Question

The thermodynamic properties for the species involved in the Zn(s)|Zn²⁺(aq)||Pb²⁺(aq)|Pb(s) cell are given below.

Substance	∆G°f kJ/mol	S° J/K*mol	∆H°f kJ/mol
Zn(s)	0	41.63	0
Pb(s)	0	64.81	0
Zn2+(aq)	-147.06	-112.1	-153.89
Pb2+(aq)	-24.43	10.5	-1.7

Calculate ∆G°, ∆S° and ∆H° for the chemical reaction occuring in the cell.Is the reaction spontaneous? Is this a favorable entropy change? Favorable energy change?

Answer 1

Zn(s) + Pb²⁺(aq) <======>Zn²⁺(aq) + Pb(s)

Substance	∆G°f kJ/mol	S° J/K*mol	∆H°f kJ/mol
Zn(s)	0	41.63	0
Pb(s)	0	64.81	0
Zn2+(aq)	-147.06	-112.1	-153.89
Pb2+(aq)	-24.43	10.5	-1.7

∆G° = ∆G°(product) - ∆G°(reactent)

= ∆G°(Zn²⁺) - ∆G°(Pb²⁺)

= -147.06 - (-24.43)

= -122.63kJ/mol.

∆H° = ∆H°(product) - ∆H°(reactent)

= ∆H°(Zn²⁺) - ∆H°(Pb²⁺)

= -153.89 - (-1.7)

= -152.19kJ/mol.

∆S° = S°(product) - S°(reactent)

= S°(Zn²⁺) + S°(Pb) - S°(Pb²⁺) - S°(Zn)

= -112.1 + 64.81 - 10.5 - 41.63

= -99.42J/K*mol.

the reaction Is spontaneous as ∆G° is -ve.

this is not a favorable entropy change as ∆S° is -ve.