Question

1.Oxygen was formed at the anode and copper was formed at the cathode.

(i) The ionic half-equation for the formation of oxygen is shown.

40H- = O2 + 2H2O + 4e-

Explain why this reaction is oxidation.

(ii) Write the ionic half-equation for the formation of copper at the cathode.

(2) The electrolysis was repeated using copper electrodes in place of carbon electrodes.

State and explain what happens to the masses of the anode and the cathode during this electrolysis.

Answer 1

1.      (i) This reaction ( anode: 40H^- = O₂ + 2H₂O + 4e^- ) is an oxidation, because the oxidation level of oxygen increases. This means that oxygen is a reducing agent and is oxidized. 2O^2- - 2e^- = 2O⁰

(ii) cathode: Cu²⁺ + 2e^- = Cu⁰

2.      Electrolysis using copper electrodes leads to the transfer of copper from the anode to the cathode. The mass of the cathode will increase, the mass of the anode will decrease. Reaction byproducts will not be released.

Anode : Сu⁰ – 2e^- = Cu²⁺, Cathode: Cu²⁺ + 2e^- = Сu⁰.