Consider the following equation for a cell
$A + B ⇌ C + D; E^{o} = X v o l t, K_{e q} = K_{1}$
$2 A + 2 B ⇌ 2 C + 2 D; E^{o} = y v o l t, K_{e q} = k_{2}$

X = y, K_{1} = K_{2}

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X = 2 y, {K_{1}}^{2} = K_{2}

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X = y, {K_{1}}^{2} = K_{2}

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X^{2} = y, {K_{1}}^{2} = K_{2}

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Solution

The correct option is C $X = y, {K_{1}}^{2} = K_{2}$
We have to know that $E_{c e l l}^{0}$ is an intensive property and is constant for particular cell. It doesn't depend on stoichiometric co-efficients of the reaction.So, for the given reaction $X = y$
But, $K_{e q}$ varies with varying of stoichiometric co-efficients.
$K_{e q} = \frac{[P r o d u c t s]^{b}}{[R e a c t a n t s]^{a}}$ where a and b are stoichiometric co-efficients.
For first reaction, $K_{e q} = \frac{[C] [D]}{[A] [B]} = K_{1}$
For second reaction, $K_{e q} = \frac{[C]^{2} [D]^{2}}{[A]^{2} [B]^{2}} = K_{2}$
Now, we get $K_{2} = K_{1}^{2}$
Hence, option C is correct answer.

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Similar questions

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Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
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