Question

In the electrolysis of acidulated water, $S{{O}_{4}}^{-2}$ ions do not get discharged at anode. Identify the correct reason.
(A) $S{{O}_{4}}^{-2}$ ions have lower discharge potential
(B) $S{{O}_{4}}^{-2}$ ions have higher discharge potential
(C) The concentration of $S{{O}_{4}}^{-2}$ ion is very less in the electrolyte
(D) $S{{O}_{4}}^{-2}$ ions get neutralised by ${{H}^{-}}$ ions

Answer 1

Hint: The discharge of electronegative ions whose discharge potential is low takes place at anode.
- The ion which is a better reducing agent will be responsible for the formation of oxygen gas and water during electrolysis.

Complete Solution :
Let us learn about the electrolysis of acidified water first-
The electrolysis of acidified water (dilute sulphuric acid) is done using electrolytic cells. Electrolysis is a way of splitting a compound using electric energy. The electrical energy comes from a direct current battery or power pack supply. The electrolyte should contain the compound that is being broken down. The electricity must flow through electrodes immersed into the electrolyte to complete the electrical circuit with the battery.
- When the circuit is complete, electrolysis will take place. The products of electrolysis would be collected at the electrode surfaces.
- Electrolysis involves the flow of electrons in the external wires and electrodes whereas, there is flow of ions in the electrolyte. There is always a reduction at the negative cathode electrodes (which attracts positive ions i.e. cations) and an oxidation at the positive anode electrode (which attracts negative ions i.e. anions) and these are the ions which are discharged to give the products.

Electrolysis-
The electrolyte is the dilute sulphuric acid which during the electrolysis splits into hydrogen and oxygen gases. Theoretically, into 2:1 ration by gas volume.
- This is the experimental method to show that the water is composed of hydrogen and oxygen atoms. Only water ionises to a small extent giving minute concentrations of hydrogen ions and hydroxide ions. So, the presence of high concentrations of hydrogen ions and sulphate ions from the acid, makes water a much better electrical conductor.

1. The negative cathode electrode-
The negative cathode electrode reaction is a reduction reaction (electron gain). The hydrogen ions (${{H}^{+}}$) are attached to the negative cathode which is discharged as hydrogen gas.
$2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)$

2. The positive anode electrode-
The positive anode reaction is oxidation electrode reaction (electron loss).
The negative sulphate ions ($S{{O}_{4}}^{-2}$) or the traces of hydroxide ions ($O{{H}^{-}}$) gets attached to the positive electrode. Due to the stability of sulphate ions, nothing will happen. So, hydroxide ions or water molecules are discharged at anode and oxidised to form oxygen.
(i) $2{{H}_{2}}O(l)\to 4{{H}^{+}}(aq)+{{O}_{2}}(g)+4{{e}^{-}}$
(ii) $4O{{H}^{-}}(aq)\to 2{{H}_{2}}O(l)+{{O}_{2}}(g)+4{{e}^{-}}$
There are two possible electrode equations that describe the formation of oxygen in the electrolysis of water, but in strongly acid solution equation (i) is more appropriate.

Illustration-
There is competition between the negative ions $O{{H}^{-}}$ and $S{{O}_{4}}^{-2}$ at the positive anode. $S{{O}_{4}}^{-2}$ ions have higher discharge potential. The hydroxide ions are better reducing agents. So, are preferentially released as oxygen gas and water.
$4O{{H}^{-}}(aq)\to 2{{H}_{2}}O(l)+{{O}_{2}}(g)+4{{e}^{-}}$
So, the correct answer is “Option B”.

Note: Seeing towards the option (D); ${{H}^{-}}$ can never be described as an ion in the electrolysis process. So, ignore that option from the start.