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 $K_{sp}$ of AgCl would be –
(a) $1 x 10^{-9.8}$
(b) $1 x 10^{-19.6}$
(c) $2 x 10^{-10}$
(d) $2.64 x 10^{-14}$

Answer 1

${\huge {\mathfrak {Answer :-}}}$

➡️ Option (B)

That's it..

Answer 2

Thanks for your question ✔

Option B is correct

Thanks