The equilibrium constant of the reaction:

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

$E^{\circ} = 0.46 V$ at $298 K$ is:

2.0 \times 10^{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.0 \times 10^{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.0 \times 10^{15}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.4 \times 10^{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $4.0 \times 10^{15}$
$K = a n t i l o g [\frac{n E^{\circ}}{0.059}] = a n t i l o g [\frac{2 \times 0.46}{0.059}]$

$= a n t i l o g 15.593$

$= 3.9 \times 10^{15}$

$\approx 4 \times 10^{15}$ .

Hence, the correct option is $C$

Suggest Corrections

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^{o} = 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

Join BYJU'S Learning Program