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)?? [2025]

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Solution

Q.
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) [2025]

Ans.
(2)

Electrolysis of dilute NaCl(aq) is just electrolysis of water, it does not change pH. Since in the question, pH changes, the solution is concentrated NaCl(aq).

$Final p O H = 14 - p H = 14 - 12 = 2, {so [OH]}_{_{final}}^{-} = 10^{- 2} M$

$n_{{OH}_{final}^{-}} = {[OH]}_{final}^{-} \times V_{solution} = 10^{- 2} M \times 0.6 L = 6 \times 10^{- 3} m o l$

Initially, the solution is neutral, pH = pOH = 7. So ${[OH]}_{_{initial}}^{-} = 10^{- 7} M$

$n_{{OH}_{initial}^{-}} = {[OH]}_{initial}^{-} \times V_{solution} = 10^{- 7} M \times 0.6 L = 6 \times 10^{- 8} m o l$

Increase in the amount of ${OH}^{-} = n_{{OH}_{final}^{-}} - n_{{OH}_{initial}^{-}}$

$= (6 \times 10^{- 3} - 6 \times 10^{- 8}) m o l \approx 6 \times 10^{- 3} m o l$

For electrolysis of concentrated solution of NaCl:

Reaction at anode (oxidation):

$2 {Cl}^{-} (a q) \to {Cl}_{2} (g) + 2 e^{-}$

Reaction at cathode (reduction):

$2 H_{2} O + 2 e^{-} \to H_{2} (g) + 2 {OH}^{-} (a q)$

As per stoichiometry of the reaction, moles of electrons passed = increase in moles of ${OH}^{-} = 6 \times 10^{- 3} m o l$

$Charge passed (Q) = moles of electrons passed \times charge on 1 mol electrons$

$= 6 \times 10^{- 3} \times 96487 C = 578.922 C$

$Current (I) = \frac{Q}{t} = \frac{578.922 C}{5 min} = \frac{578.922 C}{5×60 s}$

$= 1.93 s$ (nearest integer is 2)

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