Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12 . The current in Amperes used for the given electrolysis is ____. (Nearest integer).

Electrolysis of 600 mL aqueous NaCl for 5 min changes pH to 12. Find the current.

We use the relation $$pH + pOH = 14$$, so with pH = 12, $$pOH = 14 - 12 = 2$$.

Therefore $$[OH^-] = 10^{-pOH} = 10^{-2} = 0.01$$ M.

With a volume of 600 mL (0.6 L), the moles of $$OH^-$$ produced are $$n_{OH^-} = [OH^-]\times V = 0.01\times0.6 = 0.006$$ mol.

At the cathode the reaction $$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$ shows that 2 moles of electrons produce 2 moles of $$OH^-$$, so the moles of electrons transferred equal the moles of $$OH^-$$, namely 0.006 mol.

The total charge passed is $$Q = n_e \times F = 0.006 \times 96500 = 579$$ C.

Since the electrolysis runs for 5 min (300 s), the current is $$I = \frac{Q}{t} = \frac{579}{300} = 1.93 \approx 2$$ A.

The answer is 2 A.