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 $1.13\times 10^{-19}$ and -108080 J respectively.

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:

$\Delta G^o=-2\times 96500\times 0.56=-108080J$

• 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:

$-108080J=8.314J/mol.K\times 298K\times \ln K_{eq}\\\\K_{eq}=1.134\times 10^{-19}$

Hence, the $K_{eq}$ and $\Delta G^o$ of the reaction is $1.13\times 10^{-19}$ and -108080 J respectively.