The E - for the cell reaction- Cu S -2 Ag - aq - Cu 2 aq -2 Ag S is 0-46 V- What is its equilibrium constant-A- 15-6B- 4 - 1016- 4 - 1015D- 1-56 - 1015

The correct option is C 4 × 10^15E^0_cell=0.059/2log K_c Or log K_c=0.46 × 2/0.059=15.6 ∴ K_c=4 × 10^15 ...

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Gibbs Free Energy & Spontaneity

The E ∘ for t...

Question

The $E^{\circ}$ for the cell reaction: $C u (S) + 2 A g^{+} (a q) \to C u^{2} (a q) + 2 A g (S)$ is 0.46V. What is its equilibrium constant?

15.6

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4 \times 10^{16}

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4 \times 10^{15}

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1.56 \times 10^{15}

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Solution

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Similar questions

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

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

E^{0} = 0.46 V

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

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