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Question: 6.4 litres of aq. NaCl is electrolysed for 1 minute 4 sec using 19.3 amperes (current efficiency is ...

6.4 litres of aq. NaCl is electrolysed for 1 minute 4 sec using 19.3 amperes (current efficiency is 50%) at 25°C. The pH of resultant solution is, (1F=96500Cmol)\left(1F = 96500 \frac{C}{mol}\right)

Answer

11

Explanation

Solution

To determine the pH of the resultant solution after electrolysis of aqueous NaCl, we follow these steps:

  1. Identify Electrode Reactions: During the electrolysis of aqueous NaCl, the following reactions occur:

    • At Cathode (Reduction): Water is reduced in preference to Na⁺ ions due to its higher reduction potential. 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)
    • At Anode (Oxidation): Chloride ions are oxidized in preference to water due to overpotential effects for oxygen formation. 2Cl−(aq)→Cl2(g)+2e−2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- The production of OH−OH^- ions at the cathode increases the pH of the solution.

  2. Calculate Total Charge Passed (Q_total):

    • Time, t=1 minute 4 seconds=60+4=64 st = 1 \text{ minute } 4 \text{ seconds} = 60 + 4 = 64 \text{ s}
    • Current, I=19.3 AI = 19.3 \text{ A}
    • Total charge, Qtotal=I×t=19.3 A×64 s=1235.2 CQ_{total} = I \times t = 19.3 \text{ A} \times 64 \text{ s} = 1235.2 \text{ C}

  3. Calculate Actual Charge Used (Q_actual) considering Current Efficiency:

    • Current efficiency = 50% = 0.50
    • Actual charge used, Qactual=Qtotal×efficiency=1235.2 C×0.50=617.6 CQ_{actual} = Q_{total} \times \text{efficiency} = 1235.2 \text{ C} \times 0.50 = 617.6 \text{ C}

  4. Calculate Moles of Electrons (n_e):

    • Using Faraday's constant (1F=96500 C/mol e−1F = 96500 \text{ C/mol e}^-):
    • Moles of electrons, ne=QactualF=617.6 C96500 C/mol=0.0064 moln_e = \frac{Q_{actual}}{F} = \frac{617.6 \text{ C}}{96500 \text{ C/mol}} = 0.0064 \text{ mol}

  5. Determine Moles of OH⁻ Produced:

    • From the cathode reaction, 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq), it is clear that 2 moles of electrons produce 2 moles of OH−OH^- ions.
    • Therefore, moles of OH−OH^- produced = moles of electrons = 0.0064 mol0.0064 \text{ mol}

  6. Calculate Concentration of OH⁻ Ions:

    • Volume of solution = 6.4 litres6.4 \text{ litres}
    • Concentration of OH−OH^-, [OH−]=Moles of OH−Volume of solution=0.0064 mol6.4 L=0.001 M=1×10−3 M[OH^-] = \frac{\text{Moles of } OH^-}{\text{Volume of solution}} = \frac{0.0064 \text{ mol}}{6.4 \text{ L}} = 0.001 \text{ M} = 1 \times 10^{-3} \text{ M}

  7. Calculate pOH:

    • pOH = −log⁡[OH−]=−log⁡(1×10−3)=3-\log[OH^-] = -\log(1 \times 10^{-3}) = 3

  8. Calculate pH:

    • At 25°C, pH+pOH=14pH + pOH = 14
    • pH = 14−pOH=14−3=1114 - pOH = 14 - 3 = 11

The pH of the resultant solution is 11.