What would be the standard cell potential, if equilibrium constant of cell reaction is $1.9413 \times 10^{37}$ at $25^{o} C$ ?

Open in App

Solution

We know,
$△ G = - R T l n K_{f} = - n F E^{0}$
$l o g K_{f} = \frac{n F \times E^{0}}{R T \times 2.303}$
$l o g (1.9413 \times 10^{37}) = \frac{n}{0.059} \times E^{0}$
$n = 1$
$37.28 \times 0.059 = E^{0}$
$E^{0} = 2.19 V$

Suggest Corrections

Similar questions

Q. What would be the stands cell potntial, if equilibrium constant of cell reactiom is

1.9413 \times 10^{37}

25^{o} C

temperature?

(n = 2)

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

25^{o} C

. The equilibrium constant of the reaction would be
(Given,

F = 96500 C m o l^{- 1}, R = 8. 314 J K^{- 1} m o l^{- 1}

)

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?

Q. For the cell reaction:

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

, the standard cell potential is

0.36 V .

The equilibrium constant for the reaction is?

Join BYJU'S Learning Program

What would be the standard cell potential, if equilibrium constant of cell reaction is 1.9413×1037 at 25 oC?

We know,△G=−RTlnKf=−nFE0logKf=nF×E0RT×2.303log(1.9413×1037)=n0.059×E0n=137.28×0.059=E0E0=2.19V

What would be the standard cell potential, if equilibrium constant of cell reaction is $1.9413 \times 10^{37}$ at $25^{o} C$ ?

We know,
$△ G = - R T l n K_{f} = - n F E^{0}$
$l o g K_{f} = \frac{n F \times E^{0}}{R T \times 2.303}$
$l o g (1.9413 \times 10^{37}) = \frac{n}{0.059} \times E^{0}$
$n = 1$
$37.28 \times 0.059 = E^{0}$
$E^{0} = 2.19 V$