Given the following standard reduction potentials,
Hg22+(aq) + 2 e– 2 Hg (l) E° = +0.789 V
Hg2Cl2(s) + 2 e– 2 Hg (l) + 2 Cl-(aq) E° = +0.271 V
determine Ksp for Hg2Cl2(s) at 25 °C.
Anode electrode dissolves while cathode electrode plates out.
It changes chemical energy into electrical energy.
Anode half-cell accumulates positive charge and cathode half-cell accumulates negative charge.
Half reaction with more negative reduction potential attracts electrons and undergoes reduction.