The EMF of the cell: Ag, AgCl in 0.1 M – KCl \|\| satd. NH4NO3 \|\| 0.1 M – AgNO3, Ag is 0.42 V at 25°C. 0.1 M – KCl is 50% dissociated and 0.1 M – AgNO3 is 40% dissociated. The solubility product of

Question

The EMF of the cell: Ag, AgCl in 0.1 M – KCl \|\| satd. NH4NO3 \|\| 0.1 M – AgNO3, Ag is 0.42 V at 25°C. 0.1 M – KCl is 50% dissociated and 0.1 M – AgNO3 is 40% dissociated. The solubility product of AgCl is (2.303RT/F = 0.06)

Prashant Kotian · Accepted Answer

Correct answer: D. RHS [Ag^+] = 0.1 * 0.40 = 0.04 M. E_cell = 0.06 * log([Ag^+]_RHS / [Ag^+]_LHS). Substituting values: 0.42 = 0.06 * log(0.04 / [Ag^+]_LHS). log(0.04 / [Ag^+]_LHS) = 7, so [Ag^+]_LHS = 0.04 / 10^7 = 4 × 10^-9 M. In the LHS, [Cl^-] = 0.1 * 0.50 = 0.05 M. Ksp of AgCl = [Ag^+][Cl^-] = (4 × 10^-9) * 0.05 = 2.0 × 10^-10. Therefore, correct answer is D.

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