To determine which reactant is oxidized and which is reduced you will need to generate oxidation numbers for each element. Oxidation numbers can be determined using the following steps:

1. When summed together the value of oxidation numbers for the atoms of a molecule will equal the net charge of the molecule (for example, the net charge of CH₄ is 0 while the net charge of PO₄^2- is -2)

2. Some elements have constant oxidation numbers;

Atoms in an element is 0

H is +1

O is -2 (except in peroxides- O₂^2-, it is -1)

Alkali metals (eg Na, K) are +1

Halogens are -1 in ionic configurations (eg HCl, NaBr)

Polyatomic ions are equal to charge (eg. SO₄^2- is -2, NH₄⁺ is +1)

3. Oxidation is indicated by a loss of electrons, that is, the oxidation number for an atom increases from the reactant to the product side of the reaction.

Reduction is the gaining of electrons and is indicated by a decrease in oxidation number.

Remember: An oxidizing agent oxidizes another compound and is subsequently reduced, while a reducing agent is oxidized while reducing another compound

For the reaction CuO + H₂ → Cu +H₂O oxidation numbers are:

CuO = +2 for Cu and -2 for oxygen, summed charge = 0

H₂ = as an element, 0

Cu = 0

H₂O = +1 for each H and -2 for the O, summed charge = 0

Looking from reactant to products the oxidation number for Cu goes from +2 to 0, a gain of electrons and H goes from an oxidation number of 0 to +1, a gain of electrons. From this it can be surmised that

• Cu is reduced
• H is oxidized
• CuO is the oxidizing agent

Using the above steps, determine the oxidation numbers for the atoms,

Products

Na = 0

Cl₂ = 0

Reactants

NaCl = Na is +1, Cl is -1, summed oxidation state is 0

Na loses an electron (0 → +1)  and is oxidized

Cl gains an electron (0 → -1) and is reduced

Oxidizing agent is Cl₂

Oxidation numbers:

Products

H₂C₂O₄ = H is 2 @ +1, O is 4@-2, C is 2@+3, summed oxidation state is 0

KMnO₄ = K is 1 @ +1, O is 4@ -2, Mn is 1@+7, summed oxidation state is 0

H₂SO₄ = H is 2@+1, SO₄ is 1@-2, summed oxidation state is 0

Reactants

CO₂ = O is 2@-2, C is 1@+4, summed oxidation state is 0

MnSO₄ = SO₄ is 1@-2, Mn is 1@+2, summed oxidation state is 0

K₂SO₄ = SO₄ is 1@-2, K is 2@+1, summed oxidation state is 0

H₂O = H is 2@+1, O is 1@-2, summed oxidation state is 0

C loses an electron (+3 →+4) and is oxidized

Mn gains 5 electrons (+7 → +2) and is reduced

KMnO₄ is the oxidizing agent

Spectator ions are those which do not participate in the reaction process and therefore do not undergo a change in oxidation number. To identify the spectator ions, determine the oxidation numbers for each atom, eliminate those that do not change, write out the reduction half reaction and the oxidation half reaction and add them together

Reactants

Al = 0

HCl = H is +1, Cl is -1

Products

AlCl₃ = Al is +3, Cl is -1

H₂ = H is 0

Al loses 3 electrons and is oxidized

H gains an electron and is reduced

Cl does not participate and is the spectator ion.

Oxidation half reaction:

Al(s) Al³⁺

Reduction half reaction:

3 H⁺ 1.5 H₂

Added together:

Al_(s) + 3 H⁺ Al³⁺+ 1.5 H₂

Reactants

NaCl = Na is +1, Cl is -1

H₂O = H is 2@ +1, O is -2

Products

NaOH = Na is +1, O is -2, H is +1

H₂ = H is 0

Cl₂ = Cl is 0

H is gains an electron and is reduced

Cl loses an electron and is oxidized

Na is the spectator ion

Oxidation half reaction

2 Cl^- → Cl₂

Reduction half reaction

2 H₂O → OH^- + H₂

Added together

2 Cl^- + 2 H₂O → Cl₂ + H₂ + OH^-

Reactants

Fe₂(SO₄)₃ = SO₄ is 3@ -2, Fe is 2@ +3

NaI = Na is +1, I is -1

Products

FeSO₄ = SO₄ is -2, Fe is +2

Na₂SO₄ = SO₄ is -2, Na is 2@+1

I₂ = I is 0 

Fe is reduced (+3 → +2)

I is oxidized (-1 → 0)

Oxidation half reaction:

2 I^- → I₂

Reduction half reaction

2 Fe³⁺ → 2 Fe²⁺

Added together:

2 I^- + 2 Fe³⁺ → I₂ + 2 Fe²⁺

The following steps can be used to balance redox reactions using oxidation numbers

1. Identify the oxidation number of each element on each side of the equation. Determine which has undergone oxidation and which has undergone reduction

2. Use coefficients so that the total increase in oxidation number equals the total decrease

Products

H₂S = H is 2@+1, S is -2

HClO₃ = H is +1, O is 3@-2, Cl is +5

Reactants

H₂SO₄ = H is 2@+1, O is 4@-2, S is +6

HCl = H is +1, Cl is -1

S loses 8 electrons (-2 → +6) and is oxidized,

Cl gains 6 electrons (+5 → -1) and is reduced

To make the number of electrons lost equal to those gained, use coefficients

3S^2- → 3 H₂SO₄  results in 24 electrons lost (3 x 8)

4 HClO₃ → 4 HCl results in 24 electrons gained (4 x 6)

filling the coefficients in:

3H₂S + 4HClO₃ → 3H₂SO₄ + 4HCl

Reactants

 NH₃ = H is 3@+1, N is -3

ClO₂ = O is 2@-2, Cl is +4

Products

N₂O = O is -2, N is 2@+1

HCl = H is +1, Cl is -1

H₂O = H is 2@+1, O is -2 

N loses 4 electrons (-3 →+1) and is oxidized

Cl gains 5 electrons (+4 → -1) and is reduced

5 NH₃ → 5/2 N₂O results in 20 electrons lost

4 ClO₂ → 4 Cl^- results in 20 electrons gained

Adjusting the coefficients to give integer values and balance all atoms gives:

10 NH₃ + 8 ClO₂ → 5 N₂O + 8 HCl + 11 H₂O

Oxidation/reduction half reactions are balanced in the following manner;

1. Identify the species that is reduced and that which is oxidized by assigning oxidation states

2. Write separate reactions for reduction and oxidation

3. For each half reaction:

• balance all elements except hydrogen and oxygen
• balance oxygen using H₂0
• balance hydrogen using H⁺
• balance the charge using electrons

4. If necessary, multiply one or both of the half reactions by an integer to equalize the number of electrons transferred in the two half reactions

5. Add the half reactions and cancel identical species

6. Check that all elements and charges are balanced

follow the above steps

1. Identify the species that is reduced and that which is oxidized by assigning oxidation states

Reactants

C₂H₅OH = H is 6@+1, O is 1@-2, C is 2@-2

MnO₄^- = O is 4@-2, Mn is +7

Products

CH₃CO₂H = H is 4@+1, O is 2@ -1 (it’s a peroxide), C is 2@-1

Mn²⁺ = Mn is +2

H₂O = H is +1, O is -2

C loses an electron and is oxidized

Mn gains 5 electrons and is reduced

2. write separate reactions for reduction and oxidation

oxidation reaction

C₂H₅OH CH₃CO₂H

Reduction reaction

MnO₄^- Mn²⁺

3. For each half reaction:

Balance all elements except hydrogen and oxygen

Reactions are balanced with the exception of O and H

balance oxygen using H₂O

oxidation reaction

C₂H₅OH + H₂O CH₃CO₂H

Reduction reaction

MnO₄^- Mn²⁺ + 4H₂O

balance hydrogen using H⁺

oxidation reaction

C₂H₅OH + H₂O CH₃CO₂H + 4H⁺

Reduction reaction

MnO₄^- + 8 H⁺ Mn²⁺ + 4H₂O

balance the charge using electrons

oxidation reaction

C₂H₅OH + H₂O CH₃CO₂H + 4H⁺ + 4e^-

Reduction reaction

MnO₄^- + 8 H⁺ + 5e^- Mn²⁺ + 4H₂O

4. if necessary, multiply one or both of the half reactions by an integer to equalize the number of electrons transferred in the two half reactions

oxidation reaction

[C₂H₅OH + H₂O CH₃CO₂H + 4H⁺ + 4e-] x 5

5 C₂H₅OH + 5 H₂O 5 CH₃CO₂H + 20 H⁺ + 20e^-

Reduction reaction

[MnO₄^- + 8 H⁺ + 5e^- Mn²⁺ + 4H₂O ] x 4

4 MnO₄^- +32 H⁺ +20e^- 4 Mn²⁺ + 16 H₂O

5. Add the half reactions and cancel identical species 

5 C₂H₅OH + 4 MnO₄^- + 12 H⁺ 5 CH₃CO₂H + 4 Mn²⁺ + 11 H₂O

1. Based on the nature of the reaction, it can be seen that glucose is oxidized and oxygen is reduced (and is thus the oxidizing agent) 

2.  oxidation half reaction

C₆H₁₂O₆ C₆H₆O₄

Reduction half reaction

O₂ H₂O 

3.balance oxygen using H₂O

oxidation half reaction

C₆H₁₂O₆ C₆H₆O₄ + 2 H₂O

Reduction half reaction

O₂ 2 H₂O  

balance hydrogen using H⁺

oxidation half reaction

C₆H₁₂O₆C₆H₆O₄ + 2 H₂O + 2 H⁺

Reduction half reaction

O₂ + 4 H⁺ 2 H₂O  

balance the charge using electrons

oxidation half reaction

C₆H₁₂O₆ C₆H₆O₄ + 2 H₂O + 2 H⁺ + 2 e^-

Reduction half reaction

O₂ + 4 H⁺ + 4e^- 2 H₂O  

4. if necessary, multiply one or both of the half reactions by an integer to equalize the number of electrons transferred in the two half reactions

oxidation half reaction

[C₆H₁₂O₆ + 2 e^- C₆H₆O₄ + 2 H₂O + 2 H⁺] x 2

2 C₆H₁₂O₆  + 4 H₂O + 20 e^- 2 C₆H₆O₄ + 20 H⁺]

Reduction half reaction

O₂ + 4 H⁺ 2 H₂O + 4e 

5. add the half reactions and cancel identical species           

2 C₆H₁₂O₆ + O₂ 2 C₆H₆O₄ + 6 H₂O