Question

Reshma is a student of class 10. Her teacher was teaching the type of chemical reaction in the class. Reshma was asked to carry out the reaction of electrolysis of water in the laboratory. She set up the complete apparatus for the electrolysis of water. She took pure water to carry out this reaction. When Reshma passed electric current through this water for a considerable time, even then no chemical reaction took place. She told this problem to her classmate Deena. She asked Reshma to add a little dil.H2SO4 into pure water to make water a good conductor of electricity before passing electric current through it. When electric current was passed, then chemical reaction took place to form two gases X and Y. The volume of gas X collected over one of the electrodes was 8 ml, whereas the volume of gas Y collected over the other electrode was 16 ml.

(a) What is gas X? At which electrode gas X is evolved?
(b) What is gas Y? At which electrode gas Y is evolved?
(c) Write the balanced chemical equation for the reaction.

Answer 1

Gas X is hydrogen, collected at the cathode, and Gas Y is oxygen, collected at the anode. To conduct the electrolysis of water, dilute sulfuric acid was added to increase water's conductivity. The balanced chemical equation is 2H2O(l) → 2H2(g) + O2(g).

Step-by-step explanation:

Reshma's experiment is related to the electrolysis of water, which is a process that splits water into its constituent gases, hydrogen and oxygen, using an electric current. When pure water was used, no reaction occurred because pure water is a poor conductor of electricity.

Upon adding dilute sulfuric acid (H₂SO₄), the water became able to conduct electricity, and the electrolysis proceeded successfully.

Answers:

1. Gas X is hydrogen (H₂), which is evolved at the cathode because, during electrolysis, hydrogen forms at the cathode.
2. Gas Y is oxygen (O₂), which is evolved at the anode because oxygen gas forms at the anode during the process.
3. The balanced chemical equation for the electrolysis of water is:
   2H₂O(l) → 2H₂(g) + O₂(g)