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

Answer :

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 :

$\Delta G^o=-nRT\times \ln K$

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 = $7.0\times 10^4$

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

$\Delta G^o=-(1mole)\times (8.314J/mole.K)\times (298K)\times \ln (7.0\times 10^{4})$

$\Delta G^o=-27640.414J$

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.

$-27640.414J=-(1\times 96500\times E^o)$

$E^o=0.29V$

The standard cell potential is, 0.29 V