Electrolusis involves electronation and de-electronation at the redpective electrodes. Anode of electrolytic cell is the electrode at which de-electronation takes place whereas at cathode electronation is noticed. If two or more ions of same charge are to be electronated or de-electronated, the ion having lesser discharge potential is dischargeed. Discharge potential of an ion refers for `E_(OP)^(@)` or `E_(RP)^(@)` as the case may be. The products formed at either electrode is given in terms of Faraday's laws of electrolysis i.e., `w = (Eit)/(96500)`
During electrolysis of `CuSO_(4(aq.))`, the `pH` of solution becomes :

Options

A. `lt 7`
B. `gt 7`
C. `= 7`
D. `ge 7`

Answer

A

Explanation

Anode : `2OH^(-) rarr H_(2)O + 1//2O_(2) + 2e,`
Cathode: `Cu^(2+) + 2e rarr Cu`
`[OH^(-)]` decreases, thus `pH` decreases i.e., `lt 7`