$E^{\circ} (Cu X^{2 +} / Cu) = 0.34 V$ and $K_{s p} [Cu (OH) X_{2}] = 1.0 \times 1 0^{- 20}$ at 298 K. The standard potential of the half reaction $Cu (OH) X_{2} (s) + 2 e X^{-} Cu + 2 OH X^{-}$ is (take $2.303 RT / F = 0.0591 V$ )

Answer: A

With

[OH X^{-}] = 1 M

[Cu X^{2 +}] = K_{s p} / [OH X^{-}]^{2} = 1 0^{- 20} M

. The new standard potential is the

Cu X^{2 +} / Cu

potential at this concentration:

E^{\circ} = 0.34 + \frac{0.0591}{2} lo g (1 0^{- 20}) = 0.34 - 0.591 = - 0.251 V

. Using

n = 1

gives

- 0.842 V

; using

K_{w}

in place of

K_{s p}

gives

- 0.074 V