Answer to Question #93392 in General Chemistry for belle

Question

describe the electrolysis of dilute tetraoxosulphateVIacid

Answer 1

Component in dilute tetraoxosulphateacid are $H^+,SO_4^{2-} and H_2O$

$H_2O dissociates into H^+ and OH^-$

At cathode, reduction of $H^+$ takes place

Cathode reaction$:$ $2H^+(aq)+2e^-\to H_2(g)$

At anode, oxidation of $OH^-$ takes place as it is higher in electrochemical series than${SO_4}^{2-}$

Anode Reaction$:4OH^-(aq)\to 2H_2O(l) +O_2(g)+4e^-$

Overall reaction$:4H^+(aq) +4OH^-(aq)\to2H_2(g)+O_2(g)$