Calculate the ΔG and equilibrium constant of the reaction at 27°CMg+

1.	Calculate the ΔG and equilibrium constant of the reaction at 27°CMg+ Cu2+ at equilibrium Mg2+ +CuE°Mg2+/Mg= -2.37V, E°Cu2+/Cu=+0.34V
Answer» $\Delta G$ Answer: = - 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 $\Delta G=-2.303RTlog K$ $-523030=-2.303\times 8.314\times 300\times logK$ $K=1.12\times 10^{91}$