JEE Main 2025 Jan 28 Shift 2, Chemistry Q23: Electrolytic Cells And Electrolysis

JEE Main2025Jan 28, Shift 2Chemistry

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).

Your answer

Solution

Electrolysis of aqueous NaCl produces $C l_{2}$ at the anode and $H_{2}$ at the cathode, leaving $N a O H$ in solution:

$N a C l + H_{2} O \to N a O H + \frac{1}{2} C l_{2} + \frac{1}{2} H_{2}$ .

If the final pH is 12, then $[O H^{-}] = 1 0^{- 2}$ M. Moles of $O H^{-}$ produced in 600 mL $= 1 0^{- 2} \times 0.600 = 6 \times 1 0^{- 3}$ mol.

Each $O H^{-}$ requires the passage of 1 electron, so charge $Q = n F = 6 \times 1 0^{- 3} \times 96500 \approx 579$ C.

With $t = 5 \times 60 = 300$ s, current $I = \frac{Q}{t} = \frac{579}{300} \approx 1.93$ A.

Nearest integer $\Rightarrow I = 2$ A.

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