For a cell reaction involving two electrons- Ecell 0-0-3 V at 25- The equilubrium constant of the reaction is-A- 10-10B- 3 - 10-2C- 10D- 1010

The correct option is D 10^10At 25^∘ C (i.e. 298 K) ∵ E^∘=0.0591/nlog K or log K=E^∘×n/0.059=0.3×2/0.059 log K≈10 K=10^10 ...

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Standard XII

Chemistry

EMF

For a cell re...

Question

For a cell reaction involving two electrons, $E_{cell}^{0} = 0.3 V$ at $25^{\circ}$ . The equilubrium constant of the reaction is:

10^{- 10}

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3 \times 10^{- 2}

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10

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10^{10}

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Solution

The correct option is D $10^{10}$
At $25^{\circ} C$ (i.e. 298 K)
$∵ E^{\circ} = \frac{0.0591}{n} log K$
or $log K = E^{\circ} \times \frac{n}{0.059} = 0.3 \times \frac{2}{0.059}$
$log K \approx 10$
$K = 10^{10}$

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For a cell reaction involving two electrons, E0cell=0.3 V at 25∘. The equilubrium constant of the reaction is:

The correct option is D 1010At 25∘C (i.e. 298 K) ∵ E∘=0.0591nlogK or logK=E∘×n0.059=0.3×20.059 logK≈10 K=1010

For a cell reaction involving two electrons, $E_{cell}^{0} = 0.3 V$ at $25^{\circ}$ . The equilubrium constant of the reaction is:

The correct option is D $10^{10}$
At $25^{\circ} C$ (i.e. 298 K)
$∵ E^{\circ} = \frac{0.0591}{n} log K$
or $log K = E^{\circ} \times \frac{n}{0.059} = 0.3 \times \frac{2}{0.059}$
$log K \approx 10$
$K = 10^{10}$