Exams › JEE Main › Chemistry

Given the data at 25 °C Ag + I⁻ → AgI + e⁻ E° = 0.152 V Ag → Ag⁺ + e⁻ E° = -0.800 V What is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)

1. -37.83
2. -16.13
3. -8.12
4. +8.612

Correct answer: -16.13

Solution

AgI -> Ag+ + I- combines (AgI + e- -> Ag + I-, E = -0.152 V) with (Ag -> Ag+ + e-, E = -0.800 V), giving E(cell) = -0.952 V. log Ksp = nE/0.059 = (1)(-0.952)/0.059 = -16.13.

Related JEE Main Chemistry questions

• During electrolysis of an aqueous solution of dipotassium succinate, which gas is evolved at the electrodes?
• Which expression gives the equivalent conductance at infinite dilution of Al2(SO4)3, if Λ°Al3+ and Λ°SO4²− denote the equivalent conductances at infinite dilution of the corresponding ions?
• The standard emf of a Daniell cell is 1.10 V. What is the greatest amount of electrical work that can be extracted from this cell?
• During the recharging of a lead storage battery, which reaction takes place at the cathode?
• The limiting molar ionic conductivities of the ions X2+ and Y2− are 57 and 73, respectively. What will be the molar conductivity of the electrolyte formed by these ions?
• According to Kohlrausch’s law, at what condition does each ion contribute a fixed amount to the conductance of an electrolyte, irrespective of the other ion present?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →