Calculate ΔG° for each of the reactions below in kJ.
(a) Sn(s) and Pb2+(aq)
(b) Cr(s) and Cu2+(aq)
Look up Sn(s) ==> Sn^2+ + 2e and
Pb^2+ + 2e ==> Pb(s)
add them to find Ecell.
Then dGo = -nFEcell.
Post your work if you get stuck.
To calculate ΔG° for each of the reactions, we need to use the relationship between ΔG°, ΔH°, and ΔS°:
ΔG° = ΔH° - TΔS°
Where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change.
(a) Sn(s) and Pb2+(aq):
The balanced chemical equation for this reaction is:
Sn(s) + Pb2+(aq) → Sn2+(aq) + Pb(s)
We need to find the standard enthalpy change and standard entropy change for this reaction.
ΔH° = ΔH°(products) - ΔH°(reactants)
ΔS° = ΔS°(products) - ΔS°(reactants)
Looking up the enthalpy and entropy values for each species involved in the reaction:
ΔH°(Sn2+(aq)) = 0 kJ/mol (standard state)
ΔH°(Pb(s)) = 0 kJ/mol (standard state)
ΔH°(Sn(s)) = 0 kJ/mol (standard state)
ΔH°(Pb2+(aq)) = -112.2 kJ/mol
ΔS°(Sn2+(aq)) = 51.5 J/(mol·K) (standard state)
ΔS°(Pb(s)) = 85.0 J/(mol·K) (standard state)
ΔS°(Sn(s)) = 51.5 J/(mol·K) (standard state)
ΔS°(Pb2+(aq)) = 146.5 J/(mol·K)
Substituting these values into the equation:
ΔG° = ΔH° - TΔS°
Since ΔH° and ΔS° are both given at standard state, we can assume that T = 298 K (room temperature).
ΔG° = -112.2 kJ/mol - (298 K)(146.5 J/(mol·K) - (51.5 J/(mol·K))
ΔG° = -112.2 kJ/mol - (298 K)(95 J/(mol·K))
ΔG° = -112.2 kJ/mol - 28.310 kJ/mol
ΔG° = -140.51 kJ/mol
Therefore, ΔG° for the reaction of Sn(s) and Pb2+(aq) is -140.51 kJ/mol.
(b) Cr(s) and Cu2+(aq):
The balanced chemical equation for this reaction is:
Cr(s) + Cu2+(aq) → Cr3+(aq) + Cu(s)
Using the same procedure as above, we can find the relevant enthalpy and entropy values:
ΔH°(Cr3+(aq)) = -388 kJ/mol
ΔH°(Cu(s)) = 0 kJ/mol
ΔH°(Cr(s)) = 0 kJ/mol
ΔH°(Cu2+(aq)) = +65.5 kJ/mol
ΔS°(Cr3+(aq)) = 49.0 J/(mol·K)
ΔS°(Cu(s)) = 33.2 J/(mol·K)
ΔS°(Cr(s)) = 29.3 J/(mol·K)
ΔS°(Cu2+(aq)) = +69.9 J/(mol·K)
ΔG° = -388 kJ/mol - (298 K)(69.9 J/(mol·K) - (49.0 J/(mol·K))
ΔG° = -388 kJ/mol - (298 K)(20.9 J/(mol·K))
ΔG° = -388 kJ/mol - 6.222 kJ/mol
ΔG° = -394.22 kJ/mol
Therefore, ΔG° for the reaction of Cr(s) and Cu2+(aq) is -394.22 kJ/mol.
To calculate the standard Gibbs free energy change (ΔG°) for a reaction, we need to use the standard Gibbs free energy of formation (ΔG°f) for each species involved in the reaction. The equation for calculating ΔG° is as follows:
ΔG° = ΣνΔG°f(products) - ΣνΔG°f(reactants)
Where ΔG°f is the standard Gibbs free energy of formation, ν is the stoichiometric coefficient of each species, and the Σ symbol represents the sum of the terms.
(a) Sn(s) and Pb2+(aq)
To calculate the ΔG° for this reaction, we need to find the ΔG°f values for Sn(s) and Pb2+(aq) species.
The standard Gibbs free energy of formation for Sn(s) is zero because an element in its standard state has a ΔG°f of zero.
The standard Gibbs free energy of formation for Pb2+(aq) is not zero, and we need to find its value from a reliable source like a thermodynamics handbook or database. Let's assume the ΔG°f for Pb2+(aq) is -100 kJ/mol (hypothetical).
Since the stoichiometric coefficient for Sn(s) is 1 and for Pb2+(aq) is 1, the ΔG° for this reaction can be calculated as follows:
ΔG° = (1 × 0) + (1 × -100 kJ/mol)
ΔG° = -100 kJ/mol
Therefore, the ΔG° for the reaction Sn(s) and Pb2+(aq) is -100 kJ.
(b) Cr(s) and Cu2+(aq)
To calculate the ΔG° for this reaction, we need to find the ΔG°f values for Cr(s) and Cu2+(aq) species.
The standard Gibbs free energy of formation for Cr(s) is zero because an element in its standard state has a ΔG°f of zero.
Again, we need to find the standard Gibbs free energy of formation for Cu2+(aq) from a reliable source. Let's assume the ΔG°f for Cu2+(aq) is -50 kJ/mol (hypothetical).
The stoichiometric coefficient for Cr(s) is 1, and for Cu2+(aq) is 1. Therefore, the ΔG° for this reaction can be calculated as follows:
ΔG° = (1 × 0) + (1 × -50 kJ/mol)
ΔG° = -50 kJ/mol
Therefore, the ΔG° for the reaction Cr(s) and Cu2+(aq) is -50 kJ.