1. Describe Electrolysis of an aqueous solution
Example 1: Electrolyte : Silver bromide, AgBr solution,
Concentration : 2.0 mol dm-3 ,
Electrodes : carbon
1) Silver bromide, AgBr solution consists of Ag+ ions, H+ ions, Br - ions and OH- ions that move freely.
2) During electrolysis,
the cations Ag+ ions and H+ ions move to the cathode;
the anions Br - ions and OH- ions move to the anode.
3) At CATHODE,
Silver, Ag+ ions are selectively discharged by receiving electrons to form Ag atom.
This is because Ag+ ions has a lower position than H+ ions in the electrochemical series.
Ag+ (aq) + e- → Ag (s)
A silvery grey solid is deposited at cathode.
Silver, Ag metal is formed at the cathode.
4) At ANODE,
Bromide, Br - ions are selectively discharged by donating electrons to form Br2 molecules.
This is because Br – ion has higher concentration in the electrolyte
(even though Br - ions has a higher position than OH- ions in the electrochemical series.)
Br - (aq) → Br2 (g) + 2e-
Brown vapour is released at anode.
Bromine, Br2 gas are produced at the anode.
5) Overall equation:
{ Ag+ + e- → Ag } x2 : 2Ag2+ + 2 e- → 2Ag
2Br - → Br2 + 2 e-
2Ag+ + 2Br - → 2Ag + Br2
* Homework: Describe Electrolysis of an aqueous solution
Electrolyte : copper (II) chloride, CuCl2 solution
Concentration : 0.1 mol dm-3 ,
Electrodes : carbon

2. Example 2: Electrolyte : Copper(II) sulphate, CuSO4 solution,
Concentration : 0.1 mol dm-3 ,
Electrodes : carbon
6) Copper(II) sulphate, CuSO4 solution consists of Cu2+ ions, H+ ions, SO42- ions and OH- ions that
move freely.
7) During electrolysis,
the cations Cu2+ ions and H+ ions move to the cathode;
the anions SO42- ions and OH- ions move to the anode.
8) At CATHODE,
Copper (II), Cu2+ ions are selectively discharged by receiving electrons to form Cu atom.
This is because Cu2+ ions has a lower position than H+ ions in the electrochemical series.
Cu2+ (aq) + 2 e- → Cu (s)
A brown solid is deposited at cathode.
Copper, Cu metal is formed at the cathode.
9) At ANODE,
Hydroxide, OH- ions are selectively discharged by donating electrons to form
O2 and H2O molecules.
This is because OH- ions has a lower position than SO42- ions in the electrochemical series.
4OH- (aq) → O2 (g) + 2H2O(l) + 4 e-
Colourless gas bubbles are released at anode. (The gas relights a glowing wooden splinter.)
Oxygen, O2 gas and water, H2O are produced at the anode.
10) Overall equation:
{ Cu2+ + 2 e- → Cu } x2 : 2Cu2+ + 4 e- → 2Cu
4OH- → O2 + 2H2O+ 4 e-
2Cu2+ + 4OH- → 2Cu + O2 + H2O