REDOX
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-
Ø 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-
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 Number
The oxidation number 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.
Atoms of elements have no overall charge and are therefore given an oxidation number 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 number 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.
e.g. compound oxidation numbers
NaCl Na +1 Cl -1
CCl4 C +4 Cl -1
HBr H +1 Br -1
H2S H +1 S -2
The following general rules are useful:
All free elements (i.e. those not combined with another element) have an oxidation number of 0.
e.g. Na, Mg, Br2
In simple ions, the charge on the ion is equal to the oxidation number.
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 number in its compounds of –1.
Combined oxygen always has an oxidation number of –2, except when in combination with fluorine or in peroxides.
e.g. compound oxidation numbers
Fe2O3 Fe +3 O -2
Mn2O7 Mn +7 O -2
CrO3 Cr +3 O -2
BUT ….
H2O2 H +1 O -1
F2O O +2 F -1
Group I elements in their compounds always have an oxidation number of +1.
Group II elements in their compounds always have an oxidation number of +2.
Hydrogen in its compounds always has an oxidation number of +1, except when it has combined with a reactive metal.
e.g. compound oxidation numbers
H2O H +1 O -2
HCl H +1 Cl -1
CH4 H +1 C -4
BUT ….
NaH Na +1 H -1
The sum of the oxidation numbers 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 numbers
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 Reactions
If, during a chemical reaction, the oxidation number 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 number 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.
Example:
· work out the oxidation numbers 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 number 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 ………………………………………………………………….
OXIDATION & REDUCTION
QUESTION SHEET 2
Work out the oxidation number 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 ………………………………………………………………….
OXIDATION & REDUCTION
QUESTION SHEET 3
Work out the oxidation number 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+ ………………………………………………………………….
OXIDATION & REDUCTION
QUESTION SHEET 4
Examine each of the following redox equations. Work out the oxidation number 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
OXIDATION & REDUCTION
QUESTION SHEET 5
Examine each of the following reactions.
· For each equation, work out the oxidation number 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
When constructing half-equations for reactions which take place in aqueous solution:
· Water can be used as a source of oxygen atoms on the reactant side of the equation, the hydrogen appearing in the products as H+.
· Any extra oxygen atoms on the reactant side of the equation can be converted into water by reaction with hydrogen ions from an acid.
· The oxidation numbers of hydrogen and oxygen do not change.
Combining Half –Equations
The overall equation for a redox change can be obtained by combining the half-equations for the oxidation and reduction steps in such a way that the number of electrons donated by the reducing agent is equal to the number accepted by the oxidising agent.
Any molecules or ions which appear on both sides of the overall equation, such as H+ and H2O, then need to be cancelled down.
Examples:
Sometimes the same species oxidises and reduces itself simultaneously. This process is called disproportionation.
TOPIC 12.7:REDOX REACTIONS 6