TOPIC BTYPES OF CHEMICAL REACTIONS

REDOX • Oxidation is a loss of electrons• Reduction is a gain of electrons

• OIL RIG or LEO GER

• An oxidizing agent -- > promotes oxidation

(causes a species to lose electrons)• A reducing agent -- > promotes reduction• (causes a species to gain electrons

Common oxidizing and reducing agents

• Oxidizing agents

• Chlorine: Cl2 + 2e- 2Cl-

• Manganate (VII): MnO4- + 8H+ + 5e- Mn2+ + 4H2O

• Dichromate (VI): Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

• Oxygen: O2 + 4H+ + 4e- 2H2O

• Manganese (IV): MnO2 + 4H+ +2e- Mn2+ + 2H2O

• H+ in these reactions represents acid and is present in order to react with any excess oxygen atoms to form water.

• In general all non-metals tend to be oxidizing agents since they want to gain electrons.

• Reducing agents

• Zinc metal: Zn Zn 2+ + 2e- • Hydrogen gas: H2 + 2OH- 2H2O + 2e-

• Tin metal: Sn Sn2+ + 2e-

• In general all metals tend to be reducing agents since they want to give up electrons.

REDOX reactions• 2K + Cl2 2KCl

• This reaction can be thought of as two half equations, one involving oxidation, one involving reduction.

• Half equations include electrons.

• Oxidation: K K+ + e-

• Reduction: Cl2 + 2e- 2Cl-

• When combining two half equations it is necessary to cancel out the electrons to produce the full, balanced, REDOX equation: 2K + Cl2 2KCl

Oxidation State Definition:• … is the difference between the number of electrons

associated with an atom in a compound• … compared with the number of electrons in an atom of

the element.

• In ions – the oxidation state is the ionic charge

• In covalent compounds – the oxidation state corresponds to the formal charge

• Elements – assumed to exist in the zero oxidation state

Oxidation number concept• How do we know if a species is gaining or losing electrons

in a reaction?

• Oxidation number is the number of electrons that an atom tends to lose or gain, when it is involved in a REDOX reaction.

• gain - the oxidation number is negative. • lose – the oxidation number is positive

• No change in the number of electrons – not a part in the REDOX reaction.

Rules to simplify assignment of oxidation numbers:

• The rules should be applied in order:• 1) The oxidation number of an element when uncombined (as the

free element) is always zero.

• 2) The sum of the oxidation numbers in a neutral substance is always zero.

• 3) In an ion, the sum of the oxidation numbers of any elements present equals the ionic charge.

• 4) Some elements exhibit very common oxidation numbers in their compounds. • Group 1 - always +1, Group 2 - always, +2, F - always -1, • O - almost always -2, H - almost always +1.

• 5) In binary compounds with metals the • Group 17 elements are –1, • Group 16 are –2 and• Group 15 are –3.

• For example, calculate the oxidation number of Cr in K2Cr2O7

• K Cr O • +1 (x 2) + ? (x 2) + -2 (x 7) = 0• simple algebra shows that (Cr) = +6

• Practice Problems:

• 1. What is the oxidation number of each of the atoms listed in the following species? • (a) Sulfur in SO3

• (b) Oxygen in KO2

• (c) Nitrogen in NH4+

• (d) Strontium in SrF2

• (e) Cobalt in [CoCl6]3-

• 2. For the formation of each of the following ionic compounds:

• a) From their elements write an overall equation showing

the formation of the compound.

• b) Write two half-equations to identify the REDOX

process. • (a) calcium sulfide • (b) aluminum bromide • (c) aluminum oxide

• In aqueous solutions the key is the number of electrons produced must be the same as those required

• For reactions in acidic solution an 8 step procedure.1. Write separate half reactions2. For each half reaction balance all reactants except hydrogen and oxygen

3. Balance oxygen using H2O

4. Balance hydrogen using H+

5. Balance the charge using electrons ( e- ) 6. Multiply half-reactions by an integer to make electrons equal

7. Add half-reactions and cancel identical species

8. Check that charges and elements are balanced

Balancing Redox Equations

• 3. Where appropriate, construct half-equations for the following reactions and use them to balance the full equations.

• (a) IO3- + I- + H+ I2 + H2O

• (b) I2 + S2O32- I - + S4O6

2-

• (c) Br2 + KI KBr + I2

• (d) MnO42- + H+ MnO4

- + MnO2 + H2O

• (e) CrO42- + H+ Cr2O7

2- + H2O

• If the oxidation number becomes more positive• -- > the element has been oxidized

• If the oxidation number becomes more negative • -- > the element has been reduced.

• For example - below:• Mn has been reduced and Fe oxidized.

• MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

• This is further illustrated by considering the two half equations. • Mn7+ + 5e- Mn2+

• Fe2+ Fe3+ + e-

Classifying REDOX reactions• Disproportionation • - a simultaneous oxidation and reduction of one

species• - the disproportionation of hypochlorite ions • 3ClO- ClO3

- + 2Cl-

• Cl is simultaneously changing

Cl+1 Cl+5

Cl+1 Cl-

• Synthesis (or combination) • - a compound is formed by the reaction of simpler

elements or compounds.• - the synthesis of water from oxygen and hydrogen• 2H2(g) + O2(g) ! 2H2O(l)

• Both the elements start with an oxidation state of zero• - one is oxidized and one is reduced.

• Decomposition• - a compound is broken down into simpler substances• - the decomposition of mercury (II) oxide into its

elements • 2HgO 2Hg + O2

• The atoms in the compound start in positive or negative oxidation state and revert back to zero in the elemental state.

• It is the reverse of a synthesis reaction.

• Practice:• 1. In a simple synthesis reaction, magnesium metal is

placed in a crucible and is heated strongly in air until it has been completely oxidized by the oxygen present. The shiny, silvery colored metal turns to a white ash when the process is complete.

• (a) Suggest a formula for the white ash. • (b) After the reaction between the magnesium and the

oxygen is complete, what do you predict will happen to the mass of the crucible and its contents? Justify your answer.

• Calcium carbonate, when heated strongly in an open test tube, decomposes to yield both the corresponding metal oxide, and a non-metal oxide. The non-metal oxide is a gas, and escapes into the atmosphere. In such a reaction, the following data are collected.

• (a) What type of reaction is this? • (b) Write a balanced chemical equation to summarize the

reaction. • (c) Use the data to illustrate that matter has been conserved,

and that to justify the equation that you have written in (b).

Mass of calcium carbonate test tube before heating

30.46 g

Mass of the test tube contents after heating 27.86 g

Mass of empty test tube 24.56

• Single replacement• - an atom or ion in a compound is replaced by an

atom or ion of another element. • (i) Metal replacement • zinc metal reacting with copper (II) ions in solution

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

• Solid zinc metal atoms (oxidation state of zero) are oxidized to Zn2+(aq) ions

• These ions replace the copper (II) ions • The Cu2+(aq) ions are reduced to copper metal atoms

(oxidation state of zero)

• (ii) Hydrogen displacement from water

• - sodium metal reacting with cold water • 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

• - sodium metal (oxidation state of zero) is oxidized to Na+(aq) ions

• - hydrogen atoms in the water molecules (oxidation state +1) are reduced to hydrogen gas (oxidation state of zero).

• (iii) Hydrogen displacement from acids• - zinc metal reacting with hydrochloric acid • Zn(s) + 2HCl(aq) ! ZnCl2(aq) + H2(g)

• - zinc atoms (oxidation state of zero) lose electrons to• become Zn2+(aq) ions (oxidized)• - hydrogen ions (oxidation state of +1) gain electrons to

form hydrogen atoms (oxidation state of zero) which

form hydrogen molecules.

• Predictions can be made about the reactions that will and will not happen using the activity series.

• Metals are arranged in order of increasing ability to displace hydrogen, the most reactive at the top.• All metals above hydrogen in the series will displace it from an acid • All metals below hydrogen will not displace it from an acid or water • A metal relatively high in the series will displace one below it from a

solution of its ions but the reverse process is not possible

• (iv) Halogen displacement

• - chlorine gas replacing liquid bromine from a solution of potassium bromide.

• Cl2(g) + 2KBr(aq) 2KCl(aq) +Br2(l)

• Bromide ions (–1) lose electrons to become Br atoms (0) that pair to give bromine molecules.

• The chlorine atoms (0) gain electrons to form chloride ions (–1).

• Combustion• - a compound or element “burns” in oxygen.

• A hydrocarbon (CxHy) reacts with oxygen to form carbon dioxide, water and a large amount of energy

• - propane burning in oxygen• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) + energy

• - both carbon and hydrogen gain oxygen and are “oxidized”

• - oxygen is reduced from an oxidation state of 0 to one of -2.

• Biochemical reactions • - the metabolism of sugars, fats and proteins leads to

the production of energy (like combustion)

• - the metabolism of glucose • C6H12O6(g) + 6O2(g) 6CO2(g) + 6H2O(g) + energy