1. Electrochemistry Oxidation & Reduction
Unit 16
Electrochemistry
Oxidation & Reduction

2. Oxidation verses Reduction
Gain oxygen atoms 2 Mg + O2  2 MgO Lose electrons (e-) Mg(s)  Mg e- Lose hydrogen CH3OH  CH2O + H2
Lose oxygen atoms 2 KClO3  2 KCl + 3 O2 Gain electrons (e-) Cu e-  Cu (s) Gain hydrogen CO + 2 H2  CH3OH

3. l l l l l l l l l reduction oxidation When oxidation occurs, there is also reduction. They go “hand in hand”. Not all reactions are oxidation reduction reactions.

4. Using Appendix 6 chart: To determine the strength as an oxidizing agent, look at the left side of appendix 6. The chemical that is higher on the list (has a greater magnitude + voltage) is the better oxidizing agent. Which is better? Br2 or Sn+2 To determine the strength as a reducing agent, look at the right side of appendix 6. The chemical that is lower on the list (has the greater magnitude – voltage) is the better reducing agent. Which is better? Pb(s) or H2O2

5. The chemical that is oxidized acts as the reducing agent
The chemical that is oxidized acts as the reducing agent. The chemical that is reduced acts as the oxidizing agent. In the following, indicate which chemical is being oxidized, reduced, acts as the oxidizing agent, and acts as the reducing agent. Zn(s) Cu+2(aq)  Zn+2(aq) Cu(s)

6. Oxidation Reduction Lab
Zn (s)
Cu (s)
Pb (s)
Mg (s)
Zn+2
Cu+2
Pb+2

7. Electricity is the movement of electrons to provide energy
Electricity is the movement of electrons to provide energy. Two factors affect the amount of energy produced Voltage is the push behind the e Voltage is measured in volts Current which is how many e- pass a point each second. Current is measured in Amperes. 1 Amp = 6.0 x 1018 e-/s An electrochemical Cell (Galvanic or voltaic cell) is a chemical device to make electricity from chemical reactions. It employs oxidation and reduction. It will have a positive (+) voltage while it works.

8. Cell Potential (Eocell) or Electromotive force (emf) is the pull or “driving force” on the e-. It is measured in volts (Eo ) Standard reduction potential is the potential for the half – reaction at 1M the standard being measured off 2 H+ + 2 e-  H2 , Pt electrode = 0.00 v Remember that these are listed as Reduction Potentials. When there is reduction, there is also oxidation. One of the two half – reactions will need to be reversed (so that it is written as an oxidation)

9. To calculate the voltage of an electrochemical cell: 1
To calculate the voltage of an electrochemical cell: 1. Obtain the two half – reactions from the reduction potential chart Determine which half – reaction is the oxidation, reverse the reaction. When the reaction is reversed, the sign on the voltage is changed Balance the e- in the two half – reactions by multiplying the coefficients by some whole number. Remember that voltage is a property of the kind of chemical, not the amount. Do not multiply the voltage Add the two half – reactions to determine the net voltage for the cell.

10. Oxidation Reduction Lab
Zn (s)
Cu (s)
Pb (s)
Mg (s)
Zn+2
0.00 v No
Yes
Cu+2
Pb+2
0.00v

11. A salt bridge or porous disk connects the half cells and allows e- to flow, completing the circuit.
We use a porous cup as the barrier between the 2 half cells

12. Calculate the voltage of a Galvanic cell made with Mg and Cu used to run the clock. Mg e-  Mg(s) volts Cu e-  Cu(s) volts

13. Line Notation : A way to denote a galvanic cell
Anode: the oxidation is written first
Cathode: the reduction is written second
One line separates the phases of a half cell
Two lines separate the anode and cathode
Mg l Mg+2 ll Cu+2 l Cu

14. Chemical Acid Base Neutral
NaOH
Na2Cr2O7
NaCl
HCl
H2O
KOH
Na2CO3
KNO3
HNO3
Na3PO4
Na2C2O4
NaSCN
H2SO4
Corrosion of Iron Lab Check by each chemical if it is acid, base, or neutral

15. Corrosion is the process of returning metals to their natural state
Corrosion is the process of returning metals to their natural state. The oxidation of metals. Example: 4 Fe(s) O2 (g)  2 Fe2O3 (s) All metals corrode easily, except Au, Ag, Cu Preventing corrosion: Coating with Cd or Sb Galvanizing (coating) with Zn Stainless steel is an alloy of Fe with Cd + Ni Sacrificial metals which corrode first are Mg or Zn

16. Corrosion Lab Fe (s)  Fe+2 + 2 e- Oxidation - Blue
2 e H2O (L)  H2 (g) OH-1
Reduction - Pink
Sacrificial metal
Zn (s)  Zn e-
Oxidation - white

17. Corrosion of Iron Lab Check by each chemical if corrosion occurred
Chemical Litmus Corrosion
NaOH base
Na2Cr2O base
NaCl neutral
HCl acid
H2O neutral
KOH base
Na2CO base
KNO neutral
HNO acid
Na3PO base
Na2C2O base
NaSCN base
H2SO acid
Corrosion of Iron Lab Check by each chemical if corrosion occurred

18. A battery is a group of galvanic cells connected in series
A battery is a group of galvanic cells connected in series. (add potentials Eocell) Wet cell – contain fluid medium for e- flow Dry cell - contain a paste medium for e- flow Primary cell – one time use, not rechargeable Secondary cell - rechargeable

19. Dry Cell (Batteries) – do not contain a liquid electrolyte
Acid version
Anode reaction - oxidation
Zn  Zn e v
Cathode reaction – reduction
2NH4+ + 2MnO2 + 2e  Mn2O3 + 2NH3 + 2H2O v

20. 2MnO2 + H2O + 2 e-  Mn2O3 + 2OH- cathode
Other types
Alkaline – Zn + 2OH-  ZnO + H2O + 2e- anode
2MnO2 + H2O + 2 e-  Mn2O3 + 2OH- cathode
Silver cell – Zn anode, Ag2O cathode
Mercury cell – Zn anode, HgO cathode
Cadmium – anode, NiO2 – cathode
(secondary – rechargable)
Mercury cell

21. Pb + PbO2 + 2H2SO4  2PbSO4 + 2H2O +2.02 v
Wet cell - Lead storage battery – rechargeable Anode Pb + HSO4-  PbSO4 + H e v
Cathode PbO2 + HSO H e-  PbSO H2O v
Overall reaction
Pb + PbO2 + 2H2SO4  2PbSO H2O v

22. An Electrolytic cell is a cell with a negative voltage, so there is no reaction. An outside source of energy is used as a way to drive the reaction. This type of cell is often used as a controlled way to deposit thin layers of metals, electroplating, on another metal.

23. Power source
An electrolytic cell can be used to electroplate one metal on another metal. Pb
Anode Cu
Cathode
Cr e-  Cr
Pb e-  Pb
Balance the reaction and determine the minimum voltage that needs to be supplied to drive the reaction.