In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half-reaction. In either method H2O(l), OH?(aq), and H+(aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditions.
Acidic solution
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is
SO42?(aq)+Sn2+(aq)?H2SO3(aq)+Sn4+(aq)
Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:
SO42?(aq)+Sn2+(aq)+ ????H2SO3(aq)+Sn4+(aq)+ ???
Part A
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms.
Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).
Basic solution
Potassium permanganate, KMnO4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium fluoride:
MnO4?(aq)+F?(aq)?MnO2(s)+F2(aq)
Since this reaction takes place in basic solution, H2O(l) and OH?(aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:
MnO4?(aq)+F?(aq)+ ????MnO2(s)+F2(aq)+ ???
Part B
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH?(aq) in the blanks where appropriate. Your answer should have six terms.
Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).

We write both half-reactions (reduction and oxidation)
We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
We add electrons to balance electrically the half-reaction
We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
We add both half-reactions and take the numbers to the general equation.

Acidic solution

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Basic solution

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.