Question

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).

Hint:

Use Faraday’s laws of electrolysis to relate charge, time, and current.

Answer 1

Correct Answer: 1.93

Using Faraday's law: \[ \text{Charge} (Q) = I \times t \] Hydroxide ion (\( OH^- \)) concentration at \( pH = 12 \): \[ [OH^-] = 10^{-2} \, {M} \] Moles of \( OH^- \) in 0.6 L: \[ n = 0.6 \times 10^{-2} = 6 \times 10^{-3} \, {moles} \] Charge required: \[ Q = n \times F = 6 \times 10^{-3} \times 96500 \] \[ = 579 \, C \] Current: \[ I = \frac{Q}{t} = \frac{579}{300} \approx 1.93  \, A \]

Answer 2

Step 1: Given data.
Volume of NaCl solution = 600 mL = 0.6 L
Time of electrolysis = 5 minutes = 300 s
Final pH = 12

Step 2: Concept used.
During electrolysis of aqueous NaCl:
At the cathode: \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \)
At the anode: \( 2Cl^- \rightarrow Cl_2 + 2e^- \)

The solution becomes basic due to the production of \( OH^- \) ions.

Step 3: Relate pH and pOH.
\[ pH + pOH = 14 \Rightarrow pOH = 14 - 12 = 2. \] \[ [OH^-] = 10^{-pOH} = 10^{-2} = 0.01 \, \text{M}. \] This means 0.01 mol of \( OH^- \) are produced per liter of solution.

Step 4: Moles of \( OH^- \) produced in 0.6 L solution.
\[ n(OH^-) = 0.01 \times 0.6 = 0.006 \, \text{mol}. \]

Step 5: Relate \( OH^- \) moles to electrons transferred.
From the cathode reaction, \( 2e^- \) produce \( 2OH^- \). Hence, 1 mol of \( e^- \) produces 1 mol of \( OH^- \).
\[ n(e^-) = n(OH^-) = 0.006 \, \text{mol}. \]

Step 6: Use Faraday’s law to find current.
\[ Q = n(e^-) \times F = 0.006 \times 96500 = 579 \, \text{C}. \] \[ I = \frac{Q}{t} = \frac{579}{300} = 1.93 \, \text{A}. \]

Final Answer:
\[ \boxed{1.93 \, \text{A}} \]