E deg(Ag+/Ag) = 0.800 V and E deg(AgBr/Ag,Br-) = 0.071 V at 298 K. The value of pK_sp = -log K_sp for AgBr is closest to (take 2.303RT/F = 0.0591 V)

JEE Main levelSingle correct optionDifficulty 2 of 3ec101

$E^{\circ} (Ag X^{+} / Ag) = 0.800 V$ and $E^{\circ} (AgBr / Ag, Br X^{-}) = 0.071 V$ at 298 K. The value of $p K_{s p} = - lo g K_{s p}$ for AgBr is closest to (take $2.303 RT / F = 0.0591 V$ )

The two half reactions differ by

AgBr (s) Ag X^{+} + Br X^{-}

E^{\circ} (AgBr / Ag) = E^{\circ} (Ag X^{+} / Ag) + 0.0591 lo g K_{s p}

. So

lo g K_{s p} = \frac{0.071 - 0.800}{0.0591} = - 12.33

and

p K_{s p} = 12.3

(

K_{s p} \approx 4.7 \times 1 0^{- 13}

). Dividing by

2 \times 0.0591

(wrong

n

) gives 6.2.