Question

On passing electricity through dil. \[{{H}_{2}}S{{O}_{4}}\] solution, the amount of substance liberated at the cathode and anode are in the ratio:
A) 1:8
B) 8:1
C) 16:1
D) 1:16

Answer 1

Hint: The anode is known as the reducing electrode in electrochemical cells. It is the negative and releases the electron to the external circuit and oxidation reaction takes place here. The cathode is known as an oxidizing electron in an electrochemical cell. It is the positive and acquires the electrons to the external circuit and reduction takes place here.

Complete step-by-step answer:
When the electrolysis of dilute sulphuric acid takes place, the product formed at cathode is hydrogen gas. We can see the following equation which occurs at cathode,
\[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\]. Here the addition of electrons takes place, so it is a reduction reaction.

The product formed at anode is water molecule and oxygen. The equation for anode is,
\[4O{{H}^{-}}\to 2{{H}_{2}}O+2{{O}_{2}}+4{{e}^{-}}\]. Here the electrons are reduced, so the oxidation takes place at anode.

The overall reaction for dil. \[{{H}_{2}}S{{O}_{4}}\]is,
\[2{{H}_{2}}O\to {{O}_{2}}+2{{H}_{2}}\]
The amount of product liberated at cathode is = $\text 1 mole\,\, of\,\, hydrogen$
So the molar mass of hydrogen = $1+1=2 g/mol$
The amount of product liberated at anode is = $2 mole\,\, of\,\, oxygen$
So the molar mass of 2 mole of oxygen is = $8+8= 16g/mol$
So the ratio of the product at cathode: the product at anode is
So simplifying the ratio we get it as $ = 2 : 16 = 1: 8$

Hence the correct answer is option ‘A’.

Note: Sulphuric acid is used for the manufacturing of fertilizers such as ammonium sulphate, lime. It is also used in the production of chemicals like hydrochloric acid. Sulphuric acid is a diprotic acid. It readily absorbs moisture from air. It is a good oxidizing agent and dehydrating agent.