Chemistry
Exercise 8
8. Describe the energetics associated with the operation of a electrochemical cell and electrolytic cells.
We have picked electrochemistry as a topic since it illustrates many concepts associated with energetics, i.e. what drives processes from a thermodynamic aspect. It highlights the reversibility of reactions; the importance of entropy, and quantifies driving force as a voltage.
a) A Ni2+ solution is electrolyzed using a current of 1.25 A. What mass of Ni plates out in 30.0 min?
This is essentially a stoichiometry problem in which aqueous nickel is reduced to form nickel solid in the following reaction (which can be found in any chemistry textbook):
Ni2+(aq) + 2 e- 
Ni(s)
This problem can be solved in the following manner;
1. Use the formula C = A x t, where C is coulombs of charge, A is current in amps (or coulombs/second) and t is time in seconds
2. 1 mol of electrons carries a charge of 96 485 C of charge (this is known as the Faraday constant, F). Using the value for C above, the number of electrons that flow through the Ni2+ solution can be calculated;
3. Looking at the above chemical equation, electroplating of nickel requires 2 electrons to produce 1 equivalent of solid nickel, therefore 1 mole of electrons produce ½ mol of nickel metal
4. Solving for the mass of Ni,
massNi = 0.0116 mol x 58.69 g/mol = 0.684 g
b) In the electrolysis of a NaCl solution, what volume of Cl2(g) is produced at the same time that it takes to liberate 6.00 L of H2(g)? Assume that both gases are measured at STP.
Looking at the question, the first thing to do is identify the species that are present in this system:
Na+(aq), Cl-(aq), H2O, Na(s), Cl(s), H+(aq)
The possible reduction and oxidation half reactions for the electrolysis of NaCl are given below (found in a chemistry text)
Reduction (occurs at the cathode)
| Na+(aq) + e- Na(s) |
Eo= -2.71 |
| 2 H2O + 2 e- H2(g) + 2 OH- |
Eo= -0.83 |
At the cathode, there are two possible reactions, so to determine which is the more probable reaction, we need to look at the reduction potentials (Eo). The greater the potential, the more easily the reaction goes forward. In this case, the splitting of water has the larger potential, so it is the more likely reduction reaction for the cell.
Oxidation (at the anode)
| 2 Cl- → Cl2 + 2 e- | Eo= -1.36 |
Once the reactions are combined to given an overall cell reaction, the problem becomes a simple stoichiometric problem;
2 Cl-(aq) + 2 H2O H2(g) + 2 OH-(aq) + Cl2(aq)
At STP, 1 mol of any gas occupies a volume of 22.4 L. Calculating the moles of H2 gas produced gives;
Using stoichiometric equivalents, the moles of Cl2 produced is 1:1. this means that 0.27 mol of Cl2 is produced, which also occupies 6.0 L at STP conditions. with respect to this particular problem, it is not necessary to calculate the moles of H2. It is only necessary to know that all gases occupy the same volume for a given molar quantity at STP)
c) Use Eo values to predict whether each of the following skeleton equations represents a reaction that will occur in acid solution with all soluble substances present in 1 M concentrations. Complete and balance the equation for each reaction that is predicted to occur.
The spontaneity of electrochemical reactions can be predicted by determining the Eo value for the reaction.
An Eo greater than 0 indicates a spontaneous reaction, while a value less than 0 denotes a non-spontaneous
reaction.
Eo is calculated by summing the individual Eo values for the reduction and oxidation half reactions, remembering to change the sign of the Eo value for any reactions that are reversed. Do not multiply the Eo values to equalize the number of electrons exchanged. These values can be found in the electrochemistry section of any chemistry textbook .
i) H2O2 + Cu2+ Cu + O2
Oxidation ½ reaction
H2O2 O2 +2 H+ + 2e-, Eo= -0.68
Reduction ½ reaction
Cu2+ + 2e Cu(s), Eo= 0.34
The net Eo value is calculated as -0.68 + 0.34 = -0.34
The negative value indicates a non-spontaneous reaction
ii) H2O2 + Ag+ Ag + O2
Oxidation ½ reaction
H2O2 → O2 + 2 H+ + 2e-, Eo= -0.68
Reduction ½ reaction
Ag+ + e → Ag(s), Eo= 0.80
The net Eo value is calculated as -0.68 + 0.80 = 0.12
The positive value indicates a spontaneous reaction
iii) Ag+ + Fe2+ Ag + Fe3+
Oxidation ½ reaction
Fe2+ Fe3+ + e-, Eo= -0.77
Reduction ½ reaction
Ag+ + e Ag(s), Eo= 0.80
The net Eo value is calculated as -0.77 + 0.80 = 0.03
The positive value indicates a spontaneous reaction
d) Use electrode potentials to calculate the emf of a standard cell that uses the reaction:
Cl2(g) + 2Br-(aq) 2Cl- (aq) + Br2(1)
Oxidation ½ reaction
Cl2 + 2e 2 Cl-, Eo= 1.36
Reduction ½ reaction
2 Br- Br2 + 2 e , Eo= -1.09
net reaction
Cl2(g) + 2 Br-(aq) 2 Cl-(aq) + Br2(l), Eo=0.27
EMF = 0.27 V = 0.27 J/C
i) What is ΔGo for this reaction?
ΔGo= -nFE
ΔGo= -(2 mol e-)(96 485 C/mol e-)(0.27J/C)
ΔGo= -52.1 kJ
ii) If ΔHo = -189.44kJ, what is ΔSo?
ΔSocan be calculated using the formula,
ΔGo = ΔHo - TΔSo
Where T is 298K
ΔSo= 810.5 J
e) Suggest ways to increase the emf of a cell that is based on the reaction:
Fe(s) + 2H+(aq) Fe2+ (aq) + H2(g)
According to Le Chatelier’s principal, a reaction can be encouraged to proceed farther to the right by creating conditions that change the position of the chemical equilibrium. If a system at equilibrium is subject to changes in reactant or product concentration, temperature, volume or total pressure the equilibrium will shift to reduce the effect of the imposed change. As an example, if the concentration of a product is decreased, the equilibrium will shift to generate more product. On the other hand if the concentration of the reactant is decreased the system will shift its equilibrium to the left.
With respect to the above reaction, the emf of the cell can be increased by changing the system conditions in a way that causes the equilibrium to shift to the right. Ways to achieve this include:
- increasing concentration of [H+]
- removing Fe2+ from solution
- decreasing the pressure of H2 gas, effectively decreasing its concentration
