Question

Calculate the ΔG and equilibrium constant of the reaction at 27°C
Mg+ Cu2+ at equilibrium Mg2+ +Cu
E°Mg2+/Mg= -2.37V, E°Cu2+/Cu=+0.34V

Answer 1

Answer: $\Delta G$ = - 523030 Joules

equilibrium constant = $1.12\times 10^{91}$

Explanation:

$Mg+Cu^{2+}\rightarrow Mg^{2+}+Cu$

Here Mg undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

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

$E^0_{[Mg^{2+}/Mg]}= -2.37V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Mg^{2+}/Mg]}$

$E^0=+0.34- (-2.37V)=2.71$

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

$\Delta G$ = gibbs free energy

n= no of electrons gained or lost

F= faraday's constant

$E^0$ = standard emf

$\Delta G=-2\times 96500\times (2.71)=-523030J$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =$27^0C=27+273=300K$

K = equilibrium constant

$\Delta G=-2.303RTlog K$

$-523030=-2.303\times 8.314\times 300\times logK$

$K=1.12\times 10^{91}$