OXIDATION AND REDUCTION REACTIONS

CHAPTER 7

REDOX REACTIONS

Redox reactions: - oxidation and reduction reactions

that occurs simultaneously.

Oxidation: 1) addition of oxygen 2) loss of hydrogen 3) loss of electron 4) increase in the oxidation state /

oxidation number

Reduction: 1) loss of oxygen 2) addition of hydrogen 3) gain of electrons 4) decrease in the oxidation

state / oxidation number

Electron Transfer Reactions

• Oxidation reactions: half-reaction that involves loss of electrons

• Reduction reactions: half-reaction that involves gain of electrons

• Oxidizing agents: substance that accepts electrons (substances that are reduced)

• Reducing agents: substances that donates electrons (substances that are oxidized)

Electron Transfer Reactions

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

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

Zn is oxidizedZn Zn2+ + 2e-

Cu2+ is reducedCu2+ + 2e- Cu

Zn is the reducing agent

Cu2+ is the oxidizing agent

2) Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?

Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)

Cu Cu2+ + 2e-

Ag+ + 1e- AgAg+ is reducedAg+ is the oxidizing agent

Examples:

1)

Oxidation number

The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to the charge on the ion.Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In H2O2 and O2

2- it is –1.

4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2

-, is –½.

The Oxidation Numbers of Elements in their Compounds

O = –2

H = +1

3x(–2) + 1 + ? = –1

C = +4

Examples:

What are the oxidation numbers of all the elements in this folowing compounds or ion?

i) HCO3-

Na = +1

O = -23x(-2) + 1 + ? = 0

I = +5

ii) NaIO3

iii) IF7 iv) K2Cr2O7

v) Li2Ovi) HNO3

Titrations in Redox Reactions

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

Dark purple (MnO4-) to light pink

(Mn2+)

volume red moles red moles oxid M oxid

0.1327 mol KMnO4

1 Lx

5 mol Fe2+

1 mol KMnO4

x1

0.02500 L Fe2+x0.01642 L = 0.4358 M

M

red

rxn

coef.

V

oxid

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

16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution?

16.42 mL = 0.01642 L25.00 mL = 0.02500 L

Balancing Redox Equations

Ion electron method

Balancing Redox Equations for Reaction Takes Place in Acidic

Solution Following is the guideline for balancing

an oxidation-reduction equation in acidic solution by the ion-electron method.

Step 1:- Write the two half-reactions that contain the elements being oxidised and reduced using the entire formula of the ion or molecule.

Step 2:- Balance all the elements except oxygen and hydrogen.

Step 3:- Balance oxygen using H2O and hydrogen using H+.

Step 4:- Balance the charge using electrons. If necessary, multiplying one or both balanced half-reaction by integers to equalize the number of electrons transferred in the two half reactions.

Step 5:- Add the half-reactions and cancel identical species.- Verifying – check to be sure that the elements and charges are balanced.

Example 1:

Balance the following equation by the ion-electron method.

Cr2O72- + Cl- Cr3+ + Cl2

Solution:

Determine the oxidation number of the species involved.

2Oxn(Cr) + 7Oxn(O) = -2

2Oxn(Cr) + 7(-2) = -2

Oxn(Cr) = +6

The oxidation number of Cr is +6

The oxidation number of Cl- is -1 whereas the oxidation number of Cl2 is zero.

Step 1:The ionic equation is divided in 2 distinct half-reactions. One representing oxidation and the other representing reduction.

Cr2O72- (aq) Cr3+ (aq) ..(1)(reduction)

Cl- (aq) Cl2 ..(2)(oxidation)

Step 2 : Balance the atoms in each half- equation separately (except H and O)

Cr2O72- (aq) 2Cr3+ (aq) ……(3)

2Cl- (aq) Cl2 ……(4)

Step 3 : For reaction in acid medium, add H2O to balance the O atoms and H+ to balance the H atoms.

14H+ + Cr2O72- (aq) 2Cr3+ (aq) + 7H2O …(5)

2Cl- (aq) Cl2 …(6)

Step 4: Balance the charges of each half- reaction by adding electron(s) to either the left or right side of the equation.

If the electron(s) appears on the right, the process is an oxidation process.

If electron(s) appears on the left, it is a reduction process.

- The net charge on the left is +12. The net charge of Cr on the right is +6.

- Therefore, we add 6 electrons (-6) to the left so that the net charges on both sides are equal +6

6 e- + 14H+ + Cr2O72- (aq) 2Cr3+ (aq) + 7H2O …(7)

- For the 2nd half-equation, the net charges on the left is -2 while the net charge on the right is 0.

- Therefore, we add 2 electrons to the right so that the net charges on both sides are equal -2.

2Cl- (aq) Cl2 + 2 e- …(8)

- Equalise the number of electrons in both half-equations by multiplying one or two half-equations with appropriate coefficients.

- By looking at the 2 half equations, the 2nd half-equation is multiplied by 3 to balance the electrons on both half-reaction.

Cr2O72- (aq) + 14H+ + 6 e- 2Cr3+ (aq) + 7H2O …

(7)

2Cl- (aq) Cl2 + 2 e- …(8)

(equation (8) 3) 6Cl- (aq) 3Cl2 + 6 e- …(9)

Step 5: Then, add the 2 half-reaction equations together and balance the final equation by inspection.

The electrons on both sides should be cancelled.

Cr2O72- (aq) + 14H+ + 6 e- 2Cr3+ (aq) + 7H2O …(7)

6Cl- (aq) 3Cl2 + 6 e- …(9)

(7) + (9) Cr2O72- (aq) + 14H+ + 6Cl- (aq) 2Cr3+ (aq) + 7H2O

+ 3Cl2