The voltage of the cell given below is –0.61 V.<br>Pt \| H2 (1 bar) \| NaHSO3 (0.4 M), Na2SO3 (6.4 × — Electrochemistry Chemistry Question

💡 Solution & Explanation

E_c = -0.76 - (0.06/2)log(1/0.4) = -0.772 V. E_cell = E_c - E_a = -0.772 - (-0.06*pH) = -0.61 V. This gives 0.06*pH = 0.162, so pH = 2.7. For the buffer, pH = pKa2 + log(0.064/0.4) = pKa2 - 0.8. Thus, 2.7 = pKa2 - 0.8, meaning pKa2 = 3.5. Ka2 = 10^-3.5 ≈ 3.2 * 10^-4. Therefore, correct answer is A.