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THE ELECTRONIC THEORY OF OXIDATION & REDUCTION

1. Oxidation & Reduction

Simple definitions of oxidation and reduction are based on the loss/gain of oxygen or the loss/gain of hydrogen. Oxidation is the gain of oxygen or the loss of hydrogen; reduction is the loss of oxygen or the gain of hydrogen. These definitions can only be used when a chemical reaction involves hydrogen and oxygen, and therefore their usefulness is limited.

A more basic and more useful definition of oxidation and reduction is based on the loss/gain of electrons.

OXIDATION IS LOSS OF ELECTRONS

REDUCTION IS GAIN OF ELECTRONS

In reactions involving simple ions, it is usually easy to tell whether electrons are lost or gained, but it is less easy to tell when complex ions or covalent molecules are involved. Oxidation number is a useful concept for helping to decide in these more awkward cases.

2. Oxidation State (Number)

The oxidation state is used to express the oxidation state of an element, whether as the uncombined element or when combined in a compound; it consists of a + or – sign followed by a number, or it is zero. Oxidation states of metals are usually written in roman numerals eg Iron (III) chloride.

Atoms of elements have no overall charge and are therefore given an oxidation state of zero. When two elements combine, the atoms or ions of the more electropositive element have a positive oxidation state, and those of the more electronegative element a negative oxidation state. Elements become more electronegative the higher their Group number and the lower their Period number; therefore, the most electronegative element is fluorine.

The oxidation state of an element in a compound is equal to the charge which a particle of the element would carry in the compound, assuming the compound is ionic. This is a purely theoretical idea, and it is does not matter whether the compound in question is really ionic or covalent.

IGCSE TOPIC 10.7b: REDOX REACTIONS 1

e.g. compound oxidation state

NaCl Na +1 Cl -1CCl4 C +4 Cl -1HBr H +1 Br -1H2S H +1 S -2

The following general rules are useful:

All free elements (i.e. those not combined with another element) have an oxidation state of 0.

e.g. Na, Mg, Br2

In simple ions, the charge on the ion is equal to the oxidation state.

e.g. ion Na+ Fe3+ Br- O2-

oxidation number +1 +3 -1 -2

Since fluorine is the most electronegative element, it always has an oxidation state in its compounds of –1.

Combined oxygen always has an oxidation state of –2, except when in combination with fluorine or in peroxides.

e.g. compound oxidation numbers

Fe2O3 Fe +3 O -2Mn2O7 Mn +7 O -2CrO3 Cr +3 O -2

BUT ….H2O2 H +1 O -1F2O O +2 F -1

Group I elements in their compounds always have an oxidation state of +1.

Group II elements in their compounds always have an oxidation state of +2.

Hydrogen in its compounds always has an oxidation state of +1, except when it has combined with a reactive metal.

e.g. compound oxidation numbers

H2O H +1 O -2HCl H +1 Cl -1CH4 H +1 C -4

BUT ….NaH Na +1 H -1

The sum of the oxidation states of all the atoms in an uncharged molecule is zero: in an ion, the sum is equal to the charge on the ion.


e.g. compound/ion oxidation states

NH3 H +1 N -3 (-3+1+1+1=0)NH4

+ H +1 N -3 (-3+1+1+1+1=+1)H2SO4 H +1 S +6 O -2 (+1+1+6-2-2-2-2=0)

3. Number of Oxidation States

Many elements have several oxidation states:

e.g. sulphur H2S S SCl2 SO2 SO3+1 -2 0 +2 -1 +4 -2 +6 -2

chlorine HCl Cl2 HOCl ClF3 KClO3 KClO4+1-1 0 +1-2+1 +3-1 +1+5-2 +1+7-2

4. Redox ReactionsIf, during a chemical reaction, the oxidation state of an element increases (i.e. becomes more positive or less negative), then the element has lost electrons and has been OXIDISED.

Conversely, if the oxidation state of an element decreases (i.e. becomes less positive or more negative), then the element has gained electrons and has been REDUCED.

REDuction and OXidation occur together in what is called a REDOX reaction.


Oxidation

-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5

Reduction

Example:

work out the oxidation states for all the elements in the reaction

identify the oxidation and reduction steps

work out the total number of electrons transferred in each step: they should be equal

oxidation: loss of 2e- x 1

Mg + 2HCl MgCl2 + H2

0 +1 -1 +2 -1 0

reduction: gain of 1e- x 2 check: 2e- x 1 = 1e- x 2

Complete the equations for the remaining reactions in a similar way.

Mg + 2HCl MgCl2 + H2 0 +1 -1 +2 -1 0

2Fe2+ + Cl2 2Fe3+ + 2Cl- +2 0 +3 -1

2Fe3+ + S2- 2Fe2+ + S +3 -2 +2 0

Cu2+ + Zn Zn2+ + Cu +2 0 +2 0


OXIDATION & REDUCTION

QUESTION SHEET 1

Work out the oxidation state of each element in the following chemical formulae.

1. Cu …………………………………………………………………..

2. NaCl ………………………………………………………………….

3. CuS ………………………………………………………………….

4. Cl2 ………………………………………………………………….

5. Fe3+ ………………………………………………………………….

6. Cr2O3 ………………………………………………………………….

7. H2O ………………………………………………………………….

8. HNO3 ………………………………………………………………….

9. MnO4- ………………………………………………………………….

10. K2CO3 ………………………………………………………………….

11. NaClO3 ………………………………………………………………….

12. SO42- ………………………………………………………………….

13. NaNO2 ………………………………………………………………….

14. SOCl2 ………………………………………………………………….

15. Cu(NO3)2 ………………………………………………………………….

16. K2Cr2O7 ………………………………………………………………….

17. Al(OH)3 ………………………………………………………………….

18. K2MnO4 ………………………………………………………………….

19. Na3PO4 ………………………………………………………………….

20. Fe3O4 ………………………………………………………………….



QUESTION SHEET 2

Work out the oxidation state of each element in the following chemical formulae

1. Mg …………………………………………………………………..

2. KBr ………………………………………………………………….

3. CaS ………………………………………………………………….

4. Br2 ………………………………………………………………….

5. Ba2+ ………………………………………………………………….

6. Fe2O3 ………………………………………………………………….

7. Na2O ………………………………………………………………….

8. LiNO3 ………………………………………………………………….

9. MnO42- ………………………………………………………………….

10. Rb2CO3 ………………………………………………………………….

11. KBrO3 ………………………………………………………………….

12. IO4- ………………………………………………………………….

13. KNO2 ………………………………………………………………….

14. POCl3 ………………………………………………………………….

15. Sr(NO3)2 ………………………………………………………………….

16. K2CrO4 ………………………………………………………………….

17. Cr(OH)3 ………………………………………………………………….

18. KMnO4 ………………………………………………………………….

19. H3PO4 ………………………………………………………………….

20. CuCl2- ………………………………………………………………….

21. Pb3O4 ………………………………………………………………….



QUESTION SHEET 3

Work out the oxidation state of each element in the following chemical formulae

1. Ca …………………………………………………………………..

2. LiF ………………………………………………………………….

3. MgO ………………………………………………………………….

4. I2 ………………………………………………………………….

5. Cr3+ ………………………………………………………………….

6. Al2O3 ………………………………………………………………….

7. HF ………………………………………………………………….

8. Ni(NO3)2 ………………………………………………………………….

9. CrO42- ………………………………………………………………….

10. SrCO3 ………………………………………………………………….

11. NaClO4 ………………………………………………………………….

12. SO32- ………………………………………………………………….

13. NaIO3 ………………………………………………………………….

14. XeF4 ………………………………………………………………….

15. Pb(OH)2 ………………………………………………………………….

16. K2MnO4 ………………………………………………………………….

17. Al2(SO4)3 ………………………………………………………………….

18. NaVO3 ………………………………………………………………….

19. H3PO3 ………………………………………………………………….

20. NH4+ ………………………………………………………………….



QUESTION SHEET 4

Examine each of the following redox equations. Work out the state of each element in all the atoms, ions and molecules. Using these numbers, explain with reasons which substance is oxidised and which substance is reduced

SO2 + 2Mg 2MgO + S

C + H2O CO + H2

SnCl2 + HgCl2 Hg + SnCl4

H2 + Cl2 2HCl

2Fe3+ + 2I- 2Fe2+ + I2



QUESTION SHEET 5

Examine each of the following reactions.

For each equation, work out the oxidation state of each element in all the atoms, ions and molecules. Use these numbers to decide whether the change taking place is a redox reaction or not.

Where a redox reaction occurs, indicate, with reasons, which species is oxidised and which is reduced

Where the change is not a redox reaction, describe in one word the type of change taking place.

1. Mg(s) + Cl2(g) MgCl2(s)

2. NaCl(s) Na+(l) + Cl-(l)

3. 2Fe3+(aq) + 2I-(aq) 2Fe2+(aq) + I2(s)

4. C(s) + H2O(g) CO(g) + H2(g)

5. Ag+(aq) + Cl-(aq) AgCl(s)


6. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

7. 2H2(g) + O2(g) 2H2O(l)

8. Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)


REDOX EQUATIONS

Constructing Half –Equations

The half-equation shows either the oxidation or the reduction step of a redox change. In a half-equation:

only one element changes its oxidation state

The loss/gain of electrons responsible for the change in oxidation state is shown

the atoms of each element must balance

the total charge on both sides of the equation must be the same

the two half equations can be added together to give the full ionic equation

When sodium reacts with chlorine gas, sodium atoms lose an electron and are oxidised to sodium ions; whereas chlorine atoms gain an electron and are reduced to chloride ions. Since chlorine is a molecule (Cl2) both constituent atoms must gain an electron resulting in 2 ions.

Na Na+ + e-

Cl2 + 2e- 2Cl-

Since this latter process requires 2 electrons, two sodium atoms must also be involved hence;

2Na 2Na+ + 2e-

Now the two half equations can be added together to show the full ionic equation remembering to cancel the electrons.

Cl2 + 2e- 2Cl- 2Na 2Na+ + 2e-

Cl2 + 2Na 2Na+ + 2Cl-


If during the addition of two half equations the same ion appears on both sides of the equation i.e. has not changed in any way, these spectator ions should also be cancelled.

Eg in reaction of potassium bromide with chlorine

Cl2(s) + 2KBr(aq) 2KCl(aq) + Br2

Chlorine atoms are gaining electrons to form chloride ions as in the previous example.

Cl2 + 2e- 2Cl-

Whilst bromide ions in potassium bromide are losing electrons to turn into bromine

2K+ + 2Br - Br2 + 2e- + 2K+

So combining the half equation gives

2K+ + 2Br - Br2 + 2e- + 2K+

Cl2 + 2e- 2Cl-

Overall

Cl2 + 2Br - Br2 + 2Cl-



QUESTION SHEET 6Examine each of the following reactions.

For each equation, write the two half equations and combine them to get the overall ionic equation

1. Mg(s) + Cl2(g) MgCl2(s)

2. Br2(s) + 2KI 2KBr(aq) + I2

3. 2H2(g) + O2(g) 2H2O(l)

4. SO2 + 2Mg 2MgO + S

5. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)


Oxidising and reducing agentsAn oxidising agent oxidises another substance and is itself reduced in the process.

A reducing agent reduces another substance and is itself oxidised in the process.

In the reaction CuO + H2 Cu + H2O

CuO is losing oxygen and so is reduced. This happens when heated with hydrogen. Hydrogen has reduced CuO to copper metal and has itself gained oxygen and therefore been oxidised. You could also consider the oxidation states of each substance and would draw the same conclusion.

Potassium dichromate is an oxidising agent and can oxidise ethanol to a substance called ethanal. During this process the dichromate Cr2O7

2- (which is orange) is itself reduced to green Cr3+.

Cr2O72- Cr3+

Oxidation state +6 oxidation state +3 Reduction


Summary QuestionsTopic 7b Redox

1

2


3

4

5

Chlorine reacts with iron (II) ions as shown below

Cl2 + 2Fe2+ 2Fe3+ + 2Cl-

What type of reagent is chlorine in this reaction? Give a reason for your answer.

…………………………………………………………………………………………..

……………………………………………………………………………………………. (2)


6

7


8

(b) Use the data in the table above to calculate the oxidation number of copper in copper carbonate CuCO3.

……………………………………………………………………………………. (1)


oxidation numbers - barnard castle school€¦ · web viewoxidising and reducing agents an...

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