Oxidation-ReductionTopic 9.1

  ...etc.

1+ 2+ 3-3+ 4+/- 2- 1- 0

Memorize these!NO3

1- nitrate

NO21- nitrite

OH1- hydroxide

ClO21- chlorite

ClO31- chlorate

HCO31- hydrogencarbonat

e (bicarbonate)

SO42- sulfate

SO32- sulfite

CO3 2- carbonate

PO43- phosphate

NH41+ ammonium

Introduction

• oxidation and reduction can be considered in terms of…1. oxidation- substance gains oxygen

reduction- substance loses oxygen

2. oxidation- the loss of electrons

reduction- the gain of electrons

3. oxidation- the oxidation state/# increases

reduction- the oxidation state/# decreases

O - oxidation

I - is

L - loss of electrons

R - reduction

I - is

G - gain of electrons

Oxidation states

• the oxidation state (number) is the apparent or theoretical charge of an free element, molecule, or ion

• oxidation– a process where the number increases (more

positive because loses neg. electrons)• Ex: Mg Mg2+ (aq) + 2e-

• reduction– a process where the number decreases (less

positive/more negative because gains neg. electrons)• Ex: O + 2 e- O2- (g)

Note: oxidation states must be written with the sign in FRONT:

+2 not 2+

1 The oxidation number of an free element is always 0

O2, H2, Ne, Zn

2 The oxidation number of Hydrogen is usually +1 Metal hydrides are an exception. They are -1

HCl, H2SO4

NaH

3 The oxidation number of Oxygen is usually -2. Peroxides are an exception They are –1.

H2O, NO2, etc.-O-O- bonding

4 Group 1 metals are always +1Group 2 metals are always +2Aluminum is always +3

Li, Na…Mg, Ba..Al

Rules for assigning oxidation states

5 Fluorine is always -1Other group 17 (halogens) are often -1

HF, OF2

HI, NaCl, KBr

6 Oxidation numbers of monatomic ions follow the charge of the ion

S2-, Zn2+

7 The SUM of oxidation numbers is zero for a neutral compound.

LiMnO4…

8 For polyatomic ions, the SUM is their charge.

SO42-, NO3

1-

Practice Assigning Oxidation Numbers

NO2 O is -2 x 2 = -4N must equal +4

N2O5 O is -2 x 5 = -10N must equal +10/2 = +5

HClO3 (+1) + (x) + 3(-2) = 0;x = (+5)

HNO3 O is -2 x 3 = -6; H is +1;N must equal +5

Ca(NO3)2 O is -2 x 3 = -6 x 2 = -12; Ca is +2; N must equal +10/2 = +5

KMnO4 O is -2 x 4 = -8; K is +1; Mn must equal +7

Fe(OH)3 O is -2 x 3 = -6; Fe is +3; H is +1 x 3 = +3

K2Cr2O7 O is -2 x 7 = -14; K is +1 x 2 = +2; Cr must equal +12/2 = +6

CO32-

x + 3(-2) = -2x = +4

CN- N is -3; Charge = -1 means that C must be +2

K3Fe(CN)6 N is -3 x 6 = -18; C is +2 x 6 = +12; K is +1 x 3 =+3; Fe must be +3

CH4H is +1 x 4 = +4;C must be -4

Using Oxidation Numbers

• an increase in the oxidation number indicates that an atom has lost electrons and therefore is oxidized

• a decrease in the oxidation number indicates that an atom has gained electrons and therefore reduced

• Example

Zn + CuSO4 ZnSO4 + Cu 0 +2 +6-2 +2+6-2 0

Zn: 0 + 2 Oxidized Cu: +2 0 ReducedS and O are unchanged

Exercise

For each of the following reactions (not balanced to simplify) find the element oxidized and the element reduced

Cl2 + KBr KCl + Br2  Cu + HNO3 Cu(NO3)2

+ NO2 + H2O

HNO3 + I2 HIO3 + NO2

ExerciseFor each of the following reactions find the element oxidized and the element reducedCl2 + KBr KCl + Br2

 0 +1-1 +1-1 0

Br loses an electron -- oxidized

Cl gains an electron -- reduced

K remains unchanged at +1

Exercise For each of the following reactions find the element oxidized and the element reduced

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

0 +1+5-2 +2 +5 -2 +4 –2 +1 -2•Cu increases from 0 to +2. It is oxidized•Only part of the N in nitric acid changes from +5 to +4. It is reduced

•The nitrogen that ends up in copper nitrate remains unchanged

Exercise For each of the following reactions

find the element oxidized and the element reduced

HNO3 + I2 HIO3 + NO2

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

• N is reduced from +5 to +4. It is reduced.• I is increased from 0 to +5 It is oxidized• The hydrogen and oxygen remain unchanged.

Agents

• all redox reactions have one element oxidized and one element reduced

• the compound that supplies the electrons (is oxidized) is the reducing agent

• the compound that accepts the electrons (is reduced) is the oxidizing agent

• occasionally, the same element may undergo both oxidation and reduction. This is known as an auto-oxidation reduction

• transition metals can form more than one type of ion (i.e. lose different amounts of electrons)– Cu1+, Cu 2+

• use roman numerals to indicate the charge– Cu1+ = “Copper I” ; Cu 2+ = “Copper II”

• Exceptions: Ag1+, Zn2+, Cd2+, Al3+, Sc3+

Review of Stock nomenclature (naming) of transitional metals

• ex. copper (II) oxide (“copper two oxide”)– CuO

– oxygen has a -2 charge, so it would only take one Cu2+ to bond with Oxygen.

• ex. copper (I) oxide– copper 1+ -we would need two of these to react

with oxygen so the formula would be:• Cu2O

Examples

• lead (II) hydroxide–Pb(OH)2

• cadmium nitrate– cadmium is always +2–Cd(NO3)2

• MnO2

– manganese (IV) oxide

Balancing Redox Reactions (only in Neutral or Acid Solutions)

• many chemical reactions involving oxidations and reductions are complex and very difficult to balance by the “guess and check” methods we learned earlier

• for complicated reactions, a more systematic approach is required

Half-equations

• half equations show the changes to individual species in a redox reaction– can use charges or oxidation #’s to do this

• Fe2O3 + 2 Al 2 Fe + Al2O3

– Fe3+ + 3 e- Fe ….this is the reduction– Al Al3+ + 3 e- ….this is the oxidation

• a wide variety of half equations can be found in the data booklet

More…

Br2 + 2I- 2Br- + I2

0 -1 -1 0• 2e- + Br2 2Br-

• 2I- I2 + 2e-

• 2e- + Br2 + 2I- 2Br- + I2 + 2e-

Balancing Redox Reactions in 8 “easy” steps. Look on page 847 of the text book for another way to do these steps.1. assign oxidation states for each atom

2. deduce which species is oxidized and which is reduced

3. write half reactions for the oxidation and reduction– take compounds where “action” took place, split them and write them

as individual reactions; there will be 2 half reactions

4. balance elements other than O and H

5. balance so # of electrons lost equals the # gained by adding e-

6. add the two half equations together to write the overall redox reaction and simplify

7. total up the total charges on the reactant and product sides to see if they are the same (not opposite)

8. balance the charges by adding H+ and balance oxygens by adding H2O to the appropriate sides

Exercise• Deduce and balanced redox equation and identify

the oxidizing and reducing agents.

Fe2+ + MnO4- Fe3+ + Mn2+

• Step 1. +2 +7 -2 +3 +2• Step 2. Fe is oxidized (Mn was the reducing agent

that caused this) & Mn is reduced (Fe was the oxidizing agent that caused this)

• Step 3. Fe2+ Fe3+ + e-

MnO4- + 5e- Mn2+

• Step 4. 5Fe2+ 5Fe3+ + 5e-

MnO4- + 5e- Mn2+

Exercise• Step 5. 5Fe2+ + MnO4

- 5Fe3+ + Mn2+

• Step 6. Total charge on reactant side = 9+

Total charge on product side = 17+• Step 7. To balance the equation charges, 8H+

must be added to the reactant side.

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+

Now need to balance the hydrogens by

adding water

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

More…• Nitric acid reacts with silver in a redox

reaction.Ag(s) + NO3

–(aq) → Ag+(aq) + NO(g)

• Using oxidation numbers, deduce the complete balanced equation for the reaction showing all the reactants and products.

Exercise• Deduce and balanced redox equation and identify the

oxidizing and reducing agents.

Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)

Step 1. 0 +5 -2 +1 +2-2

Step 2. Ag is oxidized & N is reduced

Step 3. Ag Ag+ + e-

NO3- + 3e- NO

Step 4. 3Ag 3Ag+ + 3e-

NO3- + 3e- NO

Exercise

Step 5. 3Ag + NO3– → 3Ag+ + NO

Step 6. Total charge on reactant side = 1-

Total charge on product side = 3+

Step 7. To balance the equation charges, 4H+

must be added to the reactant side.

3Ag + NO3– + 4H+ → 3Ag+ + NO

Now need to balance the hydrogens by

adding water

3Ag + NO3– + 4H+ → 3Ag+ + NO + 2H2O

Balance• Cr2O7

2- + Cl- Cr3+ + Cl2

Answer• 14H+ + Cr2O7

2- + 6Cl- 2Cr3+ + 7H2O + 3Cl2

Practice #1

Balance

• Cu (s) + NO31- (aq) Cu 2+ (aq) + NO2 (aq)

Answer• 4H+ + Cu (s) + 2NO3

1- (aq) Cu 2+ (aq) + 2NO2 (aq) + 2H2O

Practice #2

Balance

• Mn 2+ + NaBiO3 Bi 3+ + MnO4 1- + Na 1+

Answer• 2Mn 2+ + 5NaBiO3 + 14H+ 5Bi 3+ + 2MnO4

1- + 5Na 1+ + 7H2O

Practice #3

Balance

• NO2- + Cr2O72- Cr 3+ + NO3

1-

Answer

• 3NO2- + Cr2O72- + 8H+ 2Cr 3+ + 3NO3

1- + 4H2O

Practice #4

Balance

• S + HNO3 H2SO3 + N2O

Answer

• 2S + 2HNO3 + H2O 2H2SO3 + N2O

Practice #5