REDOX REACTI

ONS

OXIDATION NUMBER

The oxidation state of an element in an elemental state is zero.

(O2, Fe, He)

The oxidation state of an element in a monoatomic ion is the charge of that ion.

(oxidation number of iron in Fe+3 is +3) The oxidation number of Group 1 elements in all their

compounds is +1. The oxidation number of Group 2 elements in all their

compounds is +2.

OXIDATION NUMBER

The oxidation number for fluorine in a compound is -1. The oxidation number of H is almost always +1. The oxidation of O is ordinarily -2. The sum of oxidation numbers in a molecule is zero.

H2SO4

In a polyatomic ion, the sum of the oxidation numbers is the charge of the ion.

Cr2O7-2

DEFINITIONS

Oxidation: An increase in oxidation number

Reduction: A decrease in oxidation number

Reducing agent: Whatever is oxidized

Oxidizing agent: Whatever is reduced

Zn(s)+ 2H+(aq) Zn+2

(aq) + H2(g)

Zn is oxidized (ox # 0+2) Reducing agentH+ is reduced (ox # +10) Oxidizing agent

BALANCING HALF-EQUATIONS

1) Balance the atoms of the atoms of the element being oxidized or reduced.

2) Balance the oxidation number by adding electrons.

3) Balance charge by adding H+ in acidic solution, OH- in basic solution.

4) Balance hydrogen by adding H2O molecules.

5) Check to make sure oxygen is balanced.

BALANCIN

G EQUAT

IONS IN

ACIDIC

SOLU

TION

MnO4-(aq) Mn+2

(aq)

MnO4-Mn+2

MnO4- + 5e-Mn+2

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

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

BALANCIN

G EQUAT

IONS IN

BASIC S

OLUTI

ON

Cr(OH)3(s) CrO4-2

(aq)

Cr(OH)3 CrO4-2

Cr(OH)3 CrO4-2 + 3 e-

Cr(OH)3 + 5OH-CrO4-2 + 3e-

Cr(OH)3 + 5OH-CrO4-2 + 3e- + 4H2O

BALANCIN

G REDOX

EQUATIO

NS

1) Split the equation into two half-equations.

2) Balance one of the half equations with respect to atoms and charge.

3) Balance the other half equation.

4) Combine the two half equations in such a way as to eliminate electrons.

Balance the following redox equation in acidic solution.Fe+3

(aq) + MnO4-(aq)Fe+2

(aq) + Mn+2(aq)

Oxidation: Fe+3Fe+2 Reduction: MnO4-Mn+2

Balanced half equationsFe+2 -Fe+3+ e-

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

To eliminate e-, multiply the oxidation half-equation by 5 and add to the reduction half equation.5Fe+2 + MnO4

- + 5e- + 8H+ 5Fe+3 + Mn+2 + 4H2O + 5e-

5Fe+2 + MnO4- + 8H+ 5Fe+3 + Mn+2 + 4H2O

Balance the following redox equation in basic solution.Cl2(g) + Cr(OH)3(s) Cl-

(aq)+ CrO4-2

(aq)

Cl2+ 2e- 2Cl-

Cr(OH)3 CrO4-2 + 3e- + 4H2O

Multiply reduction equation by 3 and the oxidation equation by 2.3Cl2+ 6e- + 2Cr(OH)3 6Cl- + 2CrO4

-2 + 6e- + 8H2O

3Cl2+ 2Cr(OH)3 6Cl- + 2CrO4-2 + 8H2O

STOIC

HIOMETR

Y

What volume of of 0.684 M KMnO4 solution is required to react completely with 27.50 mL of 0.250 M Fe(NO3)2?

5Fe+2 + MnO4- + 8H+ 5Fe+3 + Mn+2 + 4H2O

Moles Fe+2 = 0.02750 L x 0.250 moles Fe(NO3)2 x 1 mole Fe+2 =

1 L 1 mole Fe(NO3)2

6.88 x 10-3moles Fe+2

Moles MnO4- = 6.88 x 10-3moles Fe+2 x 1 mole MnO4

- = 0.00138 moles MnO4-

5 moles Fe+2

Moles KMnO4 = moles MnO4- = 0.00138 moles KMnO4

V = moles/M 0.00138 moles KMnO4/0.684 M KMnO4 = 2.02 x 10-2L = 2.02 mL

REDOX R

EACTION

PROBLE

MSPage 9749-5664, 66, 6870, 72