Statement I: On electrolysis of aqueous NaOH we will get H₂ at cathode and O₂ at anode

Statement II: Discharge potential of H⁺ is less than that of Na⁺

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

The answer is 1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.

Statement I is correct because the discharge potential of H⁺ is less than that of Na⁺. This means that hydrogen ions will be reduced at the cathode, before sodium ions. Statement II is also correct and it is the correct explanation of Statement I.

The discharge potential of an ion is the minimum potential that is required to reduce that ion. The lower the discharge potential, the more easily the ion will be reduced. In the case of aqueous NaOH, the discharge potentials of H⁺ and Na⁺ are -2.372 V and -2.714 V, respectively. This means that H⁺ ions will be reduced at the cathode, before Na⁺ ions.

The products of the electrolysis of aqueous NaOH are therefore hydrogen gas at the cathode and oxygen gas at the anode. This is because the discharge potential of H⁺ is less than that of Na⁺, and the discharge potential of O^2- is greater than that of OH^-.