JEE Main · 2025 · Shift-IIhardEC-137

O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B)…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

\ceO2\ce{O2} gas will be evolved as a product of electrolysis of:

(A) an aqueous solution of \ceAgNO3\ce{AgNO3} using silver electrodes. (B) an aqueous solution of \ceAgNO3\ce{AgNO3} using platinum electrodes. (C) a dilute solution of \ceH2SO4\ce{H2SO4} using platinum electrodes. (D) a high concentration solution of \ceH2SO4\ce{H2SO4} using platinum electrodes.

Choose the correct answer from the options given below:

Options
  1. a

    (B) and (C) only

    ✓
  2. b

    (A) and (D) only

  3. c

    (B) and (D) only

  4. d

    (A) and (C) only

Correct Answera

(B) and (C) only

Detailed Solution

Strategy: Calculate [H+][H^+] from the molarity and α\alpha, then find pH. For the potential, consider the cell: \cePt∣H2∣H+∣∣Standard\ce{Pt | H2 | H^+ || Standard}. Potential differen\ce =0.059⋅pH= 0.059 \cdot \mathrm{pH}.

Calculation: α=0.5\alpha = 0.5% = 0.005. [\ceH+]=Calpha=0.1×0.005=5×10−4 M[\ce{H^+}] = Calpha = 0.1 \times 0.005 = 5 \times 10^{-4} \text{ M}. pH=−log(5×10−4)=4−0.7=3.3\mathrm{pH} = -log(5 \times 10^{-4}) = 4 - 0.7 = 3.3. Potential E=−0.059×3.3≈−0.19 VE = -0.059 \times 3.3 \approx -0.19 \text{ V}.

Value =19×10−2= 19 \times 10^{-2}.

Answer: 19\boxed{\text{Answer: 19}}

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