1. Definitions of Oxidation-Reduction  Loss/Gain of electrons  Increase/Decrease of oxidation number  Determining oxidation numbers

2. Remember: A redox reaction is any reaction involving a transfer of electrons. In all redox reactions, oxidation and reduction happen at the same time. Oxidation is loss of electrons/ increase in oxidation number. Reduction is gain of electrons/decrease in oxidation number. Oxidising agents (oxidants) are themselves reduced. Reducing agents (reductants) are themselves oxidised.

3. Rules for oxidation numbers 1. Oxidation number for elements is zero. N 2, O 2, O 3, Cu, S 8 2.Oxidation number of monoatomic ions is the same as their charge Al 3+, Zn 2+, Cd 2+, Ag + 0 0 0 0 0 +3 +2 +2 +1 3. In polyatomic ions (NO 3 - ), the sum of oxidation numbers equals the charge of the ion. In compounds (HNO 3 ) the sum of the oxidation numbers equals zero.

4. 4.Oxidation number of oxygen in most compounds is –2. Exceptions: H2O2,H2O2, (peroxides) –1 5. Oxidation number of hydrogen is +1 Exceptions: bonded to metals LiH O 2 2- 6. Fluorine is always –1. Other halogens are also –1, except when they are bonded to O, then they are positive. Rules for oxidation numbers

5. Assign oxidation numbers to all of the elements: Li 2 OLi = O = PF 3 P = F = HNO 3 H = N = Cr 2 O 7 2- Cr = O = +1-2 MnO 4 - Mn =O = +3 +1-2+5 -2 +7 +6

6. Electrochemical cells:  Their properties  Electrode potentials defined as standard electrode potentials, E o (unit: Volts, V) Cell diagrams:  Use of the symbols “/” (phase boundary) “,” (same phase)and “//” (salt bridge)  Half cells  Order of notation

7.  Electrochemical cell: a cell in which oxidation and reduction occur, often in separate compartments  Half cell: a single electrode in an solution containing ions  Electrode: the conductor placed in cells that transfer charge between the external circuit and the electrolyte

8.  Anode: electrode where oxidation occurs (negative electrode)  Cathode: electrode where reduction occurs (positive electrode)  Electrolyte: substances in the salt bridge (usually liquids) that transfer charge by moving ions  Electrolytic cell: a cell that uses a supply of electricity to bring about a non spontaneous chemical reaction (year 12)

9.  Electromotive force, EMF, or E º cell : the potential difference across a voltage source when no current is following  Standard reduction (electrode) potential E º standard electrode potential measured in volts under standard conditions (25 o C, 1molL -1, 1 atm), which indicates the ability of a species to gain electrons

10. The salt bridge allows the movement of ions between the two half cells so that charges can be balanced. It completes the electric circuit. If the voltmeter (that restricts the current flowing) is replaced by a wire, the reactions will take place more quickly. Here the copper electrode would gain mass, the zinc electrode would loose mass. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ galvan5.swf

11. Give the half cell diagram Pt/H 2 /H + //

12. Electrochemical cells:  Calculations related to Electrochemistry  Spontaneity of oxidation-reduction reactions  Applications involving electrochemical cells (details of particular cells, eg dry cells, will be provided as required)

13. The emf of an electrochemical cell is calculated using the following formula: E o cell = E o (RHE) – E o (LHE) note: do not change the sign of the standard potentials If the emf is positive: The electron flow is from left to right and the oxidation takes place in the left half cell. If the emf is negative: The electron flow is from right to left and the oxidation takes place in the right half cell.

14. greater E o : Strongest Oxidant, Reduction reaction lower E o : Strongest Reductant, Oxidation reaction

15. To predict whether reactions happen spontaneously, the emf is calculated. Does Zn react with Fe 3+ ? Zn is the possible loser of electrons Zn /Zn 2+ // Fe 3+, Fe 2+ /C E o cell = 0.77 – ( - 0.76) = 1.53 V The emf is positive, therefore the electron flow is from left to right and oxidation takes place in the left half cell. That means that Zn reacts spontaneously with Fe 3+. Does Zn react with Fe 3+ ? Zn is the possible loser of electrons Zn /Zn 2+ // Fe 3+, Fe 2+ /C E o cell = 0.77 – ( - 0.76) = 1.53 V The emf is positive, therefore the electron flow is from left to right and oxidation takes place in the left half cell. That means that Zn reacts spontaneously with Fe 3+. Does Zn 2+ react with Fe 2+ ? Fe 2+ is the possible loser of electrons C/ Fe 2+, Fe 3+ // Zn 2+ /Zn E o cell = - 0.76 -0.77 = -1.53 V The emf is negative, therefore the electron flow is from right to left and oxidation takes place in the right half cell. That means that Zn 2+ does not react spontaneously with Fe 2+. Does Zn 2+ react with Fe 2+ ? Fe 2+ is the possible loser of electrons C/ Fe 2+, Fe 3+ // Zn 2+ /Zn E o cell = - 0.76 -0.77 = -1.53 V The emf is negative, therefore the electron flow is from right to left and oxidation takes place in the right half cell. That means that Zn 2+ does not react spontaneously with Fe 2+. Fe 2+ /Fe 3+ = 0.77 Zn 2+ /Zn = - 0.76

16. When the battery is charged, lead (II) ions ( Pb 2+ ) in lead sulfate are reduced to Pb and oxidised to lead (IV) ions ( Pb 4+ ) in lead oxide. Observation: A build up of lead at the anode and a build up of PbO 2 at the cathode. 2PbSO 4 + 2H 2 O  Pb + PbO 2 + 2H 2 SO 4 When the battery is discharged (providing energy to the car), the reaction is reversed and PbSO 4 is produced. This will build up on cathode and anode. If any PbSO 4 falls off the plate (which happens after long use), then it can not react and the battery needs replacing.

17. You do not need to know the details about the Dry Cell as there are different types and in the exam a different example may be chosen. A dry cell is very compact, so it may be difficult to identify cathode, anode and the half cell reactions. The electrolyte used is a paste made up of alkaline or acidic salts. For the Lechlanche cell above identify: Cathode: Anode: Oxidation: Reduction: graphite Zinc case Zn to Zn 2+ Mn 4+ to Mn 3+ Exam Questions: They may involve discussing the suitability of a redox pair for the construction of a dry cell. The production of a gas indicates no suitability!

18. Redox reactions:  Appearance and state of common oxidants and reductants  Calculations involving mole ratios (titrations) → go through examples in your book

19. Reduced formOxidised form Cu brownsolid Cu 2+ blueaq SO 2 gas SO 4 2- aq Mn 2+ aq H + /MnO 4 - purpleaq H2O2H2O2 liquid O2O2 gas H2OH2O liquid H2O2H2O2 Cr 3+ blue/greenaq Cr 2 O 7 2- orangeaq Fe 2+ pale greenaq Fe 3+ orangeaq Cl - aq Cl 2 pale greengas Br - aq Br 2 red/orangeliquid H2H2 gas H+H+ aq

20. Reduced formOxidised form MnO 2 brownsolidH 2 O/MnO 4 - purpleaq MnO 4 2- greenaqOH - /MnO 4 - purpleaq I-I- I 2 in I - = I 3 - brownaq I 2 in I - = I 3 - brownaq IO3-IO3- H2SH2S gas S yellow/whitesolid Pb 2+ aq PbO 2 brownsolid NO 2 browngas NO 3 - aq C 2 O 4 2- aq CO 2 gas S 2 O 3 2- aq S 4 O 6 2- aq Br 2 red/orangeliquid BrO 3 - aq

21. In previous exam papers, students struggled to achieve because:  they did not read the question properly  they were not able to show the direction of electron flow  could not assign oxidation numbers  did not give V as the unit for E o cell, or used the wrong sign  could not write standard cell diagrams, forgot inert electrodes  did not know that a salt bridge completes the circuit, allows ion flow  did not know what happens when the voltmeter is replaced by a wire  used the term “dissolve” incorrectly when referring to the decrease of mass of an electrode  could not identify the oxidised and reduced form  did not know the colours of species  did not identify strongest reductant/oxidant (quoted incorrectly a redox pair)