CHAPTER 5

ANALYSING OXIDANTS AND REDUCTANTS

REDOX REACTIONS

• Redox reactions involve complementary processes of oxidation and reduction, and can be identified on the basis of one or more of four definitions of oxidation and reduction.

• The most commonly used definition refers to electron transfer. (OIL RIG)

• Other definitions relate to oxygen transfer, hydrogen transfer and changes in oxidation numbers.

OXYGEN TRANSFER

• Reduction is the loss of oxygen• Oxidation is the gain of oxygen

CuO(s) + H2(g) → Cu(s) + H2O(l)

REDUCTION

OXIDATION

Oxidant – CuOReductant - H2

HYDROGEN TRANSFER

• Reduction is the gain of hydrogen• Oxidation is the loss of hydrogen

Cl2(g) + 2HI(aq) → 2HCl(aq) + I2(s)

REDUCTION

OXIDATION

Oxidant – Cl2Reductant - HI

ELECTRON TRANSFER

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

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

REDUCTION

OXIDATION

Oxidant – Cu2+

Reductant - Zn

OXIDATION NUMBERS

• A key stage in the production of sulphuric acid is the conversion of SO2 to SO3 according to the equation2SO2(g) + O2(g) → 2SO3(g)

• This is a redox reaction but is not readily identified as such by the previous definitions of oxidation and reduction. To overcome this difficulty numbers called oxidation numbers can be used.

OXIDATION NUMBERS OR OXIDATION STATES – RULE 1

• Are assigned according to relatively simple rules• The oxidation number of the atoms of free

(uncombined) elements is zero• Eg. Fe in elemental iron and N in elemental nitrogen (N2)

both have an oxidation number of zero

RULE 2

• The oxidation number of a monatomic ion is the same as the charge on the ion.• In MgF2, which contains Mg2+ and F- ions the oxidation

number of magnesium is +2 and the oxidation number of fluorine is -1

RULE 3

• The oxidation number of oxygen in most of its compounds is -2• Exceptions to this rule include hydrogen peroxide, H2O2

(where oxygen has an oxidation number of -1) as well as the compound F2O where oxygen is assigned an oxidation number of +2 (and fluorine -1) because of fluorine’s higher electronegativity.

RULE 4

• The oxidation number of hydrogen in the vast majority of its compounds is +1• Exceptions to this rule are metal hydrides such as KH and

MgH2 in which hydrogen has an oxidation number of -1

RULE 5

• The most electronegative element in a compound has the negative oxidation number.• CF4 – fluorine is more electronegative than carbon so it is

assigned a negative oxidation number (-1)

RULE 6 AND 7

• The sum of the oxidation numbers in a neutral compound is zero• Na2O, (Na = +1) (O = -2)

• The sum is 2 x (+1) + -2 = 0

• The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion• CO3

2- the sum is equal to -2

WORKED EXAMPLES

• Page 52• There are 3.• There are a few steps to follow when assigning

oxidation numbers• 1. Write down the formula• 2. Assign any known oxidation numbers, use x for the

unknown oxidation number• 3. Work out the total then use some mathematics

YOUR TURN

• Page 54• Question 4 and 5

OXIDATION NUMBERS

• Oxidation numbers can change during a reaction. • This allows us to determine if a redox reaction is

occurring.• An increased oxidation number means the

element has been oxidised (oxidation is occurring)

• A decrease in oxidation number means reduction

YOUR TURN

• Page 54• Question 6

WRITING HALF EQUATIONS

• All redox reactions involve some form of electron transfer.

• This is fairly evident in the equations for redox reactions such as the reaction of zinc with dilute acids to produce hydrogen gas.

Overall: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Oxidation: Zn(s) → Zn2+(aq) + 2e-

Reduction: 2H+(aq) + 2e- → H2(g)

WRITING HALF EQUATIONS

• For more complex half equations H2O, H+ and electrons (e-) may be utilised as part of the balancing procedure.

• Remember: in any equation• The number of atoms of each element is equal on both

sides• The total charge is equal on both sides

STEPS FOR BALANCING HALF EQUATIONS

1. Balance all elements except hydrogen and oxygen in the half equation2NO3

- → N2O

2. Balance the oxygen atoms by adding water2NO3

- → N2O + 5H2O

STEPS FOR BALANCING HALF EQUATIONS

3. Balance the hydrogen atoms by adding H+ ions 2NO3

- + 10H+ → N2O + 5H2O

4. Now balance the charges by adding electrons2NO3

- + 10H+ + 8e- → N2O + 5H2O

3. Finally put in the states2NO3

-(aq) + 10H+(aq) + 8e- → N2O(g) + 5H2O(l)

WORKED EXAMPLE 5.3

• Page 55• Your Turn • Page 55• Question 7 and 10

VOLUMETRIC ANALYSIS

• Volumetric analysis is not only for acid-base reactions. We can use it for redox reaction too.

• For redox titrations we react an oxidant with a reductant.

• Like acid-base titrations one solution is usually pipetted into a conical flask and the other is dispensed from a burette.

• Some redox titrations will need an indicator but for some a colour change will occur due to the reacting solutions

VOLUMETRIC ANALYSIS

• Worked Example 5.4 on age 56

YOUR TURN

• Page 59• Question 25 and 26