Satu liter larutan NaOH dielektrolisis dengan arus sebesar 965 coulomb, tentukan pH larutan yang terjadi

katoda : 2H2O + 4e → H2 + 4OH^-
anoda: 4OH^- → 2H2O + O2 + 4e

jumlah mol elektron :
n = 965/96500
n = 0.01 mol

mol OH^- = mol elektron = 0.01 mol
[OH^-] = n/V
[OH^-] = 0.01/2

pH = 14 - pOH
pH = 14 - (- log (0.01/2))
pH = 14 + log (5*10^-3)
pH = 14 - 3 + log 5
pH = 11 + log 5

reaksi elektrolisis
katoda : 2H₂O + 4e⁻ → H₂ + 4OH⁻
anoda: 4OH⁻ → 2H₂O + O2 + 4e⁻

ev = 1
Q = 965 coloumb
mol elektron = Q/96500 = 965/96500 = 0.01 mol
mol OH⁻ = mol elektron
              = 0.01 mol
[OH⁻] = mol/volume
          = 0.01/1 = 0.01 Molar
pOH = -log [OH⁻]
        = - log 0.01
        = 2
pH = 14 - pOH
     = 14 - 2 = 12