The cell, $\mathrm{Zn}|\mathrm{Zn}^{2+}(1 \mathrm{~M}) || \mathrm{Cu}^{2+}(1 \mathrm{~M})|\mathrm{Cu}$ ($\mathrm{E}_{\text{cell}}^{0}=1.10 \mathrm{~V}$), was allowed to be completely discharged at 298 K. The relative concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}$ $\left[\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}\right]$ is

JEE Main · 2007 · 29 Apr 2007 · Offline · Redox Reactions and Electrochemistry

The cell,

Zn ∣ Zn^{2 +} (1 M) ∣∣ Cu^{2 +} (1 M) ∣ Cu

(

E_{cell}^{0} = 1.10 V

), was allowed to be completely discharged at 298 K. The relative concentration of

Zn^{2 +}

Cu^{2 +}

[\frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]}]

Answer: (C)

At complete discharge the cell can drive no more current, so its emf has fallen to zero. The Nernst equation for this two-electron cell reads

E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}

. Setting

E = 0

with

E^{\circ} = 1.10

V gives

lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = \frac{2 \times 1.10}{0.0591} \approx 37.3

, so the ratio of zinc-ion to copper-ion concentration is about

1 0^{37.3}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option