Answer:
C - choice
Explanation:
In pure water at the negatively charged cathode, a reduction reaction takes place, with electrons (e−) from the cathode being given to hydrogen cations to form hydrogen gas. The half reaction, balanced with acid, is: Reduction at cathode: 2 H+(aq) + 2e− → H2(g) ... Oxidation at anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e.
