Question

When the electrolysis of acidified water is carried out, the equation for the discharge of ions at the cathode is:
A.\[2\mathop H\nolimits_2 O\xrightarrow{{}}\mathop O\nolimits_2 + 4{H^ + } + 4{e^ - }\]
B.\[2\mathop H\nolimits_2 O\xrightarrow{{}}\mathop O\nolimits_2 + 2{H_2}\]
C.\[4\mathop H\nolimits_2 O + 4{e^ - }\xrightarrow{{}}4{H_2} + \mathop O\nolimits_2 \]
D.\[4\mathop H\nolimits_2 O + 4{e^ - }\xrightarrow{{}}2{H_2} + O{H^ - }\]

Answer 1

Hint: Try to recall that during electrolysis, decomposition of electrolyte solution by passage of electric current resulting into deposition of metals or liberation gases at electrodes is known as electrolysis.

Complete step by step solution:
We already know that electrolysis is a way of decomposition of a compound using electrical energy. Here, acidified water means water is added with few drops of sulphuric acid to make water a better electrolyte.
It has been seen that when electrolysis of acidified water takes place then at anode, hydroxyl radical \[{\left( {OH} \right)^ - }\] undergoes oxidation and oxygen is liberated whereas at cathode, hydrogen ions \[{\left( H \right)^ + }\] undergo reduction to liberate hydrogen gas.
When electrolysis of acidified water is carried out in an electrolytic cell, then following electrode reactions takes place:
At negative cathode electrode:
We know that at cathode electrodes, reduction takes place. It has been seen that hydrogen ions \[{\left( H \right)^ + }\] are attracted to the negative cathode and are discharged as hydrogen gas. The cathodic reaction is:
\[2\mathop H\nolimits^ + \left( {aq} \right) + 2\mathop e\nolimits^ - \to \mathop H\nolimits_2 \left( g \right)\] or,
\[4\mathop H\nolimits_2 O + 4{e^ - }\xrightarrow{{}}2{H_2} + O{H^ - }\].
At positive anode electrode:
We know that at anode electrode, oxidation takes place and it has been seen that both negative sulphate ions and hydroxide \[{\left( {OH} \right)^ - }\] ions are attracted to the positive anode. But the sulphate ion \[{\left( {SO{}_4} \right)^{2 - }}\] is too stable and it does not get discharged. Instead, hydroxide ions \[{\left( {OH} \right)^ - }\] or water molecules are discharged and oxidized to form oxygen gas.

The anodic reaction is:
\[2O\mathop H\nolimits^ - \left( {aq} \right) \to \mathop H\nolimits_2 O\left( l \right) + \frac{1}{2}\mathop O\nolimits_2 \left( g \right) + 2\mathop e\nolimits^ - \]or,
\[2\mathop H\nolimits_2 O\xrightarrow{{}}\mathop O\nolimits_2 + 4{H^ + } + 4{e^ - }\]

            So, overall equation for electrolysis of acidified water: \[\mathop {2H}\nolimits_2 O\left( l \right) \to 2\mathop H\nolimits_2 \left( g \right) + \mathop {2O}\nolimits_2 \left( g \right)\]

Therefore, from above we can now say that option D is the correct answer.

Note:
It should be remembered that as the electrolysis of acidified water precedes further, the dilute sulphuric acid gets more concentrated as only water is removed from the electrolyte as hydrogen and oxygen.