Question

Explain with equation electrolysis of acidified water with both the equations (water and suplhuric acid)

Answer 1

Water is weak electrolyte and do not ionize very much to conduct the electrons so some amount of H2SO4 ia added to make it conducting. So when current is passed through the above mentioned solution, water breaks down and release H2 gas at cathode and O2 at anode.In presence of H2SO4 water can be ionized as :

H2O → H⁺ + OH^-

and sulfuric acid will ionize like:

H2SO4 → 2H⁺ + SO4^2-

During the electrolysis cations moves toward cathode.So here hydrogen ions moves toward cathode receive electrons to produce hydrogen gas.

2H+(aq) + 2e– --------> H2(g)

Now in solution there is OH^- and SO4^2- but SO4^2- remains in solution and OH^- moves toward anode.

At anode, hydroxide ions give up electrons to form oxygen gas.

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

Overall reaction:

2H+(aq) + 2e– --------->H2(g) (1)

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

equating the number of electrons in above two equation we get

(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)

Overall equation: 2H2O(l) ------> O2(g) + 2H2(g)