The cell Pt | H2 (1 bar) | H+ (pH = x) || H+ (1 M) | H2 (1 bar) | Pt has EMF 0.236 V at 298 K. The value of x (nearest integer) is.

JEE Main levelNumerical answerDifficulty 2 of 3ec22

The cell $Pt ∣ H X_{2} (1 bar) ∣ H X^{+} (pH = x) ∣∣ H X^{+} (1 M) ∣ H X_{2} (1 bar) ∣ Pt$ has EMF $0.236 V$ at 298 K. The value of $x$ (nearest integer) is.

Numerical answer type.

Hydrogen concentration cell,

n = 1

E = 0.0591 lo g \frac{[ H X ^{+} ] _{right}}{[ H X ^{+} ] _{left}} = 0.0591 lo g \frac{1}{1 0 ^{- x}} = 0.0591 x

x = 0.236/0.0591 = 3.99 \approx 4

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