Chapter 4

Reduction-Oxidation Reactions

Redox Reactions

Sodium chloride

Sodium chloride

Na Na+ + e Cl2 + 2e 2Cl

4

Oxidation–Reduction ReactionsInvolves 2 processes:

Oxidation = Loss of Electrons

Na Na+ + e Oxidation Half-Reaction

Reduction = Gain of electronsCl2 + 2e 2Cl Reduction Half-Reaction

Net reaction:

2Na + Cl2 2Na+ + 2Cl

– Oxidation & reduction always occur together

– Can't have one without the other

Oxidation Reduction ReactionOxidizing Agent - Substance that accepts e's

– Accepts e's from another substance– Substance that is reduced

Cl2 + 2e 2Cl–

Reducing Agent - Substance that donates e's

– Releases e's to another substance– Substance that is oxidized

Na Na+ + e–

Your Turn!

Which species functions as the oxidizing agent in the following oxidation-reduction reaction?

Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)

A. Pt(s)

B. Zn2+(aq)

C. Pt2+(aq)

D. Zn(s)

E. None of these, as this is not a redox reaction.

6

Redox Reactions Very common

– Batteries—car, flashlight, cell phone, computer

– Metabolism of food

– Combustion Chlorine Bleach

– Dilute NaOCl solution

– Cleans through redox

reaction

– Oxidizing agent

– Destroys stains by oxidizing them

7

Redox ReactionEx. Fireworks displays

Net: 2Mg + O2 2MgO

Oxidation:

Mg Mg2+ + 2e – Loses electrons = Oxidized– Reducing agent

Reduction:

O2 + 4e 2O2

– Gains electrons = Reduced– Oxidizing agent

8

Redox Reaction?

S + O2 SO2

• Combustion: Oxidation in old sense, reaction with oxygen

• But n no ions in SO2

• How can we decide which loses and which gains!!

• Oxidation number (state)!

• If compound were ionic, what would the charges have been.

9

Rules for Oxidation States (Numbers)

1. The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion.

2. The oxidation state of elements is zero.

3. Oxidation state for monoatomic ions are the same as their charge.

4. In its compounds fluorine is always –1.

5. Hydrogen is assigned the oxidation state +1.

6. Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide).

7. If two rules conflict, apply higher rule.

Oxidation States Assign the oxidation states to each element

in the following. CO2

NO3-

H2SO4

Fe2O3

Fe3O4

Cr2O72-

O2F2

H2O2

LiH BaO2

Oxidation-ReductionTransfer electrons, so the oxidation states change.

11

2

00

ClNaClNa

Ox

Red

22

2

00

OMgOMg

Ox

Red

Oxidation

Increase in

Oxid. number

Reduction

decrease in

Oxid. number

2

2

4

2

00

OSOS

Ox

Red

2

2

1

2

24

2

0

4

14

OHOCOHC

Ox

RedRedC (CH4) oxidized

→

CH4 reducing agent

O2 reduced →

O2 oxidizing agent

Ox

Red

PbS has been oxidized, PbS is the reducing agent.

O2 has been reduced, O2 is the oxidizing agent.

Ox

Red

PbO has been reduced, PbO is the oxidizing agent.

CO has been oxidized, CO is the reducing agent.

Identify the 1) Oxidizing agent2) Reducing agent3) Substance oxidized4) Substance reduced in the following reactions

Fe (s) + O2(g) ® Fe2O3(s)

Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g)

SO3- + H+ + MnO4

- ® SO4- + H2O + Mn+2

Balancing Redox Reactions

Ion-Electron Method – Acidic Solution

1. Divide equation into 2 half-reactions

2. Balance atoms other than H & O

3. Balance O by adding H2O to side that needs O

4. Balance H by adding H+ to side that needs H

5. Balance net charge by adding e–

6. Make e– gain equal e– loss; then add half-reactions

7. Cancel anything that is the same on both sides

Balance in Acidic Solution

Cr2O72– + Fe2+ Cr3+ + Fe3+

1. Break into half-reactions

Cr2O72 Cr3+

Fe2+ Fe3+

2. Balance atoms other than H & O

Cr2O72 2Cr3+

– Put in 2 coefficient to balance Cr

Fe2+ Fe3+

– Fe already balanced

3. Balance O by adding H2O to the side that needs O.

Cr2O72 2Cr3+

• Right side has 7 O atoms• Left side has none • Add 7 H2O to left side

Fe2+ Fe3+

• No O to balance

+ 7 H2O

4. Balance H by adding H+ to side that needs H

Cr2O72 2Cr3+ + 7H2O

• Left side has 14 H atoms• Right side has none • Add 14 H+ to right side

Fe2+ Fe3+

• No H to balance

14H+ +

5. Balance net charge by adding electrons.

14H+ + Cr2O72 2Cr3+ + 7H2O

– 6 electrons must be added to reactant side

Fe2+ Fe3+

– 1 electron must be added to product side

Now both half-reactions balanced for mass & charge

6e +

+ e

Net Charge = 2(+3)+7(0) = 6

Net Charge = 14(+1) (–2) = 12

6. Make e– gain equal e– loss; then add half-reactions

6e + 14H+ + Cr2O72– 2Cr3+ + 7H2O

Fe2+ Fe3+ + e

7. Cancel anything that's the same on both sides

6[ ]

6e + 6Fe2+ + 14H+ + Cr2O7

2

6Fe3+ + 2Cr3+ + 7H2O + 6e

6Fe2+ + 14H+ + Cr2O7

2

6Fe3+ + 2Cr3+

+ 7H2O

Practice The following reactions occur in acidic

aqueous solution. Balance them:

MnO4- + Fe2+ ® Mn2+ + Fe3+

Cu + NO3- ® Cu2+ + NO(g)

Pb + PbO2 + SO42- ® PbSO4

Mn2+ + NaBiO3 ® Bi3+ + MnO4-

Cr2O72- + C2H5OH ® Cr3+ + CO2

Ion-Electron method in Basic Solution

The simplest way to balance an equation in basic solution

Use steps 1-7 above, then

8. Add the same number of OH– to both sides of the equation as there are H+.

9. Combine H+ & OH– to form H2O

10. Cancel any H2O that you can from both sides

Basic Solution

Ag + CN- +O2 ® Ag(CN)2-

Cr(OH)3 + OCl- + OH- ® CrO42- + Cl- + H2O

CrI3 + Cl2 ® CrO4- + IO4

- + Cl-