The cell Zn|Zn^2+(1M)||Cu^2+(1M)|Cu. (E_cell^° = 1.10 V) was allowed…

The correct answer is Option (4) → \(10^{37.3}\)

As the Daniell cell discharges, \( \ce{[Zn^{2+}]} \) rises and \( \ce{[Cu^{2+}]} \) falls until the cell voltage becomes zero. The reaction is then at equilibrium, and the ratio \( \ce{[Zn^{2+}]} \)/\( \ce{[Cu^{2+}]} \) equals the equilibrium constant K_c. From the Nernst equation, at equilibrium (E_cell = 0) and 298 K:

    \( E^\circ_{cell} = \displaystyle\frac{0.059}{n}\log K_c \)

For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2:

    \( \log K_c = \displaystyle\frac{2 \times 1.10}{0.059} = 37.29 \approx 37.3 \)

    \( \displaystyle\frac{[Zn^{2+}]}{[Cu^{2+}]} = K_c = 10^{37.3} \)

(about 2 × 10³⁷, so practically all the \( \ce{Cu^{2+}} \) has been replaced by \( \ce{Zn^{2+}} \).)

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Equilibrium Constant from Nernst Equation – “If the circuit in Daniell cell (Fig. 2.1) is closed then we note that the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) takes place and as time passes, the concentration of Zn²⁺ keeps on increasing while the concentration of Cu²⁺ keeps on decreasing. … After some time, we shall note that there is no change in the concentration of Cu²⁺ and Zn²⁺ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. … But at equilibrium, [Zn²⁺]/[Cu²⁺] = K_c for the reaction 2.1 and at T = 298K the above equation can be written as”

    \( \displaystyle E^\circ_{cell} = \frac{0.059\ \text{V}}{2}\log K_C = 1.1\ \text{V} \)

    \( \displaystyle \log K_C = \frac{1.1\ \text{V} \times 2}{0.059\ \text{V}} = 37.288 \)

    K_C = 2 × 10³⁷ at 298K.