The Daniell cell reaction Zn + Cu^2+ -> Zn^2+ + Cu has E deg = 1.10 V at 298 K. Taking (2.303RT)/(F) = 0.0591 V, the equilibrium constant is of the order of

JEE Main levelSingle correct optionDifficulty 2 of 3ec067

The Daniell cell reaction $Zn + Cu X^{2 +} Zn X^{2 +} + Cu$ has $E^{\circ} = 1.10 V$ at 298 K. Taking $\frac{2.303 RT}{F} = 0.0591 V$ , the equilibrium constant is of the order of

lo g K = \frac{n E ^{\circ}}{0.0591} = \frac{2 \times 1.10}{0.0591} = 37.2

, so

K \approx 1 0^{37}

. Such a huge value means the reaction is practically complete; essentially no

Cu X^{2 +}

remains at equilibrium. Option B uses

n = 1

and option C uses

n = 4

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