The EMF of the cell: Hg(l) \| Hg2Cl2(s), KCl sol. (1.0N) \| Quinhydrone \| Pt, is 0.210 V at 298 K. What is the pH of the quinhydrone solution, the potential of the normal calomel electrode is 0.279 V

Question

The EMF of the cell: Hg(l) \| Hg2Cl2(s), KCl sol. (1.0N) \| Quinhydrone \| Pt, is 0.210 V at 298 K. What is the pH of the quinhydrone solution, the potential of the normal calomel electrode is 0.279 V and E° for the quinhydrone electrode is 0.699 V, both at the same temperature. [2.303RT/F = 0.06]

Prashant Kotian · Accepted Answer

Correct answer: A. E_cell = E_cathode - E_anode. The Quinhydrone electrode (cathode) potential is E_Q = E°_Q - 0.06*pH = 0.699 - 0.06*pH. The Calomel anode is 0.279 V. 0.210 = (0.699 - 0.06*pH) - 0.279 = 0.420 - 0.06*pH. Solving for pH gives 0.06*pH = 0.210, so pH = 3.5. Therefore, correct answer is A.

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