Given the half cell reactions- Cu - aq - e - rightarrow Cu s - E - 0-0-52 - V lCu -2- aq - e - -rightarrow -Cu - aq - E - 0-0-165 - V The equilibrium constant for the disproportionation reaction at 298 K is-2 Cu - aq - Cu s - Cu 2- aq A- 106B- 10-6C- 104D- 108

The correct option is A 10^62Cu^+(aq) → Cu(s) + Cu^2+(aq) For this reaction E^0 = 0.52 0.165 = 0.355 V log K_eq = nE^00.059 = 1× 0.3550.059 = 6.01≈ 6 ∴ K_eq= ...

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Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

Given the hal...

Question

Given the half cell reactions:
$C u^{+} (a q) + e^{-} \to C u (s); E^{0} = + 0.52 V C u^{2 +} (a q) + e^{-} \to C u^{+} (a q); E^{0} = + 0.165 V$
The equilibrium constant for the disproportionation reaction at 298 K is:
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$

10^{4}

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10^{8}

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10^{6}

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10^{- 6}

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Solution

The correct option is C $10^{6}$
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$
For this reaction
$E^{0} = 0.52 - 0.165 = 0.355 V log K_{e q} = \frac{n E^{0}}{0.059} = \frac{1 \times 0.355}{0.059} = 6.01 \approx 6 ∴ K_{e q} = 10^{6}$

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Similar questions

Q. Consider the following half cells

C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}; E^{o} = - 0.325 V

C u_{(a q)}^{+} + e^{-} \to C u_{(s)}; E^{o} = 0.50 V

The

E^{o}

for the reaction

2 C u_{(a q)}^{+} \to C u_{(a q)}^{2 +} + C u_{(s)}

Q. Consider the following half cells

C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}; E^{o} = - 0.325 V

C u_{(a q)}^{+} + e^{-} \to C u_{(s)}; E^{o} = 0.50 V

The

E^{o}

for the reaction

2 C u_{(a q)}^{+} \to C u_{(a q)}^{2 +} + C u_{(s)}

$E^{0}$ for the cell, $Z n | Z n {^{2 +}}_{(a q)} | | C u {^{2 +}}_{(a q)} | C u$ is $1.10 V$ at $25^{0} C$ . The equilibrium constant for $Z n + C u {^{2 +}}_{(a q)} \to Z n {^{2 +}}_{(a q)} + C u$ the cell reaction is of the order of :

Q. 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

298 K

is:

Q. The electrode potentials for:

C u^{2 +} (a q) + e^{-} \to C u^{+} (a q)

C u^{+} (a q) + e^{-} \to C u (s)

are + 0.15 V and 0.50 V respectively. The value of

E^{0} c u^{2} / C u

will be :

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Given the half cell reactions: Cu+(aq)+e−→Cu(s);E0=+0.52 VCu2+(aq)+e−→Cu+(aq);E0=+0.165 V The equilibrium constant for the disproportionation reaction at 298 K is: 2Cu+(aq)→Cu(s)+Cu2+(aq)

The correct option is C 1062Cu+(aq)→Cu(s)+Cu2+(aq) For this reaction E0=0.52−0.165=0.355 VlogKeq=nE00.059=1×0.3550.059=6.01≈ 6∴Keq=106

Given the half cell reactions:
$C u^{+} (a q) + e^{-} \to C u (s); E^{0} = + 0.52 V C u^{2 +} (a q) + e^{-} \to C u^{+} (a q); E^{0} = + 0.165 V$
The equilibrium constant for the disproportionation reaction at 298 K is:
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$

The correct option is C $10^{6}$
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$
For this reaction
$E^{0} = 0.52 - 0.165 = 0.355 V log K_{e q} = \frac{n E^{0}}{0.059} = \frac{1 \times 0.355}{0.059} = 6.01 \approx 6 ∴ K_{e q} = 10^{6}$