Question

Following cell has EMF 0.7995V.
Pt | H₂ (1 atm) | HNO₃ (1M) || AgNO₃ (1M) | Ag
If we add enough KCl to the Ag cell so that the final Cl⁻ is
1M. Now the measured emf of the cell is 0.222V. The Ksp of
AgCl would be –
(a) 1 × 10⁻⁹.⁸ (b) 1 × 10⁻¹⁹.⁶
(c) 2 × 10⁻¹⁰ (d) 2.64 × 10⁻¹⁴

Answer 1

Now the measured emf of the cell is 0.222V. The Ksp of

AgCl would be –

(a) 1 × 10⁻⁹.⁸

Answer 2

The Ksp of AgCl would be – 1 × 10⁻⁹.⁸ (10 to the power -9.8)

Hence, option ( A ) is correct answer.

• First calculate concentration of Ag+ ion using Nernst equation.

• After substituting the values from the question into Nernst equation , the concentration of Ag+ can be calculated that is 10⁻⁹.⁸

• Ksp of AgCl = [Ag+][Cl−]=(10⁻⁹.⁸)×(1)=1 × 10⁻⁹.⁸