A 200-mL solution of KCl(aq) was electrolyzed for 10.0 minutes, producing Cl2 gas at the anode and H2 gas at the cathode. If the pH of the final solution was 12.40, what was the average current (in amps) used?

Answer 1

pOH= 14- 12.4 = 1.6
[OH-] = 10^-1.6 = .025 M OH-
mols of OH- = .025 M * .2L =.005 mols
.005 mols OH- * (2 mols e-/ 2mols )OH-) * (96500 C / 1 mole e-) = 484.8 C
amps = 484.8 C / (10 min*60) = .81 A