1. electrolysis
[20.9]

2. Nonspontaneous reactions
Redox reactions that do not take place spontaneously (Ɛo >0; ∆G<0) can be made to happen with an exterior source of energy
Consists of anode & cathode in electrolyte with a electron pump (battery) to push electrons into one electrode and pull them from the anode

3. Sodium NaCl is electrolyzed in a Downs cell.
Gaseous Cl2 allowed to disperse
Molten Na siphoned off

4. Aluminum
In the Hall process, Al2O3 is dissolved in molten cryolite (Na2AlF6), and Al3+ is reduced to molten Al.

5. Copper
Active metal impurities oxidized at anode, but don’t plate out at cathode.
Cu2+ more easily reduced
Less active metals deposit as sludge below anode.

6. electroplating
Use of electrolysis to deposit a thin layer of one metal on another to improve beauty or resistance

7. Rules 4 cathode rxns
Salts with transition metal cations, Cu+2, Ag+,Ni+2, which are easier to reduce than water, will plate out at the cathode
Cation is reduced to elemental metal
Cu e- Cu
Yet it the cation is a representative metal, Na+, Mg+2, Al+3, the water molecules will be easier to reduce and H2 gas will evolve at the cathode.
Electrolysis of water will make the solution basic, OH-
2H2O  H OH-

8. Rules 4 anode rxns
Salts containing I,Br,Cl, these are oxidized before water.
The nonmetal is formed at the anode (I2,Br2,Cl2)
Anion less easily oxidized than water (F-, SO42-,etc.) water will be oxidized at the anode
Oxygen gas is generated and the pH drops
H2O  ½ O2 + 2H+ + 2e-
now let's that AP problem

9. Problem solving
SAMPLE EXERCISE Aluminum Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

10. Solve: First, we calculate the coulombs of electrical charge that are passed into the electrolytic cell:
Second, we calculate the number of moles of electrons that pass into the cell:
Third, we relate the number of moles of electrons to the number of moles of aluminum being formed, using the half-reaction for the reduction of Al3+:

11. SAMPLE EXERCISE 20.14 continued
Thus, three moles of electrons (3 F of electrical charge) are required to form 1 mol of Al:
Finally, we convert moles to grams:
Because each step involves a multiplication by a new factor, the steps can be combined into a single sequence of factors:

12. SAMPLE EXERCISE 20.15 Calculating Energy in Kilowatt-hours
Calculate the number of kilowatt-hours of electricity required to produce 1.0  103 kg of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V.
Solve: First, we need to calculate nF, the number of coulombs required:
We can now calculate w. In doing so, we must apply several conversion factors, including Equation 20.21, which gives the conversion between kilowatt-hours and joules:

13. SAMPLE EXERCISE 20.15 continued
Comment: This quantity of energy does not include the energy used to mine, transport, and process the aluminum ore, and to keep the electrolysis bath molten during electrolysis. A typical electrolytic cell used to reduce aluminum ore to aluminum metal is only 40% efficient, with 60% of the electrical energy being dissipated as heat. It therefore requires on the order
of 33 kWh of electricity to produce 1 kg of aluminum. The aluminum industry consumes about 2% of the electrical energy generated in the United States. Because this is used mainly to reduce aluminum, recycling this metal saves large quantities of energy.

14. Chromium Oxidized by HCl or H2SO4 to form blue Cr2+ ion.
Cr2+ oxidized by O2 in air to form green Cr3+.
Cr also found in +6 state as in CrO42− and the strong oxidizer Cr2O72−.

15. Iron Exists in solution in +2 or +3 state.
Elemental iron reacts with non-oxidizing acids to form Fe2+, which oxidizes in air to Fe3+.

16. Iron Brown water running from a faucet is caused by insoluble Fe2O3.
Fe3+ soluble in acidic solution, but forms a hydrated oxide as red- brown gel in basic solution.

17. Copper In solution exists in +1 or +2 state.
+1 salts generally white, insoluble.
+2 salts commonly blue, water-soluble.