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Oxidation-Reduction Chapter 16

Oxidation-Reduction Chapter 16

Tro, 2nd ed.

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Oxidation NumberThe oxidation number (oxidation state) of an atom represents the

number of electrons lost, gained, or unequally shared by an atom. (IT IS NOT THE SAME AS CHARGE!!)

Oxidation numbers can be zero, positive or negative.

An oxidation number of zero means the atom has the same number of electrons assigned to it as there are in the free neutral atom, WHICH ARE ALWAYS ZERO.

A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom.

A negative oxidation number means the atom has more electrons assigned to it than in the neutral atom.

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* Rule 2a: F is always -1, except when it is F2.

Apply rules from top down; first one you reach has priority.

Similar to your textbook, Tro, page 563, but more general.

*

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Oxidation NumberFor the seven diatomic elements which exist

as covalent molecules: each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity.

H-H and Cl-Cl: both atoms in each molecule have an oxidation number of 0.

Other elements: Cu, Ne, C, et., will all be assigned a 0.

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Oxidation NumberThe oxidation number of an atom that

has gained or lost electrons to form a monatomic ion is the same as the positive or negative charge of the ion.

NaCl

Na+ has an oxidation number of +1.

Cl- has an oxidation number of -1.

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Oxidation NumberGiven a compound like HCl, use the rules to assign

oxidation number. Rule 2 says H is +1 unless with a metal: therefore H is +1.

Rule 6 says the sum of the oxidation numbers must be zero, therefore Cl will be assigned a -1.

In binary compounds, halogens are often a -1, like their monatomic ions would be. However, if a halogen is part of a polyatomic ion, then its ON must be determined.

HCl +1,-1

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N2 N2O NO N2O3 NO2 N2O5

N oxidation number

0 +1 +2 +3 +4 +5 +5

Many elements have multiple oxidation numbers

-3NO

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Notice that H and group I metals have O.N. of +1

Notice that group II metals have O.N. of +1

Halogens are typically assigned O.N. of -1

Oxygen is usually -1

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Step 1 Write the oxidation number of each known atom below the atom in the formula.

Step 2 Multiply each oxidation number by the number of atoms of that element in the compound.

Step 3 Write an expression indicating the sum of all the oxidation numbers in the compound. Remember: The sum of the oxidation numbers in a compound must equal zero.

Rules for Determining the Oxidation Number of an Element Within a Compound

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Determine the oxidation number for sulfur in sulfuric acid.

H2SO4Step 1 -2+1

Write the oxidation number of each known atom below the atom in the formula.

Multiply each oxidation number by the number of atoms of that element in the compound.

Step 2 4(-2) = -82(+1) = +2

Step 3 +2 + S + (-8) = 0

Write an expression indicating the sum of all the oxidation numbers in the compound.

Step 4 S = +6 (oxidation number for sulfur)

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Determine the oxidation number for carbon in the oxalate ion.

Step 1 -2

Write the oxidation number of each known atom below the atom in the formula.

Multiply each oxidation number by the number of atoms of that element in the compound.

Step 2 4(-2) = -8

Step 3 2C + (-8) = -2 (the charge on the ion)

Write an expression indicating the sum of all the oxidation numbers in the compound.

2-2 4C O

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ASSIGNING OXIDATION NUMBERS

PRACTICE: FeCl2, MnO2, F2, Al, KI, IF5, H2O, OF2, SO2, SO4

2-, OH-, HCl, ClO-, ClO4

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+2 and -1, +4 and -2, zero, zero, +1 and -1, +5 and -1, +1 and -2, +2 and -1, +4 and -2, +6 and -2, -2 and +1, +1 and -1, +1 and -2, +7 and -2

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Oxidation-ReductionOxidation-reduction (redox) is a chemical

process in which the oxidation number of an element is changed.

Redox may involve the complete transfer of electrons to form ionic bonds or a partial transfer of electrons to form covalent bonds.

Loss of electrons = oxidationGain of electrons = reductionLEO says GER

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Oxidation-ReductionOxidation occurs when the oxidation

number of an element increases as a result of losing electrons.

Reduction occurs when the oxidation number of an element decreases as a result of gaining electrons.

In a redox reaction oxidation and reduction occur simultaneously, one cannot occur in the absence of the other.

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Oxidation-ReductionOxidizing agent The substance that causes

an increase in the oxidation state of another substance.The oxidizing agent is reduced in a redox

reaction.

Reducing agent The substance that causes a decrease in the oxidation state of another substance.The reducing agent is oxidized in a redox

reaction.

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Oxidation-ReductionElectron-transfer reactions:

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

Zn: from atom to ion; gives up e-s to HH: from ion to atom; accepts e-s from H

by electron transfer!Zn atom becomes Zn2+ ion and is oxidized.H+ ion becomes H2 atoms = reduced.Zn is the reducing agent and H+ is the oxidizing

agent.You must always have both oxidation and reduction

in the reaction.

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SCALE OF OXIDATION NUMBERS AND MEANING OF OXIDATION AND REDUCTION

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Redox ReactionsPractice: find oxid # for all elements in these

reactions to see if electron-transfer occurs.1. 2 Mg + O2 2 MgO

2. 2 KClO3 2 KCl + 3 O2

3. Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

4. HCl + NaOH H2O + NaCl

5. CH4 + O2 CO2 + 2 H2O

1. YES 2. YES 3. YES4. NO 5. YES

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Balancing Oxidation-Reduction Equations

Change-In-OxidationNumber Method (we’ll try this first)

Half-Reaction Method (more commonly used)

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oxidation number of tin increases

oxidation number of nitrogen decreases

Balance the equation Sn + HNO3 → SnO2+ NO2+H2O

Step 1 Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced. Write the oxidation numbers below each element to avoid confusing them with ionic charge.

Sn + HNO3 → SnO2 + NO2 + H2O

0 +1 +5 -2 +4 -2 +4 -2 +1 -2

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Balance the equation Sn + HNO3 → SnO2+ NO2+H2O

Step 5 Balance the remaining elements that are not oxidized or reduced to give the final balanced equation.

Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O2

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Balancing Ionic Redox Equations: The Ion-Electron Method

(acidic solution)Step 1: Determine the oxidation number of every atom/ion in the reaction.

Figure out which is being oxidized and which reduced. Connect them with arrows and determine the number of electrons lost or gained per atom. Then write the two half-reactions that contain the elements being oxidized and reduced.

Step 2: Balance the elements other than hydrogen and oxygen. (This is the step that students forget!)

Step 3: Balance oxygen by adding H2O, then balance H by adding H+.

Step 4: Add electrons (e-) to each half-reaction to bring them into electrical balance.

Step 5: Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number (using least common denominator) to make the number of electrons the same in each half-reaction.

Step 6: Add the two half-reactions together, canceling electrons and any other identical substances (like H+ and H2O) that appear on opposite sides of the equation. In ionic redox equations both the numbers of atoms and the electrical charge on both sides of the equation must be the same.

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Balance the equation

Step 1: MnO4- + S2- Mn2+ + S

+7, -2 -2 +2 0

Mn reduced, gains 5 e-/Mn; S oxidized, loses 2 e-s/S

- 2- 2+ o4MnO + S Mn + S (acidic solution)

2- oS S

- 2+4MnO Mn

Balance elements other than oxygen and hydrogen. (Step 2 is unnecessary, since these elements are already balanced).

Step 2

Oxidation half-reaction

Reduction half-reaction

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Balance the equation

Step 6 Add the two half-reactions together, canceling the 10e- from each side, to obtain the balanced equation.

o -2-5S 5S e 10

- 2+4

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- + + 2MnO 10e + 16 2Mn +H 8H O- 2- 2+4

+ o21 + 2MnO + 5S 2Mn + 5S +H 6 8H O

- 2- 2+ o4MnO + S Mn + S (acidic solution)

The charge on both sides of the balanced equation is +4

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Balancing Ionic Redox Equations: The Ion-Electron Method

(basic solution)

For reactions in alkaline (basic) solutions, first balance as thought the reactions were in an acid solution, using Steps 1-6 as before.

Add OH- to each side equal to the number of H+ on one side. Where both OH- and H+ appear on same side of reaction, combine them to make H2O.

Rewrite the equation, canceling equal numbers of water molecules that appear on opposite side of the equation.

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PRACTICE WITH BASIC SOLUTION

___Cl2 + ___IO3- ___Cl- + ___IO4

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After steps 1-6:Cl2 + IO3

- + H2O 2 Cl- + IO4- + 2 H+

Add 2 OH- to both sides:Cl2 + IO3

- + H2O + 2 OH- 2 Cl- + IO4- + 2 H+ + 2 OH-

Combine H+ and OH- to make water:Cl2 + IO3

- + H2O + 2 OH- 2 Cl- + IO4- + 2 H2O

Cancel one water on each side for finished equation:

Cl2 + IO3- + 2 OH- 2 Cl- + IO4

- + H2O

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AND THAT’S ALL WE NEED FROM CHP 16!

There is a lab on redox and practice handouts and homework! Do it!