The cell Pt | H2 (1 bar) | H+ (1 M) || Cu^2+ (x M) | Cu shows an EMF of 0.290 V at 298 K. Given E deg(Cu^2+/Cu) = 0.34 V and 2.303RT/F = 0.0591 V, the value of x in millimol per litre is (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ec085

The cell $Pt ∣ H X_{2} (1 bar) ∣ H X^{+} (1 M) ∣∣ Cu X^{2 +} (x M) ∣ Cu$ shows an EMF of $0.290 V$ at 298 K. Given $E^{\circ} (Cu X^{2 +} / Cu) = 0.34 V$ and $2.303 RT / F = 0.0591 V$ , the value of $x$ in millimol per litre is (nearest integer).

Numerical answer type, answer in mM, tolerance 1.

The left electrode is the SHE, so

E_{cell} = E (Cu X^{2 +} / Cu) = 0.34 + \frac{0.0591}{2} lo g x

. Then

lo g x = \frac{0.290 - 0.34}{0.02955} = - 1.692

x = 1 0^{- 1.692} = 0.0203 M = 20.3 mM \approx 20 mM

. The EMF is below

E^{\circ}

because

[Cu X^{2 +}] < 1 M