Question

Determine the direction in which the following reaction is spontaneous at 25oC: Mg2+(aq) + K(s) <---> Mg(s) + K+(aq) Determine the equilibrium constant K and Go using the cell potential for this reaction and which will be the anode and cathode?

Answer 1

Answer: The
`K_(eq)` and
`\Delta G^o` of the reaction is
`1.13* 10^(-19)` and -108080 J respectively.

Step-by-step explanation:

For the given cell reaction:

`Mg^(2+)(aq.)+K(s)\rightarrow Mg(s)+K^+(aq.)`

The half reaction follows:

Oxidation half reaction:
`K(s)\rightarrow K^++e^-;E^o_(K^+/K)=-2.93V`

Reduction half reaction:
`Mg^(2+)+2e^-\rightarrow Mg(s);E^o_(Mg^(2+)/Mg)=-2.37V`

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)=-2.37-(-2.93)=0.56V`

• To calculate Gibbs free energy, we use the equation:

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

where,

`\Delta G^o` = Standard Gibbs free energy of the reaction = ?

n = number of electrons exchanged = 2

F = Faraday's constant = 96500

`E^o_(cell)` = standard electrode potential of the cell = 0.56 V

Putting values in above equation, we get:

`\Delta G^o=-2* 96500* 0.56=-108080J`

• To calculate the equilibrium constant of the reaction, we use the equation:

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

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:

`-108080J=8.314J/mol.K* 298K* \ln K_(eq)\\\\K_(eq)=1.134* 10^(-19)`

Hence, the
`K_(eq)` and
`\Delta G^o` of the reaction is
`1.13* 10^(-19)` and -108080 J respectively.