For the given data
$A + B ⇌ C + D; E^{o} = a v o l t; K_{c} = K_{1}$
$2 A + 2 B ⇌ 2 C + 2 D; E^{o} = b v o l t; K_{c} = K_{2}$
Select the correct choices:

a = b

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K_{2} = K_{1}^{2}

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a \neq b

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b = a^{2}

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Solution

The correct options are
A $a = b$
B $K_{2} = K_{1}^{2}$
When a cell reaction is multiplied by 2, the value of the cell potential remains constant. It is not multiplied by 2. Hence, $a = b$ .
Thus, the option A is correct and the options C and D are incorrect.
When a cell reaction is multiplied by 2, the value of the equilibrium constant of the resulting reaction is the square of the value of the equilibrium constant of the original reaction.
Thus, $K_{2} = K_{1}^{2}$ . Thus, the option B is correct.

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Similar questions

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}

A + B ⇌ C + D

E^{o} = x

K_{e q} = K_{1}

2 A + 2 B ⇌ 2 C + 2 D, E^{o} = y

K_{e q} = K_{2}

If $A + B ⇌ C + D$ ; $K_{C} = K_{1}$ and $E^{⊖} = a V$ ; $2 A + 2 B ⇌ 2 C + 2 D$ ; $K_{C} = K_{2}$ and $E^{⊖} = b V$ then:

A + B \to C + D; E^{o} = x v o l t, Δ G = Δ G_{1}

2 A + 2 B \to 2 C + 2 D; E^{o} = x v o l t, Δ G = Δ G_{2}

2 A + 2 B ⇌ 2 C + 2 D

K_{c} = 1 / 6

K_{c}

C + D ⇌ A + B

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