The standard e.m.f. of a cell involving one electron change is found to 0.591 V at 25 $^{o}$ C. The equilibrium constant of the reaction is:

10^{30}

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10^{5}

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10^{10}

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10^{1}

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Solution

The correct option is C $10^{10}$
According to Nernst equation,
$E = E^{o} - 0.0591 l o g (Q / n)$ ,
At eqm, $E = 0$
$E_{c e l l}^{o} = \frac{0.0591}{n} l o g K_{c}$
$∴ 0.591 = \frac{0.0591}{1} l o g K_{c}$
$K_{c} = 10^{10}$

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Similar questions

The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at $25^{0}$ C. The equilibrium constant of the reaction is (F = 96500Cmol $^{- 1}$ , R = 8.314JK $^{- 1}$ mol $^{- 1}$ ) :

Q. The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at

25^{o} C

. The equilibrium constant of the reaction is

(F = 96, 500

Cmol

^{- 1}, R = 8.314 J K^{- 1}

mol

^{- 1})

Q. The standard emf of a cell involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is

[Given:

F = 96500 C m o l^{- 1}

]

Q. 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:

Q. The standard emf of a cell, involving one electron change is found to be

0.591 V

25

. The equilibrium constant of the reaction is

(F = 96500 C {m o l}^{- 1}, R = 8.314 K^{- 1} {m o l}^{- 1})

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The standard e.m.f. of a cell involving one electron change is found to 0.591 V at 25oC. The equilibrium constant of the reaction is:

The correct option is C 1010According to Nernst equation,E=Eo−0.0591log(Q/n),At eqm, E=0Eocell=0.0591nlogKc∴0.591=0.05911logKcKc=1010

The standard e.m.f. of a cell involving one electron change is found to 0.591 V at 25 $^{o}$ C. The equilibrium constant of the reaction is:

The correct option is C $10^{10}$
According to Nernst equation,
$E = E^{o} - 0.0591 l o g (Q / n)$ ,
At eqm, $E = 0$
$E_{c e l l}^{o} = \frac{0.0591}{n} l o g K_{c}$
$∴ 0.591 = \frac{0.0591}{1} l o g K_{c}$
$K_{c} = 10^{10}$