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ExamsJEE AdvancedChemistry

A galvanic cell at 25 deg C is set up as follows: Pt | H2(g, 1 bar) | H2CO3(aq, 0.2 M, 100 mL) + NaHCO3(aq, 0.2 M, 100 mL) || CH3COOH(aq, 0.1 M, 100 mL) | H2(g, 1 bar) | Pt Given: Ka1(H2CO3) = 10⁻⁷, Ka2(H2CO3) = 10⁻¹¹, Ka(CH3COOH) = 10⁻⁵, and 2.303RT/F = 0.06 V. Calculate the initial (baseline) EMF of this cell before any electrolyte is added.

  1. 0.36 V
  2. 0.24 V
  3. 0.30 V
  4. 0.12 V

Correct answer: 0.24 V

Solution

Left electrode pH = 7 (buffer at equal concentrations of H2CO3 and NaHCO3 with pKa1 = 7), right electrode pH = 3 (CH3COOH: [H+] = sqrt(10⁻⁵ * 0.1) = 10⁻³). EMF = 0.06 * (pH_left - pH_right) = 0.06 * (7 - 3) = 0.24 V.

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