Aqueous Reactions and Solution Stoichiometry

(continuation)

1. Oxidation-Reduction Reactions

2. The Activity Series

3. Balancing REDOX Equations

Terminology for Redox Reactions

• OXIDATION—loss of electron(s) by aspecies; increase in oxidation number.

• REDUCTION—gain of electron(s);decrease in oxidation number.

• OXIDIZING AGENT—electron acceptor; species are reduced.

• REDUCING AGENT—electron donor; species are oxidized.

- reaction where electrons are exchanged.2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

2Na0(s) Na+(aq) + 2 e-

oxidation – loss of electrons

2 H+(g) + 2 e- H20(g)

reduction – gain of electrons

Oxidation-Reduction Reactions

These processes occur simultaneously –

you cannot have an oxidation without a reduction and vice-versa!!!

GeneralRules for Assigning an Oxidation Number (O.NO.N.)

1) For an atom in its elemental neutral form (Na, O2 Cl2, etc.): O.N. = 0O.N. = 0

2) For a monoatomic ion: O.N. = ion chargeO.N. = ion charge

3) The sum of O.N. values for the atoms in a compoundequals zerozero.

4) The sum of the O.N. values for the atoms in apolyatomic ion equals the ion’s chargeequals the ion’s charge.

Rules for specific atoms or periodic table groups:

1) For Group IA(1): O.N. = +1 in all compounds2) For Group IIA(2): O.N. = +2 in all compounds3) For hydrogen: O.N. = +1 in combination with nonmetals

O.N. = -1 in combination with metals and boron4) For fluorine: O.N. = -1 in all compounds5) For oxygen: O.N. = -1 in peroxides

O.N. = -2 in all other compounds (except with F )6) For Group VIIA(17) O.N. = -1 in combination with metals, nonmetals

(except O), and other halogens lower in the group.

Highest and Lowest Oxidation Numbers of Chemically Reactive Main-Group Elements

1

+1-1

H

non-metals

metalloids

metals

F

Cl

Br

I

At

ONCB

1A

+1

2A

+2

2

3

4

5

6

7

Li Be

3A

+3 +4-4+44A 5A 6A 7A

+5 +6 +7-3 -2 -1

S

Se

Te

Po

P

As

Sb

Bi

Si

Ge

Sn

Pb

Al

Ga

In

Tl

Na Mg

K Ca

Rb Sr

Cs Ba

RaFr

Peri

od

Transition MetalsPossible Oxidation States

+2,+1

IIIB IVB VB VIB VIIB IB IIBVIIIB

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

HgAuPtIrOsReWTaHfLa

+3

+3

+3

+2

+2+1

+3,+1

+2

+4,+2

+4,+2

+3,+2+3,+2+7,+6+4,+3

+2+6,+3+2

+5,+4+3+2

+5,+4+2 +4,+

3

+4,+3+1

+8,+5+4,+3

+7,+5+4

+6,+5+4,+3

+5,+4+3

+6,+5+4

+7,+5+4

+8,+6+4,+3

+2

+2,+1

+4,+3+2

+4,+3

+4,+3

Determining the Oxidation Number of an Element in a Compound

Problem: Determine the oxidation number (O.NO.N.) of each element in sulfuric acid

H2SO4: The O.N. of H is +1, so the SO42- group must sum to

-2.

The O.N. of each O is -2 for a total of -8. So the Sulfur atom is +6.

Recognizing Oxidizing and Reducing Agents

NiO(s) + CO(g) → Ni(s) + CO2 (g)

NiO(s) + CO(g) Ni(s) + CO2 (g)

+2-2

+2-2 0 +4 -2

Ni[+2] Ni[0]Ni is Reduced

C[+2] C[+4]C is oxidized

CO is the reducing agent and NiO is the oxidizing agent.

Types of Redox Reactions

Redox Combination Reactions (Intermolecular)-when reducing and oxidizing agents are in different compounds

P4(s) + 6 Cl2(g) → 4 PCl3(l)

0 0 +3 -1

P4(s) + 10 Cl2(g) → 4 PCl5(l)

0 0 +5 -1

Fe(s) + O2 (g) → Fe2O3 (s)

0 0 +3 -2

Rusting

Redox Decomposition Reactions (Intramolecularreactions) -when both reducing and oxidizing agents are in the same compound

2 HgO (s) → 2 Hg(l) + O2(g)

2 Hg2O(s) → 4 Hg(l) + O2(g)

0 0

0 0

+2 -2

+1 -2

2 H2O2 (l) → 2 H2O(l) + O2(g)+1 -2 0+1 -1

Oxygenation and Hydrogenation reactions

4 Li (s) + O2(g) → 2 Li2O(s)0 0 +1 -2

2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)+2 -2 0 +4 -2+2 -2

2 Na(l) + H2(g) → 2 NaH(s)0 0 +1 -1

Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g)0 0 +1 -2+3 -2

Disproportionation reactions

(self-reduction/self oxidation)

2 H2O2 (l) → 2 H2O(l) + O2(g)+1 -2 0+1 -1

2 Cu+(aq) → Cu2+(aq) + Cu(s) (net ionic equation)

+1 0+2

Displacement Reactions

2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s) (overall equation)0 0+2 +5 -2+1 +5 -2

2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) (net ionic equation)0 0+2+1

Cl2(g) + 2 KI(aq) → I2(s) + 2 KCl(aq) (overall equation)+1 -1 +1 -100

-1 -100Cl2(g) + 2 I-(aq) → I2(s) + 2 Cl-(aq) (net ionic equation)

+

+

+

+

+

-

-

electrode electrolyte

-

-

-

Redox potential depends on the atomic structure and hydratationtendency of the element

The Activity Series is based on the abilityof element to lose electrons in aqueous solutions

- Some metals are more easily oxidized than others.Activity series - a list of metals arranged in decreasing ease of oxidation

- The higher the metal on the activity series, the more active that metal.

- Any metal can be oxidized by the ions of elements below itHydrogen is in the activity series even though it is a

nonmetal

EN<1.4

1.4< EN <1.9

1.9< EN <2.54most can be dissolved byoxidizing acids (HNO3)

Tips on Balancing Equations

• Never add O2, O atoms, or O2- to balance oxygen.

• Never add H2 or H atoms to balance hydrogen.

• Be sure to write the correct charges on all the ions.

• Check your work at the end to make sure mass and charge are balanced.

Balancing REDOX Equations:The Oxidation Number (O.N.)Method

Step 1) Assign oxidation numbers to all elements in the equation.

Step 2) From the changes in oxidation numbers, identify the oxidized andreduced species.

Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes.

Step 4) Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients.

Step 5) Complete the balancing by inspection, adding states of matter.

REDOX Balancing Using O. N.

Fe+2(aq+ MnO4

-(aq)+ H3O+

(aq) Fe+3(aq)+ Mn+2

(aq)+H2O(aq

-1e-+2 +3

+7 +2+5 e-

Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the manganese.

5 Fe+2(aq) + MnO4

-(aq) + H3O+

(aq) 5 Fe+3(aq) + Mn+2

(aq) + H2O(aq)

Make four water molecules from the 4 oxygen atoms from the MnO4- and

protons from the acid, and this will require 8 protons, or 8 hydroniumions. This will give a total of 12 water molecules formed.

5 Fe+2(aq) + MnO4

-(aq) + 8 H3O+

(aq) 5 Fe+3(aq) + Mn+2

(aq) + 12 H2O(aq)

5 FeCl2 + KMnO4 + 8HCl=5 FeCl3 + MnCl2 +KCl + 4 H2O

Balancing Redox Equations by half-reactions method

•Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product•Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.•Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation•Step 4 Balance hydrogen.

–For half-reaction in acidic solution, add H+ on to the side deficient in hydrogen. –For a half-reaction in basic solution, add H2O to the side that is deficient in hydrogen and an equal amount of OH- to the other side

•Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.•Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication•Step 7 Check for balance

Example: Balancing Redox Reactions

Dithionate ion reacting with chlorousconditionsacidicacid in aqueous

S2O62-(aq) +HClO2(aq) --> SO4

2-(aq) + Cl2(g)

Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the

species that is reduced and its product

S2O62- --> SO4

2- HClO2 --> Cl25+ 6+ 3+ 0

OXIDATION REDUCTION

Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and

hydrogen equal on the two sides of each half-equation.

S2O62- --> 2 SO4

2-

2 HClO2 --> Cl2

Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-

equationS2O6

2- --> 2 SO42-

(8Os)(6Os)2 H2O +

(Need 2 Os)

2 HClO2 --> Cl2(4Os)

+ 4 H2O(Need 4 Os)

Step 4 Balance hydrogen. For half-reaction in acidic solution, add H+ on to the side deficient in hydrogen. For a half-reaction in basic solution, add H2O to the

side that is deficient in hydrogen and an equal amount of OH- to the other side

Need 4 Hs+ 4H+2H2O + S2O6

2- → 2SO42-

(4 Hs)

2HClO2 → Cl2 + 4H2ONeed 6 Hs (2 Hs) (8 Hs)6H+ +

Step 5 Balance charge by inserting e-

(electrons) as a reactant or product in each half-reaction.

2H2O + S2O62- --> 2SO4

2- + 4H+ + 2 e-

Oxidation reaction, electrons are lost on the reactant side (gained on the product side)

6H+ + 2HClO2 + 6 e- --> Cl2 + 4H2O

Reduction reaction, electrons are gained on the reactant side

Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by

the reduction.

Oxidation Reaction has lost 2 electrons

Reduction reaction has gained 6 electrons

∴ Multiply the oxidation reaction by 3.

This will give 6 electrons on both sides of the

reaction

6H2O + 3S2O62- → 6SO4

2- + 12H+ + 6e-

Step 6 Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication

6H2O + 3S2O62- --> 6SO4

2- + 12H+ + 6e-+

6H+ + 2HClO2 + 6e- --> Cl2 + 4H2O

2H2O + 3S2O62- + 2HClO2 --> 6SO4

2- + 6H+ + Cl2

Step 7 Check for balance

2H2O + 3S2O62- + 2HClO2 --> 6SO4

2- + 6H+ + Cl2

Left side

H 6

O 24

S 6

Cl 2

Right side

H 6

O 24

S 6

Cl 2