Question

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Answer 1

Answer: The
`K_(eq)` of the reaction is
`1.73* 10^(16)`

Step-by-step explanation:

For the given half reactions:

Oxidation half reaction:
`Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(Zn^(2+)/Zn)=-0.76V`

Reduction half reaction:
`Co^(2+)+2e^-\rightarrow Co(s);E^o_(Co^(2+)/Co)=-0.28V`

Net reaction:
`Zn(s)+Co^(2+)\rightarrow Zn^(2+)+Co(s)`

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(cathode)-E^o_(anode)`

Putting values in above equation, we get:

`E^o_(cell)=-0.28-(-0.76)=0.48V`

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

`\Delta G^o=-nfE^o_(cell)`

and,

`\Delta G^o=-RT\ln K_(eq)`

Equating these two equations, we get:

`nfE^o_(cell)=RT\ln K_(eq)`

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

`E^o_(cell)` = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction =
`25^oC=[273+25]=298K`

`K_(eq)` = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

`2* 96500* 0.48=8.314* 298* \ln K_(eq)\\\\K_(eq)=1.73* 10^(16)`

Hence, the
`K_(eq)` of the reaction is
`1.73* 10^(16)`