For a cell reaction involving a $C u / C u^{2} / / A g^{+} / / A g$ . The standard Emf of the cell is found to be 0.46 $V$ at 298 k . Calculate the equilibrium constant of the reaction at $25^{0} C$ .

10^{15 / 6}

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10^{26 / 3}

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10^{5 / 2}

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10^{46 / 3}

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Solution

The correct option is D $10^{46 / 3}$
$E_{\infty 1}^{\cdot} = \frac{0.06}{n} l o g K_{c}$

$46 / 3 = l o g K_{c}$

$K_{c} = 10^{46 / 3}$

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Similar questions

Q. The standard e.m.f. of a galvanic cell involving cell reaction with n=2 is found to be 0.295 V at

25^{0} C

. The equilibrium constant of the reaction would be?

For a cell reaction involving a two-electron change, the standard emf of the cell is found to be $0.295 V$ at $25^{o} C$ . The equilibrium constant of the reaction at $25^{o} C$ will be :

Q. At

25^{\circ} C

, the standard emf of a cell having reaction involving two electrons change is found to be

0.295 V

. The equilibrium constant of the reaction is_____________.

Q. At

298 K

, given that:

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s)

E_{c e l l}^{o} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s)

E_{c e l l}^{o} = 1.10 V

Then, the

E_{c e l l}

for the following reaction at

298 K

will be:

Z n (s) | Z n^{2 +} (0.1 M) | | A g^{+} (1. 0 M) | A g (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 :

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For a cell reaction involving a Cu/Cu2//Ag+//Ag. The standard Emf of the cell is found to be 0.46 V at 298 k . Calculate the equilibrium constant of the reaction at 250C.

The correct option is D 1046/3E⋅∞1=0.06nlogKc 46/3=log Kc Kc=1046/3

For a cell reaction involving a $C u / C u^{2} / / A g^{+} / / A g$ . The standard Emf of the cell is found to be 0.46 $V$ at 298 k . Calculate the equilibrium constant of the reaction at $25^{0} C$ .

The correct option is D $10^{46 / 3}$
$E_{\infty 1}^{\cdot} = \frac{0.06}{n} l o g K_{c}$

$46 / 3 = l o g K_{c}$

$K_{c} = 10^{46 / 3}$