The EMF of the concentration cell Zn | Zn^2+ (0.001 M) || Zn^2+ (0.1 M) | Zn at 298 K, in mV, is (nearest integer).

JEE Main levelNumerical answerDifficulty 1 of 3ec03

The EMF of the concentration cell $Zn ∣ Zn X^{2 +} (0.001 M) ∣∣ Zn X^{2 +} (0.1 M) ∣ Zn$ at 298 K, in mV, is (nearest integer).

Numerical answer type, answer in mV, tolerance 1.

E = \frac{0.0591}{2} lo g \frac{0.1}{0.001} = \frac{0.0591}{2} \times 2 = 0.0591 V = 59 mV

. Zinc dissolves in the dilute compartment and deposits in the concentrated one.

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