E deg(Ag+/Ag) = 0.80 V and E deg(AgCl/Ag,Cl-) = 0.22 V. The value of -log_10 K_sp(AgCl) at 298 K is (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ec15

$E^{\circ} (Ag X^{+} / Ag) = 0.80 V$ and $E^{\circ} (AgCl / Ag, Cl X^{-}) = 0.22 V$ . The value of $- lo g_{10} K_{s p} (AgCl)$ at 298 K is (nearest integer).

Numerical answer type.

The two half reactions differ by

AgCl Ag X^{+} + Cl X^{-}

E^{\circ}

for dissolution

= 0.22 - 0.80 = - 0.58 V

n = 1

lo g K_{s p} = \frac{- 0.58}{0.0591} = - 9.8

, so

- lo g K_{s p} \approx 10

(

K_{s p} \approx 1.5 \times 1 0^{- 10}