Question

In the electrolysis of water
a) Name the gas collected at cathode and anode
b) Why is the volume of one gas collected at cathode is double of anode?
c) Why few drops of dilute \[{H_2}S{O_4}\] added to water?

Answer 1

Hint: In the method of electrolysis of water, hydrogen and oxygen gas are obtained on the different electrodes. The standard electrode potential of hydrogen is considered as zero. Hydrogen gas is used as fuel in many places.

Complete step by step solution:
Electrolysis of water can be explained as the process of decomposition of water into oxygen and hydrogen gas at respective electrodes.
In this method different reactions are observed at the two electrodes namely anode and cathode.
Oxidation is the process by which the oxidation state of an element increases after the reaction. It occurs at anode. The sign convention of anode can be represented as positive, thus the negative ions formed after electrolysis move towards the anode.
Reduction is the process by which the oxidation state of an element decreases after the reaction. It occurs at cathode. The sign convention of cathode can be represented as negative, thus the positive ions formed after the electrolysis move towards the cathode.
The half-cell reactions can be represented as-
 $ \begin{gathered}
  Cathode:2{H^ + }(aq) + 2{e^ - } \to {H_2}(g) \\
  Anode:2{H_2}O(l) \to {O_2}(g) + 4{H^ + }(aq) + 4{e^ - } \\
\end{gathered} $
Now balancing the equation at cathode-
 $ Cathode:4{H^ + }(aq) + 4{e^ - } \to 2{H_2}(g) $
Thus the net reaction after adding the two half-cell reaction, the full-cell reaction in the electrolysis can be represented by-
 $ Net:2{H_2}O(l) \to 2{H_2}(g) + {O_2}(g) $
(a) Hydrogen gas at cathode and Oxygen at anode.
(b) From net reaction we can conclude that because of the presence of $ {H^ + } $ ions from acids the hydrogen gas obtained at cathode is twice as compared to oxygen gas at anode.
(c) Sulphuric acid $ ({H_2}S{O_4}) $ is a strong diprotic and it dissociates in water to give two $ {H^ + } $ ions. By adding sulphuric acid $ ({H_2}S{O_4}) $ in the electrolysis of water, the number of $ {H^ + } $ ions increases and the conduction ion ions in the solution increases. Thus acid helps in passing current through the solution initially by increasing charge carrying ions.

Note:
The electrolysis of water can also be performed at a basic medium in presence of a strong electrolyte base namely sodium hydroxide. Then the half-cell reactions will be slightly different as compared to the acidic medium reactions.