$C u (s) + 2 {A g}^{+} (a q) \to {C u}^{2 +} (A q) + 2 A g (s)$

The equilibrium constant $(K)$ for the above reaction will be:
[ Given, $E_{c e l l}^{\circ} = 0.46 V$ ]

Antilog 15.6

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Antilog 2.5

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Antilog 1.5

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Antilog 12.2

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Solution

The correct option is B Antilog 15.6
$E^{\circ} = \frac{0.059}{n} log K_{c}$

$0.46 = \frac{0.059}{2} log K_{c}$

$log K_{c} = \frac{0.46 \times 2}{0.059} = 15.6$

$K_{c} = antilog 15.6$

Hence, the correct option is $A$

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Similar questions

Calculate the equilibrium constant of the reaction.
$C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$ , $E_{c e l l}^{⊖} = 0.46 V$

C u (s) + 2 A g^{+} (a q .) \to C u^{2 +} (a q .) + 2 A g (s)

E_{c e l l}^{\circ} = 0.46 V

C u^{2 +}, E_{c e l l}^{\circ}

C u (s) + 2 {A g}^{+} (a q) ⟶ {C u}^{2 +} (a q) + 2 A g (s)

E^{o} = 0.46 V

298 K

C u (s) + 2 A g^{+} (a q) ⇌ C u^{2 +} (a q) + 2 A g (s)

E^{\circ} = 0.46 V

298 K

K_{c}

C u (s) + 2 A g^{+} (a q .) ⇌ C u^{2 +} (a q .) + 2 A g (s) (E_{c e l l}^{\circ} = 0.46 V)

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