Oxidation Reduction Reaction Equations Exam Practice

Oxidation-Reduction Reaction Equations:

Oxidation-reduction reactions (shortened to: redox) are electron transfer reactions. In these reactions one element will lose electrons to another element. The loss of electrons is called oxidation and the gain of electrons is called reduction.

The degree of oxidation can be determined by assigning an oxidation number. The rules for determining the oxidation number of an element are:

The rules for determining the oxidation number are:

1. Any free neutral element = 0.

2. Any simple monotonic ion is equal to the charge of the ion.

3. The sum of oxidation numbers of atoms in a molecule or molecular ion is equal to the charge (= 0 for a neutral molecule).

4. In molecules fluorine = -1.

5. Group I A is +1

6. Hydrogen is usually +1.

7. Oxygen is usually –2.

The rules are applied in order. If there is a conflict, the earlier rule prevails.

Examples:

Find the oxidation number of the underlined element:

a. Na2S The oxidation number of Sulfur is -2. Sodium is a group IA element, each with contribute +1, and the total must be zero, because the compound is neutral.

b. NaClO The oxidation number of Cl is +1. Sodium is +1 and Oxygen is -2, therefore to make the compound neutral Cl must be +1.

c. MnO4 The oxidation number of Mn is +7. Each oxygen is -2, for a total of -8, because the ion has a -1 charge, Mn must be +7.

Practice Problem #1 Find the oxidation number of the underlined element.

a. Cl2

b. Na2SO4

c. CO2

d. CH4

e. H2O2

f. HCl

g. HClO3

h. SiO2

i. HNO3

j. NH3

Writing and Balancing Oxidation-Reduction Reactions

Example:

If metallic zinc is added to a solution of copper sulfate, the zinc dissolves and solid copper precipitates out of solution.

The reaction equation is:

Cu+2(aq) + Zn(s) Cu(s) + Zn+2 (aq)

In this reaction, zinc loses electrons to copper. Zinc is oxidized and copper is reduced. Because the zinc causes the copper to be reduced, it is also called the reducing agent, and because the Cu+2 causes the zinc to be oxidized, it is the oxidizing agent. This reaction is balanced because zinc is stable after losing two electrons. If instead of adding metallic zinc to the copper solution we added metallic aluminum, the balanced reaction equation would be:

3Cu+2(aq) + 2Al(s) 3Cu(s) + 2Al+3 (aq)

because the charge must also be balanced on both sides of a reaction equation. Aluminum is only stable as a +3 ion, and each copper must transfer two electrons, therefore the true stoichiometric ratio is:

3Cu+2≏ 2 Al

Sometimes it is difficult to balance the charge in redox reactions, particularly when water, acid or base are part of the reaction, and a general method for balancing oxidation-reduction equations in aqueous solution has been developed.

A Method for balancing redox reactions:

1. Divide into half reactions, showing one substance being oxidized and the other reduced.

2. Balance all elements in half reactions except O, and H

3. Balance O with H2O.

4. Balance H with H+

5. Balance the charge of each half reaction by adding e

6. Multiply the half reactions to have the same number of electrons in each, to get rid of the e

7. Add half reactions together and simplify

8. If it is in basic solution add OH equal to the number of H+ to get rid of the H+ and simply again.

Example 1:

If solid aluminum, Al(s) , is mixed with aqueous hydrochloric acid, HCl(aq), Hydrogen gas, H2(g), and aqueous aluminum ion are produced.

Write the reactants an arrow and then the products.

Al(s) + H+ (aq) H2(g) + Al+3(aq)

Divide into half reactions:

Al Al+3

H+ H2

Balance the elements

Al Al+3

2H+ H2

Add electrons as negative charge to balance the charge in each half reaction.

Al Al+3 + 3e-

2H+ + 2e- H2

Multiply the half reactions to have the same number of electrons in each to get rid of the electrons.

2x(Al Al+3 + 3e-)

3x(2H+ + 2e- H2)

Add half reactions together

2Al + 6H+ +6e- 2Al+3 + 3H2 + 6e-

Simplify to get the balanced reaction equation.

2Al + 6H+ 2Al+3 + 3H2

Example 2:

If a solution containing VO+2(aq) is mixed with solution of sodium chromate, the vanadium is oxidized to VO2+ and chromate is reduced chromium (III) . Balance the reaction equation in acidic solution. Write the reactant,s an arrow and then the products.

VO+2(aq) + CrO4-2 (aq) VO2+ + Cr+3

Divide into half reactions:

CrO4-2 Cr+3

 

VO+2 VO2+

Balance the elements, V and Cr are already balanced, use water and H+ to balance oxygen, then hydrogen.

 

Chromate half reaction

Oxygen is balanced by adding water.

CrO4-2 Cr+3 + 4H2O

Hydrogen is balanced by adding H+

CrO4-2 + 8H+ Cr+ + 4H2O

Vanadium half reaction

Oxygen is balanced by adding water.

VO+2+ H2O VO2+

 

Hydrogen is balanced by adding H+

VO+2+ H2O VO2+ + 2H+

Add electrons to balance the charge in each half reaction.

CrO4-2 + 8H+ + 3e- Cr+3 + 4H2O

VO+2+ H2O VO2+ + 2H+ + 1e-

Multiply the half reactions to have the same number of electrons in each to get rid of the electrons

1x(CrO4-2 + 8H+ + 3e- Cr+3 + 4H2O)

3x(VO+2+ H2O VO2+ + 2H+ + 1e-)

Add half reactions together

CrO4-2 + 8H+ + 3VO+2 + 3H2O +3e- Cr+3 + 4H2O + 3VO2+ + 6H+ + 3e-

Simplify to get the balanced reaction equation.

CrO4-2 + 2H+ + 3VO+2 Cr+3 + H2O + 3VO2+

Example 3: Balance the following reaction in basic solution

MnO4 + SO3-2 Mn+2 + SO4-2

Divide into half reactions:

MnO4 Mn+2

 

SO3-2 SO4-2

Balance the elements, Mn and S are already balanced, use water and H+ to balance oxygen, then hydrogen.

 

permanganate half reaction

MnO4 +8H+ Mn+2 + 4H2O

sulfite half reaction

SO3-2 + H2O SO4-2 + 2H+

Add e- to balance the charge in each half reaction.

permanganate half reaction

MnO4 +8H+ + +5e- Mn+2 + 4H2O

sulfite half reaction

SO3-2 + H2O SO4-2 + 2H+ +2e-

Multiply the half reactions to have the same number of electrons in each to get rid of the e

2X(MnO4 +8H+ + +5e- Mn+2 + 4H2O)

5X(SO3-2 + H2O SO4-2 + 2H+ +2e- )

Add half reactions together

2MnO4 +16H+ + +10e- + 5SO3-2 + 5H2O 2Mn+2 + 8H2O + 5SO4-2 + 10H+ +10e-

Simplify

2MnO4 +6H+ + 5SO3-2 2Mn+2 + 3H2O + 5SO4-2

To change to basic solution add OH- on both sides of the reaction equation equal to the H+. Combine with the H+ to make H2O and simplify again.

2MnO4 +6H+ + 6OH + 5SO3-2 2Mn+2 + 3H2O + 5SO4-2 + 6OH

2MnO4 +6H2O + 5SO3-2 2Mn+2 + 3H2O + 5SO4-2 + 6OH

Final simplification gives the reaction equation balanced in basic solution.

2MnO4 +3H2O + 5SO3-2 2Mn+2 + 5SO4-2 + 6OH

Practice problem 4 Balance the redox reactions

1. NO3 + Au NO2 + Au +3

2. MnO4 + SO3-2 Mn+4 + SO4-2

3. PbO2 + Cl PbCl2 + Cl2

4. XeO3 + I Xe + I2

5. IO3 + AsO3-3I + AsO4-3

6. S(s) + F2(g) SF6(g)

7. SO2 + MnO4 MnO2 + SO4-2

8. PbS(s) + C(s) + O2(g) Pb(s) + CO(g) + SO2(g)

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