redox reactions. oxidation number the oxidation state of an element in an elemental state is zero....

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