Chapter 4: The Major Classes of Chemical Reactions

4.1 The Role of Water as a Solvent4.2 Precipitation Reactions and Acid-Base Reactions4.3 Oxidation - Reduction (Redox) Reactions4.4 Counting Reactants and Products in Precipitation, Acid-Base, and Redox Processes4.5 Reversible Reactions: An Introduction to Chemical Equilibrium

C142B Lecture NotesCampbell / Callis

The Role of Water as a Solvent: The solubility of Ionic Compounds

Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution.

Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes.

NaCl(s) + H2O(l) Na+(aq) + Cl -(aq)

When sodium chloride dissolves into water the ions become solvated,and are surrounded by water molecules. These ions are called “aqueous”and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

Fig 4.1 (P135) Electrical Conductivity of Ionic Solutions

Fig. 4.2

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I

Problem: How many moles of each ion are in each of the following:

a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 mL of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water

a) Na2CO3(s) 2 Na+(aq) + CO3-2(aq)

moles of Na+ = 4.0 moles Na2CO3 x

= 8.0 moles Na+ and 4.0 moles of CO3-2 are present

H2O

2 mol Na+

1 mol Na2CO3

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II

b) RbF(s) Rb+ (aq) + F - (aq) H2O

moles of RbF =

c) FeCl3(s) Fe+3 (aq) + 3 Cl - (aq)H2O

moles of FeCl3 = 9.32 x 1021 formula units x

moles of Cl - =

moles Fe+3 =

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II

b) RbF(s) Rb+ (aq) + F - (aq) H2O

moles of RbF = 46.5 g RbF x = 0.445 moles RbF 1 mol RbF 104.47 g RbF

thus, 0.445 mol Rb+ and 0.445 mol F - are present

c) FeCl3(s) Fe+3 (aq) + 3 Cl - (aq)H2O

moles of FeCl3 = 9.32 x 1021 formula units 1 mol FeCl3

6.022 x 1023 formula units FeCl3

x= 0.0155 mol FeCl3

moles of Cl - = 0.0155 mol FeCl3 x = 0.0465 mol Cl - 3 mol Cl -

1 mol FeCl3

and 0.0155 mol Fe+3 are also present.

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III

d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq)H2O

Converting from volume to moles:

Moles of ScBr3 = 75.0 mL x x 1 L103 mL

Moles of Br - =

e) (NH4)2SO4 (s) 2 NH4+ (aq) + SO4

- 2 (aq)H2O

Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4

+ 2 mol NH4+

1 mol(NH4)2SO4

and 7.8 mol SO4- 2 are also present.

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III

d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq)H2O

Converting from volume to moles:

Moles of ScBr3 = 75.0 mL x x = 0.042 mol ScBr3 1 L103 mL

0.56 mol ScBr3

1 LMoles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 3 mol Br -

1 mol ScBr3

0.042 mol Sc+3 are also present

e) (NH4)2SO4 (s) 2 NH4+ (aq) + SO4

- 2 (aq)H2O

Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4

+ 2 mol NH4+

1 mol(NH4)2SO4

and 7.8 mol SO4- 2 are also present.

Fig. 4.3

The Solubility of Ionic Compounds in Water

The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compoundand the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compoundsmay be several orders of magnitude less than ones that are called“soluble” in water, for example:

Solubility of NaCl in water at 20oC = 365 g/LSolubility of MgCl2 in water at 20oC = 542.5 g/LSolubility of AlCl3 in water at 20oC = 699 g/LSolubility of PbCl2 in water at 20oC = 9.9 g/LSolubility of AgCl in water at 20oC = 0.009 g/LSolubility of CuCl in water at 20oC = 0.0062 g/L

The Solubility of Covalent Compounds in Water

The covalent compounds that are very soluble in water are the oneswith -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol (C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze.

CHH

HO H Methanol = Methyl Alcohol

Other covalent compounds that do not contain a polar center, or the-OH group are considered “Non-Polar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in Gasoline and Oil. This leads to the obvious problems in Oil spills, wherethe oil will not mix with the water and forms a layer on the surface!

Octane = C8H18 and / or Benzene = C6H6

Acids - A group of Covalent molecules which loose Hydrogen ions to water molecules in solution

HI(g) + H2O(l) H+ (aq) + I - (aq)

When gaseous hydrogen Iodide dissolves in water, the attraction of theoxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an Iodide ion in solution. We can write the Hydrogen atom in solution as either H+(aq) or as H3O+(aq) they mean the same thing in solution. The presence of a Hydrogen atom that is easily lost in solution is an “Acid”and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HIwas dissolved.

HI(g) + H2O(l) H3O+(aq) + I -(aq)

HI(g) H+(aq) + I -(aq)H2O(l)

Fig. 4.4

Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of AcidsProblem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H2SO4 into sufficient water to produce 2.30 Liters of sulfuric acid solution?Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity.Solution: Two moles of H+ are released for every mole of acid:

H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq)

Molarity of H3O+ (or simply H+) =

Moles H2SO4 = = 1.58 moles H2SO4155 g H2SO41 mole H2SO4

98.09 g H2SO4

x

Molarity of SO4- 2 =

Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of AcidsProblem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H2SO4 into sufficient water to produce 2.30 Liters of sulfuric acid solution?Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity.Solution: Two moles of H+ are released for every mole of acid:

H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq)

Molarity of H+ = 2 x 0.687 mol H+ / 2.30 liters = 0.597 Molar in H+

Moles H2SO4 = = 1.58 moles H2SO4155 g H2SO41 mole H2SO4

98.09 g H2SO4

x

1.58 mol SO4-2

2.30 L solutionMolarity of SO4

- 2 = = 0.687 Molar in SO4- 2

Fig. 4.5

Precipitation Reactions: A solid product is formedWhen ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together.

Pb(NO3) (aq) + NaI(aq) Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + I- (aq)

When we add these two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide andSodium Nitrate formed, to determine which will happen we must look atthe Solubility Rules (P 141) to determine what could form. Rule 3a in Table 4.1 indicates that Lead Iodide will be insoluble, so a precipitate will form! The remainder (sodium sulfate) is soluble (Rule 4a).

Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

Precipitation Reactions: Will a Precipitate form?

If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate?

KCl(aq) + NH4NO3(aq) = K+ (aq) + Cl-(aq) + NH4+(aq) + NO3

-(aq)

By exchanging cations and anions we see that we could have PotassiumChloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1:

If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate,will we get a precipitate? Table 4.1 shows that:

Therefore we get:

Na2SO4(aq) + Ba(NO3)2(aq) BaSO4( ) + 2 NaNO3( )

Precipitation Reactions: Will a Precipitate form?

If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate?

KCl(aq) + NH4NO3(aq) = K+ (aq) + Cl-(aq) + NH4+(aq) + NO3

-(aq)

By exchanging cations and anions we see that we could have PotassiumChloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1 shows all possible products as soluble, so there is no net reaction!

KCl(aq) + NH4NO3(aq) = No. Reaction!

If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate,will we get a precipitate? Table 4.1 shows that Barium Sulfate is insoluble, therefore we will get a precipitate!

Na2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2 NaNO3(aq)

Fig. 4.6

Predicting Whether a Precipitation Reaction Occurs (see Table 4.1); Writing Equations:

a) Calcium Nitrate and Sodium Sulfate solutions are added together.

Ca(NO3)2(aq) + Na2SO4(aq) CaSO4(s) + NaNO3(aq)

Ca2+(aq) + 2 NO3-(aq) +2 Na+(aq) + SO4

-2(aq) CaSO4(s) + 2Na+(aq) + 2 NO3

-(aq)

Molecular Equation

Total Ionic Equation

Net Ionic Equation

Ca2+(aq) + SO4-2(aq) CaSO4 (s)

Spectator Ions are Na+ and NO3-

b) Ammonium Sulfate and Magnesium Chloride are added together.In exchanging ions, no precipitates will be formed, accd. To Table 4.1,so there will be no reactions occurring! All ions are spectator ions!

Acid - Base Reactions : Neutralization Rxns.An Acid is a substance that produces H+ (H3O+) ions when dissolved in water.A Base is a substance that produces OH - ions when dissolved in water.

Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium orhydroxide ions, resp.. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate only to a tiny extent (<<100%) and are weak electrolytes.

The generalized reaction between an Acid and a Base is:

HX(aq) + MOH(aq) MX(aq) + H2O(l)

Acid + Base = Salt(aq) + Water

Fig. 4.7

Table 4.2 (P 143) Selected Acids and Bases

Acids Bases

Strong (100% to H+) Strong (100% to OH-) Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH)2

Nitric acid, HNO3 Strontium hydroxide, Sr(OH)2

Sulfuric acid, H2SO4 Barium hydroxide, Ba(OH)2

Perchloric acid, HClO4

Weak (tiny % to H+) Weak (tiny % yield of OH-) Hydrofluoric, HF Ammonia, NH3

Phosphoric acid, H3PO4

Acidic acid, CH3COOH (or HC2H3O2)

Writing Balanced Equations for Neutralization Reactions - I

Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and HydroIodic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq)Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts.Solution:

a) Ca(OH)2(aq) + 2HI(aq) CaI2(aq) + 2H2O(l)

Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) Ca2+(aq) + 2 I -(aq) + 2 H2O(l)

2 OH -(aq) + 2 H+(aq) 2 H2O(l)

Writing Balanced Equations for Neutralization Reactions - II

b) LiOH(aq) + HNO3(aq)

c) Ba(OH)2(aq) + H2SO4(aq)

Writing Balanced Equations for Neutralization Reactions - II

b) LiOH(aq) + HNO3(aq) LiNO3(aq) + H2O(l)

Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)

Li+ (aq) + NO3

-(aq) + H2O(l)

OH -(aq) + H+(aq) H2O(l)

c) Ba(OH)2(aq) + H2SO4(aq) BaSO4(s) + 2 H2O(l)

Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42 -(aq) BaSO4(s) + 2 H2O(l)

Ba2 +(aq) + 2 OH -(aq) + 2 H+(aq) + SO42 -(aq) BaSO4(s) + 2 H2O(l)

Fig. 4.8

Finding the Concentration of Acid from an Acid - Base TitrationVolume (L) of base (difference inburet readings) needed to titrate

Moles of base needed to titrate

Moles of acid which were titrated

M (mol/L) of original acid soln.

M (mol/L) of base

molar ratio

volume (L) of acid

Potassium Hydrogenphthalate KHC8H4O4

C

O

O K+

C

OH

O

K+

O

C

O

O

C

O

H+

= “KHP” = a common acid used to titrate bases (M = 204.2 g/mol)

Finding the Concentration of Base from an Acid - Base Titration - I

Problem: A titration is performed between Sodium Hydroxide andKHP (204.2 g/mol) to standardize the base solution, by placing 50.00 mg of solid Potassium Hydrogenphthalate in a flask with a few drops of an indicator. A buret is filled with the base, and the initial buret reading is 0.55 mL; at the end of the titration the buret reading is 33.87 mL. What is the concentration of the base?Plan: Use the molar mass of KHP to calculate the number of moles of the acid, from the balanced chemical equation, the reactionis equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base.Solution: The rxn is: KHP(aq) + OH-(aq) --> KP-(aq) + H2O or

HKC8H4O4(aq) + OH -(aq) KC8H4O4-(aq) + H2O(aq)

Finding the Concentration of Base from an Acid - Base Titration - IImoles KHP = 50.00 mg KHP

Volume of base = Final buret reading - Initial buret reading =

one mole of H+ : one mole of OH-; therefore ____________ moles of KHP will titrate _____________ moles of NaOH.

molarity of NaOH = = moles L

molarity of base =

Finding the Concentration of Base from an Acid - Base Titration - IImoles KHP = x = 0.00024486 mol KHP50.00 mg KHP

204.2 g KHP 1 mol KHP

1.00 g1000 mg

Volume of base = Final buret reading - Initial buret reading = 33.87 mL - 0.55 mL = 33.32 mL of base

one mole of H+ : one mole of OH-; therefore 0.00024486 moles of KHP will titrate 0.00024486 moles of NaOH.

molarity of NaOH = = 0.07348679 moles per liter0.00024486 moles 0.03332 L

molarity of base = 0.07349 M

Fig. 4.9

Fig. 4.10

Fig. 4.11

Highest and Lowest oxidation numbers of Chemically reactive main-group Elements

1

+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

odFig. 4.12

Fig. 4.13

Rn

Xe

Kr

Ar

NeFONCBBeLi

HeHPeriod

1

2

3

4

5

6

IA

IIA IIIA IVA VA VIA VIIA

VIIIA Common Ox #s: Main Group Elements

Na

K

Rb

Cs

Mg

Ca

Sr

Ba

Al

Ga

In

Tl

Si

Ge

Sn

Pb

P

As

Sb

Bi

S

Se

Te

Po

Cl

Br

I

At

+1 -1

+1

+1

+1

+1

+1

+2

+2

+2

+2

+2

+3

+3

+3, +2

+3,+2+1

+3,+1

+4,+2-1,-4

+4,-4

+4,+2-4

+4,+2,-4

+4,+2

-1

+7,+5+3,+1

-1

-1+7,+5+3,+1-1+7,+5+3,+1-1+7,+5+3,+1

+2

+6,+4+2

+2

-1,-2

+6,+4+2,-2

+6,+4-2

+6,+4-2

+6,+4+2,-2

+3

+5,+3-3

+5,+3-3

+5,+3-3

all from+5 -3

Transition MetalsPossible Oxidation States

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

+4,+3+2

+4,+3

+4,+3

+2

+2

+2,+1

+2,+1

+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

Determining the Oxidation Number of an Element in a Compound

Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a) Iron III Chloride b) Nitrogen Dioxide c) Sulfuric acidPlan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge.Solution: a) FeCl3 This compound is composed of monoatomic ions. The Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3.

b) NO2

c) H2SO4

Determining the Oxidation Number of an Element in a Compound

Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a) Iron III Chloride b) Nitrogen Dioxide c) Sulfuric acidPlan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge.Solution: a) FeCl3 This compound is composed of monoatomic ions. The Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3.

b) NO2 The Ox. No. of oxygen is -2 for a total of -4. Since the Ox. No. in a compound must add up to zero, the Ox. No. of N is +4.

c) H2SO4 The Ox. No. of H is +1, so the SO42- group must sum to -2.

The Ox. No. of each O is -2 for a total of -8. So the Sulfur atom is +6.

Recognizing Oxidizing and Reducing Agents - I

Problem: Identify the oxidizing and reducing agent in each of the Rx: a) Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) 8 SO3 (g) c) NiO(s) + CO(g) Ni(s) + CO2 (g)Plan: First we assign an oxidation number (O.N.) to each atom (or ion)based on the rules in Table 4.3. A reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). Areactant is the oxidizing agent if it contains an atom that is reduced( O.N. decreased).Solution: a) Assigning oxidation numbers:

Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2 (g)0 +1

-1+2

-10

HCl is the oxidizing agent, and Zn is the reducing agent!

Recognizing Oxidizing and Reducing Agents - II

b) Assigning oxidation numbers:

S8 (s) + 12 O2 (g) 8 SO3 (g)

____ is the reducing agent and ____ is the oxidizing agent.

c) Assigning oxidation numbers:

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

___ is the reducing agent and ____ is the oxidizing agent.

Recognizing Oxidizing and Reducing Agents - II

b) Assigning oxidation numbers:

S8 (s) + 12 O2 (g) 8 SO3 (g)

0 0 +6-2

S8 is the reducing agent and O2 is the oxidizing agent

c) Assigning oxidation numbers:

S [0] S[+6]S is Oxidized

O[0] O[-2]O is Reduced

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

Balancing REDOX Equations: The oxidation number methodStep 1) Assign oxidation numbers to all elements in the equation.

Step 2) From the changes in oxidation numbers, identify the oxidized and reduced 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 Ox. No. Method - I

___ H2 (g) +___ O2 (g) ___ H2O(g)

0 +1- 1 e-

0 -2+2 e-

electrons lost must = electrons gained therefore multiply

Hydrogen reaction by 2!

and we are balanced!

2 2

REDOX Balancing Using Ox. No. Method - II

Fe+2(aq) + MnO4

-(aq) + H3O+

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

(aq) + H2O(aq)

-1e-+2 +3

+7 +2+5 e-

5 Fe+2(aq) + MnO4

-(aq) +__ H3O+

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

(aq) +__ H2O(aq)

Multiply Fe+2 & Fe+3 by ___ to correct for the electrons gained by theManganese.

__Fe+2(aq)+ MnO4

-(aq) + H3O+

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

(aq) + H2O(aq)

Next balance the # O and H atoms:

REDOX Balancing Using Ox. No. Method - II

Fe+2(aq) + MnO4

-(aq) + H3O+

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

(aq) + H2O(aq)

-1e-+2 +3

+7 +2+5 e-

5 Fe+2(aq) + MnO4

-(aq) +8 H3O+

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

(aq) +12 H2O(aq)

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

5 Fe+2(aq) + MnO4

-(aq) + H3O+

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

(aq) + H2O(aq)

Next balance the # O and H atoms: Make 4 water molecules from theoxygens of the MnO4

-, which requires 8 protons from the acid (H3O+

ions). These 8 H3O+ ions will also make 8 more water molecules, for a total of 4+8 = 12 water molecules.