Question

Is the following reaction spontaneous at 50°C? Ag/Ag'(0.0078 M)||Ni2 (0.023 M) Ni

Answer 1

Answer: The reaction is non spontaneous at
`50^0C`

Step-by-step explanation:

`E_(cell)`= +ve, reaction is spontaneous

`E_(cell)`= -ve, reaction is non spontaneous

`E_(cell)`= 0, reaction is in equilibrium

`Ag/Ag^(+)(0.0078M)//Ni^(2+)(0.023M)/Ni`

Using Nernst equation :

`E_(cell)=E^o_(cell)-(2.303RT)/(nF)\log ([Ag^(+)])/([Ni^(2+)]^2)`

where,

R = gas constant =
`8.314J/Kmol`

T= temperature in kelvin =
`50^0C=(5+273)K=323K`

n = number of electrons in oxidation-reduction reaction = 2

F= Faraday's constant = 96500 C

`E^o_(cell)` = standard electrode potential =
`E^0_(cathode)- E^0_(anode)`

Where both
`E^0` are standard reduction potentials.

`E^0_([Ag^(+)/Ag])= +0.80V`

`E^0_([Ni^(2+)/Ni])= -0.23V`

`E^0=E^0_([Ni^(2+)/Ni])- E^0_([Ag^(+)/Ag])`

`E^0=-0.23-(+0.80)=-1.03V`

`E_(cell)=-1.03 - (2.303* 8.314* 323)/(2* 96500)\log ([0.0078])/([0.023]^2)`

`E_(cell)=-1.067V`

Thus as
`E_(cell)` is negative , the reaction is no spontaneous.