In which of the following solutions will the `pH` remain at `7.0` even after electrolysis?

Options

A. `CH_(3)COONa`
B. `Na_(2)SO_(4)`
C. `CuSO_(4)`
D. `NaCl`

Answer

B

Explanation

In the electrolysis of aqueous sodium sulphate using insert electrodes, we observe the following.
(1) Gaseous `H_(2)` is produced at one electrode. The solution becomes basic around that electrode
(2) Gaseous `O_(2)` is produced at the other electrode. The solution becomese acidic around that electrode.
This is because, water is reduced in preference to `Na^(+)` at the cathode and water is also perferentially oxidized relative to the sulphate ion, `SO_(4)^(2-)`, at the anode.
`{:("cathode(reduction)" 2[2H_(2)O+2^(-)rarrH^(2)+2OH^(-)]),("Anode(oxidation)" 2H_(2)OrarrO^(2)+4H^(+)4e^(-)):}/{:("Overall cell reaction"),(6H^(2)Orarr2H^(2)+O_(2)+ubrace(4H^(+)+HOH^(-))_(4H_(2)O)):}`
Net reaction `2H_(2)O rarr 2H_(2)+O_(2)`
The net result is the electrolysis of water and hence solution remains neutral `(pH = 7.0)`. This occurs because `H_(2)O` is more readily reduced than `Na^(+)` and more readily oxidized than `SO_(4)^92-)`. The ions of `Na_(2)SO(4)` conduct the current through the solution but they take no part in the reaction. if we add bromothymol blue indicatior to the solution, the indicator turns blue in the basic solution near the cathode (where `OH^(-)` is produced) and yellow in the acidic solution near the anode (where `H^(+)` is formed).