A Daniell cell with $E_{cell}^{\circ} = 1.10 V$ is allowed to discharge at 298 K until its EMF falls to zero. At that point the ratio $[Zn X^{2 +}] / [Cu X^{2 +}]$ is (take $2.303 RT / F = 0.0591 V$ )

Answer: A

When

E = 0

the cell has reached equilibrium and

Q = K

0 = 1.10 - \frac{0.0591}{2} lo g K

, so

lo g K = \frac{2 \times 1.10}{0.0591} = 37.2

and

[Zn X^{2 +}] / [Cu X^{2 +}] = 1 0^{37.2}

. Using

n = 1

gives the distractor

1 0^{18.6}

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