The equilibrium constant of the reaction;
$C u (s) + 2 {A g}^{+} (a q) ⟶ {C u}^{2 +} (a q) + 2 A g (s)$
$E^{o} = 0.46 V$ at $298 K$

2.4 \times 10^{10}

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2.0 \times 10^{10}

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4.0 \times 10^{10}

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4.0 \times 10^{15}

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Solution

The correct option is D $4.0 \times 10^{15}$
$C u (s) + 2 {A g}^{2 +} (a q) ⟶ {C u}^{2 +} (a q) + 2 A g (s)$
$E^{o} = 0.46 V$ at $298 K$
$log K_{C} = \frac{n F E^{o}}{R T}$
$= \frac{2 \times 0.46}{0.059}$
$K_{C} = 4 \times 10^{15}$

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Similar questions

The equilibrium constant of the reaction:

$C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s); E^{\circ} = 0.46 V$ at 298K is :

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

E^{\circ} = 0.46 V

298 K

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

K_{c}

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

10 \times 10^{15}

E_{c e l l}^{o}

298 K

[2.303 \frac{R T}{F} a t 298 K = 0.059 V]

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

(K)

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

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