1. Redox Reactions The corrosion of metals, especially the corrosion of iron (rusting), has a great economic impact. Rust is one of the common causes of bridge accidents. The Kinzua Bridge in Pennsylvania was blown down by a tornado in 2003 largely because the central base bolts holding the structure to the ground had rusted away, leaving the bridge resting by gravity alone.

2. Redox Reactions The January report by independent engineers said the rust was starting to "detrimentally affect the structure". Rust fears for Harbour Bridge Auckland Harbour Bridge is rusting away at "structurally significant points", forcing engineers to step up maintenance work. And a cable - an "essential" part of the bridge's bracing system for earthquakes and high winds - has been disconnected to deal with more rust. Other cables can bear the weight when one is disconnected. One of the walkways that had been used by Bridge Climb operators is corroded, necessitating safety repairs, according to an engineering report. NZ Herald 22/03/2009

4. Redox Reactions Explain why this reaction is a redox reaction! Give more than one answer. Gain/loss of oxygen/hydrogen, gain/loss of electrons (transfer), increase/decrease in Ox.number

5. 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.

7. Redox Reactions Examples: Write the oxidation number for each of the underlined atoms: PbO 2 BrO 3 - [FeSCN] 2+ Examples: Write the oxidation number for each of the underlined atoms: PbO 2 BrO 3 - [FeSCN] 2+ +4 +5 +3

8. Redox Reactions Example: Use oxidation numbers to determine whether the following reaction is a redox reaction, and if so, which element has been oxidised, and which has been reduced: 4NH 3(g) + 3O 2(g) → 2N 2(g) + 6H 2 O (l) Example: Use oxidation numbers to determine whether the following reaction is a redox reaction, and if so, which element has been oxidised, and which has been reduced: 4NH 3(g) + 3O 2(g) → 2N 2(g) + 6H 2 O (l) -30 0 -2 It is a redox reaction. The nitrogen in ammonia has been oxidised (ON increased). The oxygen has been reduced (ON decreased). Complete worksheet #1

9. Colours of species in redox reactions

10. The colour of any species depends on the oxidation number. You are expected to know a number of species and their colours. The colour of any species depends on the oxidation number. You are expected to know a number of species and their colours.

11. 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

12. 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

13. Names of unfamiliar ions: C 2 O 4 2- Oxalate ion (from oxalic acid) S 2 O 3 2- Thiosulfate ion S 4 O 6 2- Tetrathionate BrO 3 - Bromate IO 3 - Iodate Names of unfamiliar ions: C 2 O 4 2- Oxalate ion (from oxalic acid) S 2 O 3 2- Thiosulfate ion S 4 O 6 2- Tetrathionate BrO 3 - Bromate IO 3 - Iodate Citrobacter live in the intestines of warm blooded animals, without oxygen. They can respire by converting tetrathionate into thiosulfate. Rhubarb contains oxalic acid. This makes your teeth feel furry.

14. Some species can be recognised by their colour. For example: The purple permanganate ion (MnO 4 - ) can be reduced in alkaline conditions (presence of OH - ). The product is the green manganate ion (MnO 4 2- ). Observation: the purple solution turns green. Some species can be recognised by their colour. For example: The purple permanganate ion (MnO 4 - ) can be reduced in alkaline conditions (presence of OH - ). The product is the green manganate ion (MnO 4 2- ). Observation: the purple solution turns green. Recognising species (observations):

15. Some species can not be recognised by their colour as they are colourless or do not have a distinct colour. Therefore you need to memorise some tests to detect these species: I 2 : turns blue/black with starch solution SO 4 2- : forms a white precipitate with H + /BaCl 2 Fe 3+ : forms a blood red solution with KSCN Cl 2 : turns damp starch-iodide paper blue/black Some species can not be recognised by their colour as they are colourless or do not have a distinct colour. Therefore you need to memorise some tests to detect these species: I 2 : turns blue/black with starch solution SO 4 2- : forms a white precipitate with H + /BaCl 2 Fe 3+ : forms a blood red solution with KSCN Cl 2 : turns damp starch-iodide paper blue/black Recognising species (observations): Complete worksheet #2

16. Balancing complicated half equations 1. Balance the atoms that aren’t O or H. Cr 2 O 7 2- → Cr 3+ Cr 2 O 7 2- → 2Cr 3+ 2. Balance the oxygen by adding water. Cr 2 O 7 2- → 2Cr 3+ + 7H 2 O 3. Balance the hydrogens by adding H +. Cr 2 O 7 2- + 14H + → 2Cr 3+ + 7H 2 O 4. Add electrons to the side that is more positive. Cr 2 O 7 2- + 14H + + 6e - → 2Cr 3+ + 7H 2 O

17. Balancing complicated half equations 1. Balance the atoms that aren’t O or H. SO 2 → SO 4 2- 2. Balance the oxygen by adding water. SO 2 + 2H 2 O → SO 4 2- 3. Balance the hydrogens by adding H +. SO 2 + 2H 2 O → SO 4 2- + 4H + 4. Add electrons to the side that is more positive. SO 2 + 2H 2 O → SO 4 2- + 4H + + 2e -

18. Combining the half equations The number of electrons in the two half equations needs to be the same. Therefore the half equations have to be multiplied by a factor.

19. Combining the half equations Cr 2 O 7 2- + 14H + + 6e - → 2Cr 3+ + 7H 2 O SO 2 + 2H 2 O → SO 4 2- + 4H + + 2e - Multiply whole equation by 3 3 SO 2 + 6H 2 O → 3 SO 4 2- + 12H + + 6e -

20. Cr 2 O 7 2- + 14H + + 6e - → 2Cr 3+ + 7H 2 O 3 SO 2 + 6H 2 O → 3 SO 4 2- + 12H + + 6e - Combining the half equations Cr 2 O 7 2- + 14H + + 3 SO 2 + 6H 2 O→2Cr 3+ + 7H 2 O + 3 SO 4 2- + 12H + 5. Cancel out H + and H 2 O. Cr 2 O 7 2- + 2H + + 3 SO 2 → 2Cr 3+ + H 2 O + 3 SO 4 2-

21. Balancing redox equations in acidic/neutral conditions Last year you learned the steps to balance redox reactions. One of the steps included adding H + ions. H + ions are present in small amounts in neutral conditions (same as OH - ions). In acidic conditions H + ions are present in a higher concentration than OH - ions. It is therefore ok if your redox equations contain H + ions as they are present.

22. Balancing redox equations in acidic/neutral conditions The permanganate ion can be reduced in neutral, acidic and in alkaline conditions. Each of these conditions leads to a different product. Write down the balanced half equations for the reduction or permanganate in acidic conditions (producing Mn 2+ ),

23. Answers Acidic: MnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O Now write a balanced half equation for the reduction of permanganate in neutral conditions (producing MnO 2 ).

24. Answers Neutral: MnO 4 - + 4H + + 3e - → MnO 2 + 2H 2 O H + ions are relatively scarce. Add OH - ions to both sides of the equation to neutralise. MnO 4 - + 2H 2 O + 3e - → MnO 2 + 4OH -

25. Balancing redox equations in alkaline conditions The permanganate ion can also be reduced in alkaline conditions, producing the manganate ion (MnO 4 2- ). Write a balanced half equation using the usual steps. If adding H + ions (which are scarce in the alkaline solution) is necessary, add an equal amount of OH - ions to both sides of the equation.

26. Answer Alkaline: MnO 4 - + e - → MnO 4 2- Complete worksheet #3

27. Key Learning Outcomes You should now be able to: Determine the oxidation number of any atom in a compound or ion and use oxidation numbers to identify the oxidised and reduced species in a reaction. Recall common oxidising and reducing agents, state the colours of the reagents and their products, and recall any other observations or conditions characteristic of their use. Write ion-electron equations for oxidation and reduction half-reactions and combine the half equations to give a balanced ionic equation.