Water is a poor conductor of electricity, but it does contain some hydrogen ions, H+, and hydroxide ions, OH-. These ions are formed when a small proportion of water molecules naturally dissociate. If water is acidified with a little dilute sulfuric acid:
The overall balanced equation for the process is:
2H2O(l) → 2H2(g) + O2(g)
The volume of hydrogen given off is twice the volume of oxygen given off.
The ions compete at each electrode to gain or lose electrons.
Whether hydrogen or a metal is produced at the cathode depends on the position of the metal in the metal reactivity series:
Hydrogen will be produced because sodium is more reactive than hydrogen.
Predict the product formed at the negative electrode during the electrolysis of copper(II) chloride solution.
Copper will be produced because copper is less reactive than hydrogen.
Oxygen is produced (from hydroxide ions), unless halide ions (chloride, bromide or iodide ions) are present. In that case, the negatively charged halide ions lose electrons and form the corresponding halogen (chlorine, bromine or iodine).
The table summarises the product formed at the anode during the electrolysis of different electrolytes in solution.
|Negative ion||Element given off at anode|
|Chloride, Cl-||Chlorine, Cl2|
|Bromide, Br-||Bromine, Br2|
|Iodide, I-||Iodine, I2|
|Sulfate, SO42-||Oxygen, O2|
|Nitrate, NO3-||Oxygen, O2|
Predict the product formed at the positive electrode during the electrolysis of concentrated sodium chloride solution.
Chlorine will be produced.
Predict the product formed at the positive electrode during the electrolysis of concentrated sodium sulfate solution.
Oxygen will be produced.
If a halide solution is very dilute, oxygen is given off instead of the halogen. This is because the number of halide ions in the solution is very small. Therefore, very few halide ions will travel to the anode and lose electrons to form a halogen. Instead, water molecules in the solution, some of which break down to give hydroxide ions, produce oxygen at the anode.