Question

During the electrolysis of acidified water, the substance liberated at the cathode is

• A. Water
• B. Oxygen gas
• C. Sulphuric acid
• D. Hydrogen gas

Answer 1

The correct option is D

Hydrogen gas

Pure water does not ionize to a greater extent and therefore does not allow passage of electricity. A little of dilute sulphuric acid is mixed with water, which when ionizes gives $H^{+}$ ions. These $H^{+}$ ions combine with water and form $H_3{O}^{+}$ ions.

$H_{2}S{O}_4\left(l\right)+H_{2}O\left(l\right)\to 2H_3{O}^{+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

The reaction occurring at the cathode is:

$2H_3{O}^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)+2H_{2}O\left(l\right)$

The ions moving towards positive electrode is $S{O}_4^{2-}$. On comparing $S{O}_4^{2-}$ with water, water has a higher tendency to get oxidised.

The reaction occurring at the anode is:

$2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}$

Hydrogen gas is obtained at the cathode and oxygen gas is obtained at the anode.