1

A solution is a homogeneous mixture of two or more substances.

A solution forms when a substance dissolves at the atomic level in a solvent.

If the substance being dissolved is ionic, the atomic level is ions.

If the substance being dissolved is covalent, the atomic level is molecules.

If the substance being dissolved is atomic (like a noble gas), the atomic level is atoms.

This is in contrast to a heterogeneous mixture where the solute will not break down to atomic level. I.E. the particles are larger than ions, atoms or molecules. Examples like this include suspensions and colloids.

2

A solution is a homogeneous mixture of two or more substances.

A solvent is a substance that is present in the largest amount in a solution. The solvent dissolves the other compound(s).

A solute is a substance that is present in a lesser amount in a solution. The solute dissolves in the solvent.

If a substance will dissolve in a solvent, then the substance is soluble.

If a substance will not dissolve in a solvent, then the substance is insoluble.

If a liquid substance will not dissolve in a liquid solvent, the two substances are immiscible.

If a liquid substance will dissolve in a liquid solvent, the two substances are miscible.

3

Solution formation (solid dissolving in water)

When the ions become surrounded by water molecules, they are said to be solvated.

Solvation is the process of surrounding a solute particle with solvent molecules. If the solvent is water, the process can also be called hydration.

If the intermolecular forces between the water molecules and the ions are stronger than the forces between the ions themselves, the ionic substance will dissolve.

4

a) Solution of a strong

electrolyte

b) Solution of a weak

electrolyte

c) Solution of a non-

electrolyte

like sugar

like salt

like a weak acid

An aqueous solution of ions will conduct electricity.

5

NaCl (a salt) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte.

NaCl (s) + H2O (l) → Na+1 (aq) + Cl1 (aq)

Any ionic compound that dissolves in water will produce solvated ions (aq), so all water soluble ionic compounds are strong electrolytes.

If a soluble ionic compound contains the OH ion, the ionic compound will be a strong base.

6

HCl (a strong acid) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte.

This is also the definition of a strong acid.

An acid is a substance that breaks apart (dissociates) in water to produce aqueous H+1 ions. If the acid dissociates nearly 100% it is a strong electrolyte. If an acid is a strong electrolyte, it is a strong acid.

HCl (g) + H2O (l) → H+1 (aq) + Cl1 (aq)

There are six acids that you need to recognize as strong acids:

HCl, HBr, HI, HNO3, H2SO4, HClO4

7

If an acid only dissociates to a small extent (usually less than 5% ions produced), the acid would be classified as a weak acid and would also be a weak electrolyte.

Any Acid that is not one of the six strong acids will be a weak acid. When H+1 ions are added to many of the polyatomic anions, the compound you make will usually be an acid.

Common examples of weak acids are:

HF, CH3CO2H, H3PO4, HClO, HClO2, HClO3

A substance that produces OH ions in water is a base. If a base only produces a few ions (5% or less), the base would be classified as a weak base and would also be a weak electrolyte.

Common examples of weak bases are:

NH3, Ionic compounds that contain OH but are only slightly soluble

8

Substances that do not form ions when dissolved in water ate called Nonelectrolytes. Nonelectrolytes are covalent molecules that are soluble in water.

Common examples of nonelectrolytes are:

Glucose (C6H12O6), Sucrose (C12H22O11), and ethanol (CH3CH2OH)

C6H12O6 (s) + H2O (l) → C6H12O6 (aq)

9

Concentration: describing how much solute is dissolved in a solvent

A concentrated solution has a lot of solute dissolved.

While a dilute solution has only a little solute dissolved.

Percent by Mass

Percent by Volume

Molarity

Molality

Mole Fraction

(mass of solute/mass of solution)X100

(volume of solute/volume of solution)X100

(moles of solute/liter of solution)

(moles of solute/kilogram of solvent)

(moles of solute)/(moles of solute + moles of solvent)

There are five ways that we will quantify concentration in this class.

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Molarity (M): (moles of solute)/(L of solution) Units are: mole/L

If the amount of solute is given in grams, you need to convert from grams to moles first!

Example Problem: An intravenous solution contains 5.10 g of glucose (C6H12O6) in 100.5 mL of solution. What is the molarity of the solution?

Volume of solution: 100.5 mL Mass of Solute: 5.10 g C6H12O6

Molar Mass of C6H12O6: 180.0 g/mol

( )5.10 g C6H12O6 ( )180.0 g C6H12O6

1 mol C6H12O6

( )100.5 mL ( )1000 mL

1 L

Divide the moles by the L to get molarity (M)

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Preparing solutions of a specific concentration:

Calculations Steps:

Step 1: identify the volume of solution to be made.

Step 2: identify the concentration of solution to be made.

Step 3: Use the concentration and volume information to determine the “amount” of solute needed.

Step 4: If the amount is in moles, convert it to grams using molar mass.

Step 5: If the solute is a liquid, convert the mass into volume using density.

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Examples: a) How many grams of methanol (CH3OH) would be needed

to make 500.0 mL of a 1.50 molar solution?

Volume = 500.0 mL = 0.5000 L

Concentration: 1.50 molar (M) = 1.50 mol/L

So: n = MVn = (1.50 mol/L)(0.5000 L)

= 0.750 mol CH3OH

Grams CH3OH = (0.750 mol CH3OH)(32.04 g/mol) = 24.0 g CH3OH

density = mass/volume So: volume = mass/density

volume = (24.0 g)/0.783 g/mL) = 30.7 mL

b) How many mL of CH3OH would be needed for this solution if the

density of CH3OH is 0.783 g/mL?

13

Steps Involved in the Preparation of a Standard Aqueous Solution using a solid compound.

a) Measure the mass of the correct number of grams of solid compound.b) Add some water and shake until all solid is dissolved.c) Carefully add enough water to fill the container to the volume marker.

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Steps Involved in the Preparation of a Standard Aqueous Solution using a liquid compound.

a) Measure out the correct volume of liquid compound using a pipette or burette.

b) Add some water and shake until all liquid is dissolved.c) Carefully add enough water to fill the container to the volume marker.

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Dilution Problems: If you increase the amount of solvent in a solution without changing the amount of solute, you are doing a dilution problem.

In a dilution problem the number of moles of solute stays the same while the volume and the concentration of the solution change.

Remember: So: n = MV

Now in a dilution problem, there are two sets of conditions, the initial and the final (1 and 2 respectively).

M1V1 = mol of solute and M2V2 = mol of solute

and M1V1 = M2V2

Use this form of the equation to solve dilution problems!

so n1 = n2

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Five Types of Chemical Reactions: (SOL level of definition)

Synthesis:

Combustion:

Decomposition:

Double Replacement:

Single Replacement:

Two or more reactants combine to make a more complicated product.

Any rapid reaction with oxygen (O2 gas) to produce

heat and/or light.

One complicated reactant breaking down to make two or more products.

AX + BY → AY + BX

AX + Y → AY + X

ABX → AB + X

AB + X → ABX

AB + O2 → AO + BO

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Double Replacement: AX + BY → AY + BX

The reactants are always aqueous ionic compounds.

One of the products must be a solid, a liquid (usually water), or a gas while the other product will be a new aqueous ionic compound.

Na2SO4 (aq) + BaCl2 (aq) → 2 NaCl (aq) + BaSO4 (s)

A reaction that makes a solid: (these are also called precipitation reactions)

H2SO4 (aq) + Sr(OH)2 (aq) → 2 H2O (l) + SrSO4 (aq)

A reaction that makes water: (these are also called acid-base reactions)

Na2S (aq) + 2 HCl (aq) → H2S (g) + 2 NaCl (aq)

A reaction that makes a gas:

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For double replacement reactions that produce precipitates, switch ions and write correct ionic formulas for the products-then balance the equation.

For Example Mix: Al2(SO4)3 (aq) + BaCl2 (aq)

Take the ionic compounds apart:

Al+3 ions Ba+2 ionsSO42 ions Cl1 ions

Switch the ions around-remember positive always mixes with negative:

Al+3 ions Ba+2 ions SO42 ionsCl1 ions

Write correct formulas for ionic compounds-remember the net charge is 0:

AlCl3 + BaSO4 These are the possible products

Write a skeleton equation:

Al2(SO4)3 (aq) + BaCl2 (aq) → AlCl3 (aq) + BaSO4 (s)

Balance to get the chemical equation (also called the molecular equation):

Al2(SO4)3 (aq) + 3 BaCl2 (aq) → 2 AlCl3 (aq) + 3 BaSO4 (s)

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Illustration of a precipitation reaction between Ba(NO3)2 (aq) and K2CrO4 (aq)

a) Ions present when the two solutions are first mixed.

b) Ions remaining after formation of precipitate (shown on the bottom of the beaker).

Ba(NO3)2 (aq) + K2CrO4 (aq) → BaCrO4 (s) + 2 KNO3 (aq)

The reaction in chemical equation form:

K+1 and NO31 are spectator ions

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Ba(NO3)2 (aq) + K2CrO4 (aq) → BaCrO4 (s) + 2 KNO3 (aq)

Chemical Equation or Molecular Equation (balanced)

Complete Ionic Equation (must show all aqueous ions separated)

Ba (aq) + 2 NO3 (aq) + 2 K(aq) + CrO4

(aq) →

→ BaCrO4 (s) + 2 K+1 (aq) + 2 NO3 (aq)

(reactant side)

(product side)

Net Ionic Equation (spectator ions are removed)

Ba (aq) + CrO4 (aq) → BaCrO4 (s)

Spectator ions: K+1 (aq) and NO3 (aq)

Note: the aqueous ionic product always contains the Spectator ions!

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Spectator ions are the ions that remain in the solution after the reaction.

The spectator ions do not take part in the reaction.

The spectator ions remain in solution after the precipitate formsand the ions in the precipitate are removed from the solution (settle out on the bottom)

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KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq)

1st solution 2nd solution mixing Final solution

K+1 (aq) + Cl (aq) + Ag+1 (aq) + NO3(aq) → AgCl (s) + K+1 (aq) + NO3

(aq)

Can you identify the spectator ions?

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There are three types of double replacement reactions:

1) Those that form a solid ionic precipitate.

2) Those that form water.

3) Those that produce a gas.

If switching the ions in a double replacement reaction does not do one or more of these three things, then no reaction is taking place.

These are also called precipitation reactions.

These are also called acid-base reactions.

These have no other specific name.

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Double replacement reactions that produce water (acid-base)

HBr (aq) + NaOH (aq) → H2O (l) + NaBr (aq)

H+1 (aq) + Br (aq) + Na+1 (aq) + OH (aq) → H2O (l) + Na+1 (aq) + Br (aq)

H+1 (aq) + OH (aq) → H2O (l)

The key to identifying these reactions is to recognize acids and bases.

Adding H+1 ions to most of the negative ions we have used so far will produce an acid.

The only bases we have at this point are OH and CO3

Hydrobromic acid plus sodium hydroxide

Therefore, hydroxide compounds and carbonate compounds are bases.

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Double replacement reactions that produce a gas.

2 HI (aq) + Li2S (aq) → H2S (g) + 2 LiI (aq)

2 H+1 (aq) + 2 I (aq) + 2 Li+1 (aq) + S (aq) → H2S (g) + 2 Li+1 (aq) + 2 I (aq)

2 H+1 (aq) + S (aq) → H2S (g)

The are not very many of these reactions. Common ones produce H2S or CO2.

2 HCl (aq) + Na2CO3 (aq) → H2CO3 (aq) + 2 NaCl (aq)

H2CO3 (aq) → H2O (l) + CO2 (g)

But carbonic acid is not stable and falls apart!

Therefore:

2 HCl (aq) + Na2CO3 (aq) → H2O (l) + CO2 (g) + 2NaCl (aq)

26

AgNO3 (aq) + CaBr2 (aq) → AgBr (s) + Ca(NO3) 2 (aq)

KOH (aq) + HI (aq) → KI (aq) + H2O (l)

HNO3 (aq) + Ca(OH)2 (aq) → H2O (l) + Ca(NO3) 2 (aq)

HClO4 (aq) + Na2S (aq) → H2S (g) + NaClO4 (aq)

Pb(NO3)2 (aq) + Li2SO4 (aq) → LiNO3 (aq) + PbSO4 (s)

For each of the following: Identify the type of reaction, identify the spectator ions, and write the net ionic equation. You may want to write the complete ionic equation to help you identify the spectator ions.

Spectators: Ca+2 and NO31, double replacement: precipitation

Spectators: K+1 and I1, double replacement: acid-base

Spectators: Ca+2 and NO31, double replacement: acid-base

Spectators: Na+1 and ClO41, double replacement: produces a gas

Spectators: Li+1 and NO31, double replacement: precipitation

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Rules for predicting precipitates:

Slightly soluble means that they will be a precipitate.

If a compound contains ions other than those mentioned, assume they are insoluble (form a precipitate) unless told otherwise.

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NaCl Na2CO3 K2CO3 AgCl

AgNO3 CaBr2 AgBr BaSO4

Mg3(PO4)2 PbBr2 AgBr MgSO4

Fe2(CO3)3 Li2S Al(OH)3 (NH4)3PO4

Use the rules on the previous page to predict whether the following ions will be soluble in water. If they are not soluble, then they would be precipitates in double replacement reactions.

soluble soluble soluble insoluble

soluble soluble insoluble insoluble

insoluble soluble insoluble soluble

insoluble insoluble insoluble soluble

Practice:

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Synthesis: Two or more reactants combine to make a more complicated product.

AB + X → ABX

The reactants may be compounds (ionic or covalent) or may be elements.

The product must always be a compound (ionic or covalent).

If more than one kind of product is made, the reaction is not technically a synthesis reaction.

CaO (s) + H2O (l) → Ca(OH)2 (s)

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Decomposition: One complicated reactant breaking down to make two or more products.

ABX → AB + X

The reactant must always be compounds (ionic or covalent).

The products may be compounds (ionic or covalent) or elements.

Fe2(CO3) 3 (s) → Fe2O3 (s) + 3 CO2 (g)

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Single Replacement: AX + Y → AY + X

A + BX → AX + B

or

An ionic compound is reacting with an element.

The element (charge of zero) will become an ion and replace an ion in the ionic compound.

The ion that was replaced will become an element (charge of zero)

SnBr2 (s) + Fe (s) → FeBr3 (s) + Sn (s)

The element iron (Fe) is replacing the Sn+2 ion. When it does this, iron becomes an ion (Fe+3) while the Sn+2 ion becomes an element (Sn).

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Activity Series for Single Replacement Reactions

Metals Most Reactive

Least Reactive

LiRbKCaNaMgAlMnZnFeNiSnPbCuAgPtAu

Most Reactive

Least Reactive

FClBrI

Halides (halogens)

Only a more reactive element can replace another element in a compound!

The element produced must be less reactive than the starting element!

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2 LiBr (s) + F2 (g) → 2 LiF (s) + Br2 (l)

Will the reaction below will work?

2 LiBr (s) + I2 (g) → 2 LiI (s) + Br2 (l)

Will the reaction below will work?

AgBr (s) + Li (s) → LiBr (s) + Ag (s)

Will the reaction below will work?

CaBr2 (s) + Zn (s) → ZnBr2 (s) + Ca (s)

Will the reaction below will work?

Yes because F is more reactive than Br

No because I is less reactive than Br

Yes because Li is more reactive than Ag

No because Zn is less reactive than Ca

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Notice that Halogen atoms (group 17) always replace halogen ions!

Notice that Metal atoms always replace metal ions!

2 LiBr (s) + F2 (g) → 2 LiF (s) + Br2 (l)

Here the element fluorine (F2) is replacing the ion Br1

AgBr (s) + Li (s) → LiBr (s) + Ag (s)

Here the element lithium (Li) is replacing the ion Ag1

Notice that the ion that was replaced always becomes an element!

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Pop Quiz: a) Write the complete ionic equation for the following reaction. b) Write the the net ionic equation for the following reaction. c) Identify any spectator ions.

3 Ca(NO3)2 (aq) + Au2(SO4)3 (aq) → 3 CaSO4 (s) + 2 Au(NO3)3 (aq)

2 AlCl3 (aq) + 3 Na2CO2 (aq) → 6 NaCl (aq) + Al2(CO2)3 (s)

Group A

Group B

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Acid-Base Titration

Titration is a process where the concentration of an acidic solution can be determined by reacting the acid with a measured volume of a basic solution of known concentration.

Titrations make use of indicators. An indicator is a substance that changes color when the pH of the solution changes.

A common indicator is Phenolphthalein.

Colorless in acid and Pink in base.

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The point during a titration where the moles of base added is equal to the moles of acid originally present is called the equivalence point.

In order for an indicator to be useful for a titration, the color change must occur at or very near to the equivalence point.

For strong acid-strong base titrations, the equivalence point will be at pH of 7.

So indicators that change color at or very near pH 7 will work.

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The Useful pH Ranges for Several Common Indicators

39

Why is Phenolphthalein a good indicator even though its color change occurs at a pH of about 8.5?

One: Its acidic form is colorless, so that makes the color change easier to detect.

Two: The titration curve is nearly vertical near the equivalence point, so adding one or two drops of base will cause a very large change in pH. The error associated with one or two drops of added base is small.

Calculations for the titration of 50.0 mL of 0.20 M HCl with 0.200 M NaOH:

At the start: pH = 0.70 and mole HCl = (0.200 mol/L)(0.0500 L) = 0.0100 mol

After 10.0 mL of NaOH has been added:

Mole NaOH added = (0.200 mol/L)(0.0100 L) = 0.00200 mol

Mole of HCl remaining = (0.0100 mol – 0.00200 mol) = 0.0080 mol

Concentration of HCl now = (0.0080 mol/(0.0500 L + 0.0100 L)) = 0.13 M

pH now = log(0.13) = 0.89

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After 30.0 mL of NaOH has been added:

Mole NaOH added = (0.200 mol/L)(0.0300 L) = 0.00600 mol

Mole of HCl remaining = (0.0100 mol – 0.00600 mol) = 0.0040 mol

Concentration of HCl now = (0.0040 mol/(0.0500 L + 0.0300 L)) = 0.050 M

pH now = log(0.050) = 1.30

After 49.5 mL of NaOH has been added:

Mole NaOH added = (0.200 mol/L)(0.0495 L) = 0.00990 mol

Mole of HCl remaining = (0.0100 mol – 0.00990 mol) = 0.0001 mol

Concentration of HCl now = (0.0001 mol/(0.0500 L + 0.0495 L)) = 0.001 M

pH now = log(0.001) = 3.0

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After 50.5 mL of NaOH has been added:

Mole NaOH added = (0.200 mol/L)(0.0505 L) = 0.0101 mol

Mole of HCl remaining = (0.0100 mol – 0.0101 mol) = no meaning for negative numbers

Concentration of NaOH now = (0.0101 mol 0.0100 mol) = 0.0001 mol = (0.0001 mol/(0.0500 L + 0.0505 L) = 0.001 mol/L

pOH now = log(0.001) = 3.0

pH = 14.0 – 3.0 = 11.0

42

43

A titration curve: an acid being titrated with NaOH.

44

The pH Curve for the Titration of 100.0 mL of 0.10 M of HCl with 0.10 M NaOH

45

The pH Curve for the Titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH; Phenolphthalein Will Give an End Point Very Close to the Equivalence Point of the Titration

46

Summary for predicting pH of solutions at the equivalence point of a titration.

Strong Acid + Strong Base

pH at Equivalence Point

7.00

Strong Acid + Weak Base Acidic

Weak Acid + Strong Base Basic

Actual pH depends upon Ka and Kb values of the

conjugates.

Weak Conjugate Acid

Weak Conjugate Base

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Redox: Oxidation-Reduction Reactions

An Oxidation-Reduction Reaction is any reaction where one substance gains electrons while another substance loses electrons.

2 Na (s) + Cl2 (g) → 2 Na + 2 Cl (ions in the crystal)

What happened to Na? What happened to Cl?

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Na (s) → Na + eEach Na is losing an electron to

become a cation

Cl2 + 2 e → 2 Cl Each Cl is gaining an electron to

become an anion

Oxidation is losing electrons

Reduction is gaining electrons

OIL RIG

Oxidation Is LossReduction Is Gain

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Since metals have low electronegativities, they tend to lose electrons (be oxidized) while nonmetals which have higher electronegativities tend to gain electrons (be reduced)A substance that is losing electrons is becoming more positive. This means that the oxidation state is increasing.

A substance that is gaining electrons is becoming more negative. This means that the oxidation state is decreasing.An agent is a substance that causes something to happen. If it causes an oxidation to occur, it would be an oxidizing agent. If it causes a reduction to occur, it would be a reducing agent.

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2 NaBr (aq) + Cl2 (aq) → 2 NaCl (aq) + Br2

What substance is being oxidized? What substance is being reduced?

What substance is the oxidizing agent? What substance is the reducing agent?

How many electrons are transferred in the balanced reaction?

2 Br → Br2 + 2 e

Cl2 + 2 e → 2 Cl

Na+ → Na+ (spectator ion)

Oxidation half reaction

Reduction half reaction

Since Br is oxidized, Cl2 must have caused it to happen, so Cl2 is the

oxidizing agent.Since Cl2 is reduced, Br must have caused it to happen, so Br is the

reducing agent.

Oxidation States or Oxidation Numbers are helpful in identifying what is being oxidized or reduced, and they let you determine the number of electrons gained or lost.

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Al + Fe2O3 → Fe + Al2O3

Aluminum as an element (no charge)Aluminum as an ion (+3 charge-lost electrons)

Iron as an ion (+3 charge)

Iron as an element (no charge-gained electrons)

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What is being oxidized?

What is being reduced?

What is the reducing agent?

What is oxidizing agent?

Al → Al+3

Therefore: we can write the reaction in separate parts

Fe+3 → Fe O2 → O2

Since oxygen is the same on both sides, it is technically a spectator ion.

Al Al

Fe+3 Fe+3

Notice that the substance being oxidized will be the reducing agent and the substance being reduced will be the oxidizing agent!

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General Rules for Oxidation Numbers:

An element in its normal state has oxidation number = 0

An element that becomes an ion has an oxidation number equal to the charge of the ion.

The sum of the oxidation numbers of all atoms in a compound or ion must equal the charge of the compound or ion.

In most compounds, oxygen has an oxidation number of 2. Oxygen will have an oxidation number of 1 when oxygen is a peroxide. Peroxides must have an oxygen-oxygen bond in the molecule.

Fluorine always has an oxidation number of 1 when bonded to other elements.

In compounds where hydrogen is bonded to a nonmetal, hydrogen has an oxidation number of +1, when bonded to a metal it will have an oxidation number of 1.

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Examples:

Pt S8 Mg+2

H2O NO3 Fe(NO3)3

Cr2(SO3)3 Na2O2 CH3F

CO2 NH4+ NH4NO2

LiH

Br2

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The change in oxidation number can be used to determine if a redox reaction has occurred. If nothing changes oxidation number, a redox reaction has not occurred.

The change in oxidation number can also be used to identify the substance being oxidized or reduced.

The change in oxidation number can also be used to balance a redox reaction.

In order for a redox reaction to be balanced, the number of electrons lost must equal the number of electrons gained.

Al → Al+3 + 3 e

2 e + In+3 → In

If Al and In+3 were reacting together, how many electrons must be transferred in the balanced reaction?

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(Al → Al+3 + 3 e)

(2 e + In+3 → In)

2

3

2 Al → 2 Al+3 + 6 e

6 e + 3 In+3 → 3 In

2 Al + 3 In+3 → 3 In+1 + 2 Al+3

We see that 6 electrons need to be transferred in order for the Redox reactions to be balanced.

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What could this look like if spectator ions were included?

Al + InCl3 → InCl + AlCl3

How would we balance this?

We would use the half reactions to balance the electrons and then notice that the spectator ions will be balanced as well.

2 Al + 3 InCl3 → 3 InCl + 2 AlCl3

2 Al → 2 Al+3 + 6 e

6 e + 3 In+3 → 3 In

2 Al + 3 In+3 → 3 In+1 + 2 Al+3

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Balance the following reaction using the half-reaction method

Cu + HNO3 → Cu(NO3)2 + NO2 + H2O

First: Assign oxidation numbers to all atoms.

Second: Use the changes in oxidation numbers to identify oxidation and reduction half-reactions. Ignore the elements that do not change oxidation number

Cu + HNO3 → Cu(NO3)2 + NO2 + H2O 0 1 5 -2 2 5 -2 4 -2 1 -2

Cu → Cu+2 N+5 → N+4

0 2 5 4

Going from 0 to +2 is an oxidation! Going from +5 to +4 is a reduction!

Cu → Cu+2 + 2 e

Third: Balance the electrons transferred using coefficients.

Cu → Cu+2 + 2 e

1 e + N+5 → N+4

2 (1 e + N+5 → N+4 )

2 e + 2 N+5 → 2 N+4

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Cu → Cu+2 + 2 e

Fourth: Add the reactions together.

Cu + 2 N+5 → Cu+2 + 2 NO2

Notice that N and O are not balanced. Change coefficients to balance the equation but keep in mind that you can not change the ratio of Cu and N that are reacting. Two NO3

ions react to become NO2, but there must be two more

that do nothing (spectators). Therefore, we need two more HNO3.

2 e + 2 N+5 → 2 N+4

Fifth: Put the spectator atoms back into the equation and then balance them if needed.

Cu + 2 HNO3 → Cu(NO3)2 + 2 NO2 + H2O

Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + H2O

Now N is balanced. But H and O are not.

Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + 2 H2O

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ClO4 + Br → Cl + Br2 in acidic solution

7 -2 -1 -1 0

Cl+7 + 8 e → Cl

2 Br → Br2+ 2 e 4( )

ClO4 + 8 Br → Cl + 4 Br2

Oxidation Half-Reaction

Reduction Half-Reaction

ClO4 + 8 Br → Cl + 4 Br2

Notice that oxygen can not be balanced here! Now we use the fact that the reaction is performed in aqueous acidic solution. This means that we can add water and/or H+ ions as needed to balance the equation.

+ 4 H2O8 H+ +

Check: The total charge on both sides of the equation must be equal!

(+8) + (9) = (1)(1) = (1)

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If a problem is performed in basic solution, balance it as if it were in acid and then convert to base. See the example below.

ClO4 + 8 Br → Cl + 4 Br2 + 4 H2O8 H+ +

First: Add enough OH to both sides to react with all H.

ClO4 + 8 Br → Cl + 4 Br2 + 4 H2O ++ 8 H+ + 8 OH 8 OH

Second: H+ + OH makes water. Cancel out any water that appears on both sides.

ClO4 + 8 Br → Cl + 4 Br2 + 4 H2O + 8 OH8 H2O +

ClO4 + 8 Br → Cl + 4 Br2 + 8 OH4 H2O +

Check for total charge

(9) = (9)

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Balance the following redox reaction using the half-reaction method (in base).

Mn (aq) + BiO3 (aq) → Bi (aq) + MnO4

(aq)

Mn → Mn + 5 e

Bi + 3 e → Bi

+2 +5 +2 +7

3 Mn → 3 Mn + 15 e

5 Bi + 15 e → 5 Bi

3 Mn + 5 Bi → 5 Bi + 3 Mn

3 Mn (aq) + 5 BiO3 (aq) → 5 Bi (aq) + 3 MnO4

(aq) + 3 H2O (l)6 H+ (aq) +

3 Mn (aq) + 5 BiO3 (aq) → 5 Bi (aq) + 3 MnO4

(aq)3 H2O (l) + + 6 OH (aq)

(+7) = (+7)

(+1) = (+1)