A silver electrode is dipped in 0.1 M AgNO3 at 298 K. K_sp(AgCl) = 1.8 x 10^-10, K_sp(AgBr) = 5 x 10^-13. Which statements are correct?

Adding KBr in excess lowers the electrode potential more than adding the same excess of KCl

Adding excess aqueous ammonia lowers the electrode potential

Adding KBr in excess lowers the electrode potential more than adding the same excess of KCl

Adding excess aqueous ammonia lowers the electrode potential

Adding KNO3 lowers the electrode potential by a large amount

E deg(AgCl/Ag,Cl-) is lower than E deg(AgBr/Ag,Br-)

JEE Advanced levelOne or more correct optionsDifficulty 2 of 3ec104

A silver electrode is dipped in $0.1 M$ $AgNO X_{3}$ at 298 K. $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{s p} (AgBr) = 5 \times 1 0^{- 13}$ . Which statements are correct?

AAdding KBr in excess lowers the electrode potential more than adding the same excess of KCl
BAdding excess aqueous ammonia lowers the electrode potential
CAdding $KNO X_{3}$ lowers the electrode potential by a large amount
D $E^{\circ} (AgCl / Ag, Cl X^{-})$ is lower than $E^{\circ} (AgBr / Ag, Br X^{-})$

E = E^{\circ} + 0.0591 lo g [Ag X^{+}]

, so anything that lowers free

[Ag X^{+}]

lowers

E

. AgBr is less soluble than AgCl, so KBr leaves a smaller

[Ag X^{+}]

and lowers

E

more (A). Ammonia forms

[Ag (NH X_{3}) X_{2}] X^{+}

and also lowers

E

(B).

KNO X_{3}

forms no precipitate or complex (C is wrong).

E^{\circ} (AgX / Ag) = 0.80 + 0.0591 lo g K_{s p}

; since

K_{s p} (AgCl) > K_{s p} (AgBr)

E^{\circ} (AgCl / Ag)

is higher, not lower (D is wrong).