The equilibrium constant for the equation: $E^{0} = 0.46 V$ at 298 K is:
$C u + 2 A g^{+} (a q) ⟶ {C u}^{2 +} (a q) + 2 A g$

2.0 \times 10^{10}

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3.9 \times 10^{15}

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4.0 \times 10^{10}

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2.4 \times 10^{10}

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Solution

The correct option is B $3.9 \times 10^{15}$
For equilibrium , $Δ G^{0} = - R T l n K_{e q}$
But $Δ G^{0} = - n F E^{0} = - 2 \times 96500 \times 0.46$
$Δ G^{0} = - 88780$
$l n K_{e q}$ = $\frac{88780}{8.314 \times 298}$
$l n K_{e q} = 35.833$
$K_{e q} = 3.9 \times 10^{15}$

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

(K)

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

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

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

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