - 0.40 = -0.17 V this is the standard cell potential

as you can see 2 electrons are being exchanged so we apply the equation (nernst equation):

E = Eo - (2.303*RT / nF) * log K , E is the cell potential, E0 is the standard potential, R is gas constant, T is temperature, F is faraday constant, n is number of electrons, k is equilibrium constant

At equilibrium E is equal to zero, 2.303*RT / F = 0.0591, temperature used is 298.15K, F is 96500

0 = -0.17 - (0.0591/n) * log K, n is the number of electrons that are being transferred

0 = -0.17 - 0.0591/2 * log K

0.17 = - 0.02955 * log K

-5.74 = log K

K = 10^-5.74 = 1.8 x 10^-6

*If you like this answer dont forget to rate it =)