Question

Be sure to answer all parts. What are E and AG of a redox reaction at 25°C for which n=1 and K= 7.0 x 104? AG° = OJ

Answer 1

The standard Gibbs free energy is, -27640.414 J

The standard cell potential is, 0.29 V

Explanation :

First we have to calculate the
`\Delta G_^o`.

Formula used :

`\Delta G^o=-nRT* \ln K`

where,

`\Delta G^o` = standard Gibbs free energy = ?

R = gas constant = 8.314 J/K.mole

n = number of moles = 1 mole

T = temperature =
`25^oC=273+25=298K`

k = equilibrium constant =
`7.0* 10^4`

Now put all the given values in this formula, we get:

`\Delta G^o=-(1mole)* (8.314J/mole.K)* (298K)* \ln (7.0* 10^(4))`

`\Delta G^o=-27640.414J`

The standard Gibbs free energy is, -27640.414 J

Now we have to calculate the standard cell potential.

Formula used :

`\Delta G^o=-nFE^o`

where,

`\Delta G^o` = Gibbs free energy = -27640.414 J

n = number of electrons = 1

F = Faraday constant = 96500 C/mole

`E^o` = standard e.m.f of cell = ?

Now put all the given values in this formula, we get the Gibbs free energy.

`-27640.414J=-(1* 96500* E^o)`

`E^o=0.29V`

The standard cell potential is, 0.29 V