Answer:
4OH-→ 2H²O(l) + O² + 4e- (Anode)
2H+ → H²O (l) + 2e- (Cathode)
Step-by-step explanation:
OH- ions are preferentially discharged at the anode since they have a less positive electrode potential (+0.40V) then chloride ions (+1.36)
H+ ions are preferentially discharged at the cathode since they have a less negative standard electrode potential (0.00) than Potassium (-2.92)