A-Level Redox, electrode potentials and electrolysis: worked solution

Question

Predict, with explanation, how the electrode potential of Cd2+(aq)/Cd(s) changes when aqueous sodium hydroxide is added to that half-cell.

Adding NaOH precipitates insoluble Cd(OH)2, greatly lowering [Cd2+] in the half-cell. For the equilibrium Cd2+(aq) + 2e- -> Cd(s), removing Cd2+ shifts it to the left (less tendency to be reduced). By the Nernst equation E = E-standard + (RT/2F)ln[Cd2+], a lower [Cd2+] makes E more negative. So the electrode potential becomes more negative (decreases).