Let us examine three galvanic cells: Pt(s) | H2(g) | HCl(aq) | Cl2(g) | Pt(s) Pb(s) | PbCl2(s) | HCl — Organic Chemistry Chemistry Question
THEORETICAL PROBLEM 24
Let us examine three galvanic cells:
Pt(s) | H2(g) | HCl(aq) | Cl2(g) | Pt(s)
Pb(s) | PbCl2(s) | HCl(aq) | H2(g) | Pt(s)
Pb(s) | PbSO4(s) | K2SO4(aq) || KNO3(aq) || KCl(aq) | PbCl2(s) | Pb(s)
Thermochemical data at 25 °C:
[VISUAL]
Species | ΔfHº (kJ mol–1) | Sº (J mol–1 K–1)
Cl2(g) | 0.0 | 223.1
H2(g) | 0.0 | 130.7
HCl(aq) | –167.2 | 56.5
K2SO4(aq) | –1414.0 | 225.1
KCl(aq) | –419.5 | 159.0
Pb(s) | 0.0 | 26.4
PbCl2(s) | –359.4 | 136.0
PbSO4(s) | –920.0 | 148.5
Write down the equation for the cell reactions.
Model Answer
H2(g) + Cl2(g) 2 HCl(aq)
Pb(s) + 2 HCl(aq) H2(g) + PbCl2(s)
PbCl2(s) + K2SO4(aq) PbSO4(s+ 2 KCl(aq)
Estimate the standard cell reaction potential of the galvanic cells at 25 °C based on thermochemical data.
Model Answer
For the first cell:
ΔrHº = –334.4 kJ mol–1, ΔrSº = –240.8 J mol–1 K–1, ΔrGº = –262.6 kJ mol–1
Eºcell = –ΔrGº / (n F) = 1.361 V
For the second cell:
ΔrHº = –25 kJ mol–1, ΔrSº = 127.3 J mol–1 K–1, ΔrGº = –62.9 kJ mol–1
Eºcell = 0.326 V
For the third cell:
ΔrHº = 14.4 kJ mol–1, ΔrSº = 105.4 J mol–1 K–1, ΔrGº = –17.0 kJ mol–1
Eºcell = 0.088 V
Write down the cathode and the anode reactions in the galvanic cells if the measured electromotive force equals the standard cell reaction potential.
Model Answer
For the first cell:
Anode: H2 = 2 H+ + 2 e–
Cathode: Cl2 + 2 e– = 2 Cl–
For the second cell:
Anode: Pb + 2 Cl– = PbCl2 + 2 e–
Cathode: 2 H+ +2 e– = H2
For the third cell:
Anode: Pb + SO4^2– = PbSO4 + 2 e–
Cathode: PbCl2 + 2 e– = Pb + 2 Cl–
Calculate the equilibrium constant for the cell reactions.
Model Answer
For the first cell: K = 1.02·10^46
For the second cell: K = 1.05·10^11
For the third cell: K = 952
How do the electromotive forces change with temperature?
Model Answer
If we suppose that the thermodynamic data are independent of the temperature in a limited range, then the following expression:
ΔE_cell / ΔT = ΔrSº / (n F)
shows that the temperature dependence of the cell potential is determined by the reaction entropy. For the first cell this is negative, therefore the electromotive force decreases with increasing temperature. For the other two cells the electromotive force increases with increasing temperature.
Let us define a ‘thermal efficiency’ parameter as the theoretical maximum of the ratio between the electrical work and the enthalpy change in the cell.
What are the values of this parameter for these cells? What can we conclude from these numbers?
Model Answer
For the first cell: ηmax = 78.5 %
For the second cell: ηmax = 252 %
For the third cell: ηmax = –118 %
In the first cell, the strongly exothermic reaction proceeds, although there is a significant entropy decrease resulting from the disappearance of gases. This entropy decrease during the reaction necessarily means energy dissipation and efficiency less than 100 %.
In both the other two cells, there is an entropy increase associated with the reaction (a gas forms or the number of dissolved ions increases). The maximal work is thus larger than the reaction enthalpy. The difference must be taken from the environment, meaning that both cells could extract heat from the environment and convert it into electrical work.
In the third cell an endothermic cell reaction is driven by the entropy increase. Energy from the environment for the operation of the galvanic cell dominates over the reaction enthalpy, so the efficiency parameter is ill-defined in this case.