Question

What happens to sulphate ions during electrolysis of acidified water?Explain with equation

Answer 1

Water can be written in ionic form as :

H₂O(aq) -------> H+(aq) + OH–(aq)

When an acid is added in water it dissociates like:

H2SO4 (aq) -------> 2H+(aq) + (SO4)2–(aq)

Reactions at cathode and anode

At cathode, hydrogen ions receive electrons, they undergo reduction and discharge to form hydrogen gas.

2H+(aq) + 2e– --------> H2(g)

At anode hydroxide ions give up electrons, they undergo oxidation and discharge to form oxygen gas because sulphate ion remain in solution and it does not give up the electron.

4OH-(aq) ------> O2(g) + 2H2O(l) + 4e–

Overall reaction:

2H+(aq) + 2e– --------->H2(g) (1)

4OH-(aq) ------> O2(g) + 2H2O(l) + 4e– (2)

In equation 2 there is 4e and in equation 1 there is 2 e so we balance the no. of electrons by multiplying the equation 1 by 2

(1)x2: 4H+(aq) + 4e– ------> 2H2(g) (3)

(2)+(3): 4OH-(aq) + 4H+(aq) ------> O2(g) + 2H2O(l) + 2H2(g)

4H2O(l) --------> O2(g) + 2H2O(l) + 2H2(g)

Canceling the 2 H2O from both side we get.

2H2O(l) ------> O2(g) + 2H2(g)