A silver concentration cell has $0.1 M$ $AgNO X_{3}$ in the right compartment. The left compartment contains $0.1 M$ $AgNO X_{3}$ to which enough NaCl was added so that, after precipitation of AgCl, the free chloride concentration is $0.1 M$ . $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $2.303 RT / F = 0.0591 V$ , $lo g 1.8 = 0.255$ . The cell EMF at 298 K is closest to

Answer: A

In the left compartment,

[Ag X^{+}] = K_{s p} / [Cl X^{-}] = 1.8 \times 1 0^{- 10} /0.1 = 1.8 \times 1 0^{- 9} M

. This is the dilute side, so it is the anode.

E = 0.0591 lo g \frac{0.1}{1.8 \times 1 0 ^{- 9}} = 0.0591 \times (8 - 1 - 0.255) = 0.0591 \times 7.745 = 0.458 V

. Taking

[Ag X^{+}] = K_{s p}

directly (forgetting to divide by

[Cl X^{-}]

) gives 0.517 V.