At $25$ $^{o}$ C, the standard e.m.f. of cell having reactions involving two electron change is found to be $0.295 V$ . The equilibrium constant of the reaction is:

29.5 \times 10^{- 2}

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10

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10^{10}

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29.5 \times 10^{10}

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Solution

The correct option is C $10^{10}$
As we know,
At eqm, $E_{c e l l} = 0$ & given that $E_{c e l l}^{o} = 0.295 V$

From Nearst equation,
$E_{c e l l}^{o} = \frac{0.059}{n} l o g K_{c}$

$l o g K_{c} = n E^{o} / 0.0591$
$= 2 \times 0.295 / 0.0591 \approx 10$

$K_{c} = 10^{10}$
Option C is correct.

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At 25oC, the standard e.m.f. of cell having reactions involving two electron change is found to be 0.295V. The equilibrium constant of the reaction is:

The correct option is C 1010As we know,At eqm, Ecell=0 & given that Eocell=0.295VFrom Nearst equation,Eocell=0.059nlogKclogKc=nEo/0.0591 =2×0.295/0.0591≈10Kc=1010Option C is correct.

At $25$ $^{o}$ C, the standard e.m.f. of cell having reactions involving two electron change is found to be $0.295 V$ . The equilibrium constant of the reaction is:

The correct option is C $10^{10}$
As we know,
At eqm, $E_{c e l l} = 0$ & given that $E_{c e l l}^{o} = 0.295 V$

From Nearst equation,
$E_{c e l l}^{o} = \frac{0.059}{n} l o g K_{c}$

$l o g K_{c} = n E^{o} / 0.0591$
$= 2 \times 0.295 / 0.0591 \approx 10$

$K_{c} = 10^{10}$
Option C is correct.