1. Chapter Nine Electrochemistry Applications

2. Batteries and Fuel Cells
We’ve seen examples of batteries in our examination of electrochemical cells. Following are common examples that employ such methods.

3. A Flashlight Battery Consists of a Zinc Cup Anode and a Cathode Made of an Inert Carbon Electrode Immersed in a Paste containing MnO2 and Mn2O3
Zn(s) MnO2(s) + H2O  Zn2(aq) Mn2O3(s) OH-(aq)

4. The Atomic-level View of the Reactions Involved in a Car Battery Shows Lead's Oxidation to PbSO4 at the Anode and Reduction of PbO2 to PbSO4 at the Cathode
Pb(s)  PbSO4(s)
PbO2(s)  PbSO4(s)
Balance this reaction

5. A Hydrogen-Oxygen Fuel Cell in an Environmentally Friendly Method for Producing Electrical Energy. The Only By-Product is Water

6. In the Fuel Cell, Electron Transfer Is Indirect
O2(g) H2O(l) e-  4 OH-(aq) Eo = V
2 H2O(l) e-  H2(g) OH-(aq) Eo = V
Combine these and compute the cell voltage

7. Breathalyzer Fuel Cell
Cathode
O2(g) + 2 H2O(l) e-  4 OH-(aq)
Anode:
C2H5OH(g) + 4 OH-(aq)  HC2H3O2(g) + 3 H2O + 4 e-
Overall:
C2H5OH(g) + O2(g)  HC2H3O2(g) + H2O(l)

8. Electrolysis
As the name defines, “lysis” is to break apart, thus electro-lysis is the breaking of materials using electricity
An electrolytic cell is a battery run in reverse. This process regenerates the battery in your car while the engine is running.
Question: Can one produce sodium and chlorine gas from a salt water solution?

9. Water Electrolysis In water, the possible reactions are:
2 H2O(l) e-  H2(g) OH-(aq) Eo = V
O2(g) H+(aq) e-  2 H2O (l) Eo = V
It appears at first that when the threshold of 1.23 V – (-0.83 V) = 2.06 V is achieved, water electrolysis will begin. But this isn’t true for water. These values presume 1 M [H+] and [OH-] .

10. Water Electrolysis

11. Water Electrolysis Overall Reaction:
4 H2O (l) e- → 2 H2(g) OH-(aq) EO = V
2 H2O(l) → O2(g) H+(aq) e- Eo = V
Note that the cathode reaction is reversed now.
Overall: 6 H2O(l) → 2 H2(g) O2(g) H2O(l) or
2 H2O(l) → 2 H2(g) O2(g) Ecell = V
For reasons of consistency, I’ve adopted the following strategy for determining electrolytic cell reactions;
As before, subtract the smallest voltage from the largest to produce a positive Ecell. Ecell = V – ( V) = V
Know that we are running the cell in reverse from the electrochemical (or battery) cell. Therefore, we switch the sign of the cell voltage.
Eelectroly = V We need V to start the electrolysis.
If you have a battery, change the direction of the smaller voltage (anode) reaction. If we have an electrolytic cell, change the direction of the larger voltage (cathode) reaction.

12. Salt Water Electrolysis
In a solution of NaCl, we have Na+, Cl- and H2O. The possible reactions are:
Na+(aq) + e-  Na(s) Eo = V
Cl2(g) e-  2 Cl-(aq) Eo = V
2 H2O(l) e-  H2(g) OH-(aq) Eo = V
O2(g) H+(aq) e-  2 H2O (l) Eo = V
Electrolysis will begin when the minimum threshold voltage is achieved.
To produce Na and Cl2 : 1.36 V – (-2.71 V) = 4.07 V are required.
However, as we’ve seen, when the threshold of 1.23 V is achieved, water electrolysis will begin first! So, it is not possible to produce Na and Cl2 this way.

13. Salt Water Electrolysis
In a solution of NaCl, we have Na+, Cl- and H2O. The possible reactions are:
Na+(aq) + e-  Na(s) Eo = V
Cl2(g) e-  2 Cl-(aq) Eo = V
2 H2O(l) e-  H2(g) OH-(aq) Eo = V
O2(g) H+(aq) e-  2 H2O (l) Eo = V
One might ask: Isn’t the smallest beginning potential: 1.36 V – 1.23 V = 0.13 V? Why doesn’t electrolysis begin at that potential? Recall that when we compute a cell potential for an electrolytic cell, we change the direction of the cathode reaction and not the anode reaction. In our anode reaction, the reactants are O2(g) and H+(aq). So the overall reaction would be: 4 Cl-(aq) + O2(g) + 4 H+(aq)  2 Cl2(g) But we have no O2(g) in the system! We have only Na+(aq), Cl-(aq) and H2O(l). So this reaction would not occur.

14. Production of Metals
The production of metals usually involves electrolysis of the molten salt. Here, NaCl is kept above the melting point of 800oC

15. Schematic of the Apparatus for the Electrolytic Production of Aluminum Showing Molten Aluminum Sinking to the Bottom of the Tank

16. Corrosion Prevention
Corrosion of metals is of great concern in everything from power lines and utility poles to bridges and buildings.
Active metals used in construction should be protected in some form.

17. Sacrificial Anodes
Metal such as iron is optimal for many reasons including it’s low cost and availability, conductive properties, and metallic behavior.
However, it suffers from being a fairly active metal. How does one prevent the iron from decomposing?
Answer: Place the iron in contact with another, more active metal. Being more active, the more active metal will oxidize preferentially to iron. Such metals are called “sacrificial anodes”

18. Underground Steel Pipes
Here, Mg is more active than the pipe and will corrode first. Attack on the pipe is prevented by the transfer of electrons by the magnesium metal. Copper is very inactive and will not corrode.

19. A Steel Utility Pole Is Connected to a Magnesium Stake
A Steel Utility Pole Is Connected to a Magnesium Stake. The Magnesium Feeds Electrons to the Iron in the Steel Pole via the Conducting Connector, Preventing Oxidation of the Iron

20. Oxide Coating Protection
Aluminum is an active metal, however, when exposed to air, it forms a tough Al2O3 coating that protects the interior metal.
Iron nails are often “galvanized” by coating with a layer of zinc oxide, ZnO which protects the interior metal.

21. Oxide Coating Protection
A measure of the effectiveness of an oxide coat is the Pilling-Bedworth ratio of P-B ratio defined as:
𝑃−𝐵 𝑟𝑎𝑡𝑖𝑜= 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟 𝑚𝑒𝑡𝑎𝑙 𝑎𝑡𝑜𝑚 𝑖𝑛 𝑜𝑥𝑖𝑑𝑒 𝑚𝑒𝑡𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
A P-B ratio between 1 and 2 forms a tough non-porous coating.
A P-B ratio > 2 causes flaking because oxygen stresses the lattice.
A P-B ratio < 1 is porous because the oxide is smaller than the lattice leaving spaces.

22. Oxide Coating Protection
Densities :
Al: 2.07 g/mL (27.0 g/mol) Zn: 7.13 g/mL (65.4 g/mol)
Al2O3: 4.0 g/mL (102 g/mol) ZnO: 5.6 g/mL (81.4 g/mol)
𝑃−𝐵 𝑟𝑎𝑡𝑖𝑜= 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟 𝑚𝑒𝑡𝑎𝑙 𝑎𝑡𝑜𝑚 𝑖𝑛 𝑜𝑥𝑖𝑑𝑒 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑡𝑎𝑙
Al molar volume = (27.0 g/mol)/(2.07 g/mL) = 13.0 mL Al/mol
Al2O3 molar volume = (102 g/mol)/(4.0 g/mL) = 25.5 mL Al2O3/mol = 12.8 mL Al/mol
P-B ratio = 12.8/13.0 = 0.98.
A similar analysis for Zn/ZnO gives a P-B ratio of 0.63.
So Al2O3 protects Al very efficiently while ZnO protects Zn less so as the coating is porous.
Fe2O3 gives a ratio of Thus, the oxide is larger and stresses the metal causing flaking of the rust.