Question

A 1 M solution of H₂SO4 is electrolysed. Select correct statement in respect of products obtain at anode and cathode respectively:

Given :
2SO₄²⁻→S₂O₈²⁻+2e−,E∘=−1.23V
H₂O(l)→2H⁺(aq)+1/2O₂(g)+2e−,E∘=−1.23V

A. concentration of H₂SO₄ remain constant ,H₂,O₂
B. concentration of H₂SO₄ increases,O₂,H₂
C. concentraion of H2SO4 decreasesO₂,H₂
D. concentration of H₂SO₄ remains constant,S₂O₈²⁻,H₂

Answer 1

During the electrolysis of 1 M H₂SO₄, hydrogen gas is produced at the cathode and oxygen gas at the anode. This process increases the concentration of H₂SO₄. The correct statement about the products at anode and cathode is 'B. concentration of H₂SO₄ increases, O₂, H₂'.

Step-by-step explanation:

During the electrolysis of a 1 M H₂SO₄ solution, the products obtained at the anode and cathode can be determined based on their respective electrode potentials and reaction favorability.

At the cathode, the reduction of water is more likely to occur over the reduction of sulfate ions due to its more favorable standard electrode potential. The reduction of water at the cathode is described by the equation:

2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)

Hydrogen gas is evolved at the cathode as a result. Since hydroxide ions are produced, they will react with any free H⁺⁺ ions present from the dissociation of sulfuric acid, which will form water and in turn increase the concentration of sulfate ions (SO₄²⁻), effectively increasing the acidity of the solution, or the concentration of H₂SO₄.

At the anode, oxygen evolution is typically favored over the production of persulfate due to practical electrochemical conditions. The oxidation of water at the anode is described by the equation:

2H₂O(l) → O₂(g) + 4H⁺⁺(aq) + 4e⁻

Oxygen gas is evolved at the anode. This reaction also produces protons, contributing to the increased concentration of sulfuric acid.

Therefore, the correct statement in respect of products obtained at anode and cathode respectively is:

B. concentration of H₂SO₄ increases, O₂, H₂