Predicating the products of electrolysis: What do you expect to be the half reactions in the electrolysis of aqueous `Na_(2)So_(4)` solution.
Strategy: Before we look at the electrode reations, we should consider the following facts: (1) Since `Na_(2)So_(2)` does not hydrolyze in water, the `pH`of the solution is close to `7. (2)` The `Na^(+)` ions are not reduced at the cathode and the `SO_(4)^(2-)` ions are not oxidized at the anode. These conclusions are drawn from the electrolysis of water in the presence of sulphuric acid in aqueous sodium chloride solution.

Explanation

The electrode reaction are
Anode: `2H_(2)(l) rarr O_(2)(g) + 4H^(+)(aq.)+4e^(-)`
Cathode: `2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-)`
The overall reaction, obtained by doubling the cathode reaction coefficients and adding the result to the anode reaction, is
`6H_(2)O(l) rarr 2H_(2)(g)+O_(2)(g)+4H^(+)(aq.)+4OH^(-)(aq.)`
If the `H^(+)` and `OH^(-)` ions ar allowed to mix, then
`4H^(+)(aq.)+4OH^(-)(aq.) rarr (4H_(2))(l)`
and the overall reaction becomes
`2H_(2)O(l) rarr 2H_(2)(g)+O_(2)(g)`