Inorganic Chemistry IReduction-Oxidation (REDOX)

Kun Sri Budiasih

Oxidation-Reduction Reactions

• Reactions that involve the transfer of electrons are called oxidation-reduction or redox reactions

• Oxidation is the loss of electrons by a reactant

• Reduction is the gain of electrons by a reactant

• Oxidation and reduction always occur together

• The total number of electrons lost by one substance is the same as the total number of electrons gained by the other

• For a redox reaction to occur, something must accept the electrons that are lost by another substance

• The substance that accepts the electrons is called the oxidizing agent

• The substance that lost the electrons is called the reduction agent

• Note that the oxidizing agent is reduced and the reducing agent is oxidized

• For example:2 Na + Cl2 2 NaCl

– Na is the reducing agent because it lost electrons and was oxidized

– Cl2 is the oxidizing agent because it gained electrons and was reduced

• Oxidation numbers provide a way to keep track of electron transfers :

1) The oxidation number of any free element is zero.2) The oxidation number of any simple, monoatomic

ion is equal to the charge on the ion.3) The sum of all oxidation numbers of the atoms in

a molecule or polyatomic ion must equal the charge on the particle.

4) In its compounds, fluorine has an oxidation number of –1.

5) In its compounds, hydrogen has an oxidation number of +1.

6) In its compounds, oxygen has an oxidation number of –2.

• If there is a conflict between two rules apply the rule with the lower number and ignore the conflicting rule

• In binary ionic compounds with metals, the nonmetals have oxidation numbers equal to the charges on their anions

Example: What is the oxidation number of Fe in Fe2O3?ANALYSIS: This binary compound is ionic.

Apply rule 3 and 6

Fe: 2x

O: 3(-2) = -6

0 = 2x + (-6) or x = +3 = ox. number of Fe

• Note that fractional values of oxidation numbers are allowed

• In terms of oxidation numbers:– Oxidation is an increase in oxidation number– Reduction is a decrease in oxidation number

• This provides a simple way to follow redox reactions

• Many redox reactions take place in aqueous solution

• A procedure called the ion-electron method provides a way to balance these equations

• The oxidation and reduction are divided into equations called half-reactions

• The half-reactions are balanced separately, then combined into the fully balanced net ionic equation

• Both mass and charge must be balanced

• Charge is balanced by adding electrons to the side of the equation that is more positive or less negativeExample: Balance the following skeleton

equation

)3Cu( )(2Al )(3Cu )Al( 2

combining 6, isfactor common least The

)Cu( 2e )(Cu :Reduction

3e Al )Al( :Oxidation

:SOLUTION

reaction.redox a is This :ANALYSIS

)Cu( )(Al )(Cu )Al(

32

-2

-3

32

saqaqs

saq

s

saqaqs

• Many reactions occur in either acidic or basic solutions

The Ion-Electron Method in Acidic Solution:1) Divide the equation into two half-reactions.

2) Balance atoms other than H and O.

3) Balance O by adding water.

4) Balance H by adding hydrogen ion.

5) Balance net charge by adding electrons.

6) Make electron gain and loss equal: add half-reactions.

7) Cancel anything that’s the same on both sides of the equation.

• The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution:Additional Steps for Basic Solutions

can.you that OHany Cancel 10)

O.H form toOH and H Combine 9)

.H are thereas ionsOH of

number same heequation t theof sidesboth toAdd 8)

2

2-

-

Example: Balance the following in basic solution:

2223

-4

--242

2223

-4

-2

-242

2

223

-4

-2

-242

-

223

-42

-242

22--

4

--232

-242

-232

-242

-4

2MnOO2H6CO2MnO4OHO3C

Simplify

2MnOO4H6CO2MnO4OHO2HO3C

OH Form

2MnO)OH4(H6CO2MnO4OHO2HO3C

OH Add

2MnO4H6CO2MnOO2HO3C :ionicNet

O2H MnO 3e 4H MnO

2e 4H 2CO O2H OC

:SOLUTION

convert"."hen solution t acidicin if as Balance :ANALYSIS

CO MnO OC MnO

• Metals more active than hydrogen (H2) dissolve in oxidizing acids

O4H)S(H 8e10HSO

agent) reducing strong with conc.,(hot

O2H)(SO3e4HSO conc.) (hot, SOH

O3HNH8e10HNO

agent) reducing strong with dilute,(very

O4H)NO(3e4HNO (dilute)

OH)(NOe2HNO (conc.) HNO

ReactionReduction Acid Ox.

acids. organicmost and ),(POH

),(SOH dilute cold ),HCl( :acids ngNonoxidizi

22--

4

22--2

442

24--

3

2--

3

22--

33

43

42

g

g

g

g

aq

aqaq

• Some examples:

• More active metals will displace a less active metal from its compound

• This often occurs in solution and is called a single replacement reaction

O2H)(SO)(CuSO)(SO2H)Cu(

:SOH edconcentratHot

O4H)2NO()()Cu(NO3)(8HNO)3Cu(

:HNO Dilute

O2H)(2NO)()Cu(NO)(4HNO)Cu(

:HNO edConcentrat

22442

42

2233

3

22233

3

gaqaqs

gaqaqs

gaqaqs

• An activity series arranges metals according to their ease of oxidation

• They can be used to predict reactions

Zinc is a more active metal than copper. Copper ions (blue) collide with zinc metal (gray) picking up electrons. Copper ions become copper atoms (red-brown) and stick to the zinc surface. Zinc ions (yellow) replace the copper ions in solution.

Activity Series for Some Metals and Hydrogen

• A given element will be displaced from its compounds by any element below it in the table

Cs Cesium activeMost

Na Sodium

H Hydrogen

Cu Copper

Ag Silver

Au Gold activeLeast

ProductOxidation Element

2

3

(See Table 6.2 for a more extensive list.)

• Oxygen reacts with many substances

• The products depends, in part, on how much oxygen is available

• Combustion of hydrocarbons

• Organic compounds containing O also produce carbon dioxide and water

O2HCOCH :limited very O

O4HCO23O2CH :limited O

O2HCO2OCH :plentiful O

2242

2242

22242

O3H2CO3OOHHC 22252

• Organic compounds containing S produce sulfur dioxide

• Many metals corrode or tarnish when exposed to oxygen

• Most nonmetals react with oxygen directly

• Redox reactions are more complicated than most metathesis reactions

222252 2SOO6H4CO9OSHH2C

O2AgO4Ag :silver of Tarnishing

O2Fe3O4Fe :iron ofCorrosion

22

322

CO2O2C :limited O

COOC :plentiful O

22

222

• In general, it is not possible to balance a redox reaction by inspection

• This is especially true when acid or bases are involved in the reaction

• Once balanced, they can be used for stoichiometric calculations

• Redox titrations are common because they often involve dramatic color changes

• Mole-to-mole ratios are usually involved

Example: A 0.3000 g sample of tin ore was dissolved in acid solution converting all the tin to tin(II). In a titration, 8.08 mL of 0.0500 M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample?

ANALYSIS: This is a redox titration in acidic solution.

SOLUTION:

• Form skeleton equation and use the ion-electron method to produce a balanced equation

O4H2MnO3Sn8H2MnO3Sn 224-

42

• Use the balanced equation to define equivalence relations and determine the mass of Sn in the original sample

• Convert to percentageSn g 0.0719

sol MnO L 0.00808

Sn mol 1.000Sn g 118.7

Sn mol 1Sn mol 1

MnO mol 2Sn mol 3

sol MnO L 1.00

MnO mol 0.0500 -4

2

-4

2

-4

-4

orein Sn %0.24%100 Sn % g 0.3000g 0.0719