Oxidation and ReductionOxidation and Reduction

Definitions of oxidation and reductionDefinitions of oxidation and reduction

Oxidation numbersOxidation numbers

Redox equationsRedox equations

Oxidation - reductionOxidation - reduction

Oxidation is loss of electronsOxidation is loss of electrons Reduction is gain of electronsReduction is gain of electrons

Oxidation is always accompanied by reductionOxidation is always accompanied by reduction• The total number of electrons is kept constantThe total number of electrons is kept constant

Oxidizing agents oxidize and are Oxidizing agents oxidize and are themselves reducedthemselves reduced

Reducing agents reduce and are Reducing agents reduce and are themselves oxidizedthemselves oxidized

Follow the electronsFollow the electrons

Oxidation numbersOxidation numbers

Metals are typically considered Metals are typically considered more 'cation-like' and would more 'cation-like' and would possess positive oxidation possess positive oxidation numbers, while nonmetals are numbers, while nonmetals are considered more 'anion-like' and considered more 'anion-like' and would possess negative would possess negative oxidation numbers.oxidation numbers.

Oxidation number is the number of electrons Oxidation number is the number of electrons gained or lost by the element in making a gained or lost by the element in making a compoundcompound

Predicting oxidation numbersPredicting oxidation numbers

Oxidation number of atoms in element is zero in all casesOxidation number of atoms in element is zero in all cases Oxidation number of element in monatomic ion is equal to Oxidation number of element in monatomic ion is equal to

the chargethe charge sum of the oxidation numbers in a compound is zero sum of the oxidation numbers in a compound is zero sum of oxidation numbers in polyatomic ion is equal to the sum of oxidation numbers in polyatomic ion is equal to the

chargecharge F has oxidation number –1F has oxidation number –1 H has oxidn no. +1; except in metal hydrides where it is –1H has oxidn no. +1; except in metal hydrides where it is –1 Oxygen is Oxygen is usually usually –2. Except:–2. Except:

O is –1 in hydrogen peroxide, and other peroxidesO is –1 in hydrogen peroxide, and other peroxides O is –1/2 in superoxides KOO is –1/2 in superoxides KO22

In OFIn OF22 O is +2 O is +2

Position of element in periodic table Position of element in periodic table determines oxidation number determines oxidation number

G1A is +1G1A is +1 G2A is +2G2A is +2 G3A is +3 (some rare exceptions)G3A is +3 (some rare exceptions) G5A are –3 in compounds with metals, H or with NHG5A are –3 in compounds with metals, H or with NH4+4+. .

Exceptions are in compounds to the right; in which case Exceptions are in compounds to the right; in which case use rules 3 and 4.use rules 3 and 4.

G6A below O are –2 in binary compounds with metals, G6A below O are –2 in binary compounds with metals, H or NHH or NH4+4+. When they are combined with O or with a . When they are combined with O or with a lighter halogen, use rules 3 and 4.lighter halogen, use rules 3 and 4.

G7A elements are –1 in binary compounds with metals, G7A elements are –1 in binary compounds with metals, H or NHH or NH4+4+ or with a heavier halogen. When combined or with a heavier halogen. When combined with O or a lighter halogen, use rules 3 and 4.with O or a lighter halogen, use rules 3 and 4.

Identifying reagentsIdentifying reagents

Those elements that tend to give up Those elements that tend to give up electrons (metals) are typically categorized electrons (metals) are typically categorized as reducing agents and those that tend to as reducing agents and those that tend to accept electrons (nonmetals) are referred to accept electrons (nonmetals) are referred to as oxidizing agents.as oxidizing agents.

Iron can reduce CuIron can reduce Cu2+2+ to Cu to Cu

The iron nail reduces the CuThe iron nail reduces the Cu2+2+ ions and ions and becomes coated with metallic Cu. At the becomes coated with metallic Cu. At the same time, the intensity of the blue color same time, the intensity of the blue color diminishes due to loss of Cudiminishes due to loss of Cu2+2+ ions from ions from solution.solution.

Any element can be both an oxidizer and Any element can be both an oxidizer and reducer depending on relative positions reducer depending on relative positions

in the activity seriesin the activity series Fe reduced CuFe reduced Cu2+2+, but Cu can reduce Ag, but Cu can reduce Ag+ +

(lower activity(lower activity FeFe2+2+ is reduced by Zn is reduced by Zn

More active metals are strongly More active metals are strongly reducingreducing

Predicting results of displacement Predicting results of displacement reactionsreactions

In this reaction the element metal A In this reaction the element metal A displaces the ion metal B from its compounddisplaces the ion metal B from its compound

This will only occur if A lies above B in the This will only occur if A lies above B in the activity seriesactivity series

Displacement reaction exercisesDisplacement reaction exercises

)()()()( aqAXsBaqBXsA

Nuggets of redox processesNuggets of redox processes

Where there is oxidation there is Where there is oxidation there is alwaysalways reductionreduction

Oxidizing agentOxidizing agent Reducing agentReducing agent

Is Is itselfitself reduced reduced Is Is itselfitself oxidized oxidized

GainsGains electrons electrons LosesLoses electrons electrons

Causes oxidationCauses oxidation Causes reductionCauses reduction

Identify redox by change in oxidation Identify redox by change in oxidation numbers: follow the flow of electronsnumbers: follow the flow of electrons Reducing agent increases its oxidation numberReducing agent increases its oxidation number Oxidizing agent decreases its oxidation numberOxidizing agent decreases its oxidation number

In reaction with metals, nonmetals In reaction with metals, nonmetals are always oxidizersare always oxidizers

Reactions of elements are always redoxReactions of elements are always redox The nonmetal gains electrons, becomes a The nonmetal gains electrons, becomes a

negative ionnegative ion The metal loses electrons, becomes a The metal loses electrons, becomes a

positive ionpositive ion Identification is harder when there are no Identification is harder when there are no

elements involved: oxidation numbers must elements involved: oxidation numbers must be usedbe used

Balancing redox equations: Balancing redox equations: systematic methodssystematic methods

Oxidation number method – tracking Oxidation number method – tracking changes in the oxidation numberschanges in the oxidation numbers

Half-reaction method – tracking changes in Half-reaction method – tracking changes in the flow of electronsthe flow of electrons

Same principles, different emphasisSame principles, different emphasis Use is a matter of choice, but familiarity with Use is a matter of choice, but familiarity with

both is importantboth is important

Oxidation number methodOxidation number method

What goes up must come down…What goes up must come down… Sum of the changes in oxidation numbers in Sum of the changes in oxidation numbers in

any process is zeroany process is zero

Six habits of the redox equation Six habits of the redox equation balancerbalancer

Walking through the stepsWalking through the steps

STEP 1: write the unbalanced net ionic equationSTEP 1: write the unbalanced net ionic equation

In an acid solution, permanganate is reduced by In an acid solution, permanganate is reduced by bromide ion to give Mnbromide ion to give Mn2+2+ ion and bromine ion and bromine

)()()()( 22

4 aqBraqMnaqBraqMnO

STEP 2: Balance the equation for STEP 2: Balance the equation for elements other than O and Helements other than O and H**

We need to double the bromide ions We need to double the bromide ions on the L.H.S. to balance the on the L.H.S. to balance the equationequation

**O and H can remain unbalanced O and H can remain unbalanced because we will top up with water because we will top up with water and hydronium ions laterand hydronium ions later

)()()(2)( 22

4 aqBraqMnaqBraqMnO

STEP 3: Assign oxidation numbersSTEP 3: Assign oxidation numbers

Use the rules of oxidation numbersUse the rules of oxidation numbers Element is zero Element is zero Monatomic ion: oxidation number is same as Monatomic ion: oxidation number is same as

chargecharge Oxide is -2Oxide is -2

)()()(2)( 22

4 aqBraqMnaqBraqMnO

0+2-1+7

STEP 4: Identify oxidized and STEP 4: Identify oxidized and reducedreduced

Mn is reduced from +7 to +2 Mn is reduced from +7 to +2 Net gain of 5 electronsNet gain of 5 electrons

Br is oxidized from -1 to 0Br is oxidized from -1 to 0 Net loss of 1 electronNet loss of 1 electron

)()()(2)( 22

4 aqBraqMnaqBraqMnO

0+2-1+7

STEP 5: Balance the oxidized and STEP 5: Balance the oxidized and reduced speciesreduced species

For every Mn reduced (decrease in For every Mn reduced (decrease in oxidation number of 5), need five Broxidation number of 5), need five Br-- oxidized (increase in oxidation number of 1)oxidized (increase in oxidation number of 1)

Equation becomesEquation becomes

Redox is now complete but material balance Redox is now complete but material balance is notis not

)()()(2)( 22

4 5252 aqBraqMnaqBraqMnO

STEP 6: Material balance with HSTEP 6: Material balance with H22O O

and Hand H++

Strategy: add HStrategy: add H22O to the side that lacks for O O to the side that lacks for O

and add Hand add H+ + (the reaction is in acid solution) to (the reaction is in acid solution) to the other sidethe other side

In basic solution we add OHIn basic solution we add OH-- and H and H22O O

instead of Hinstead of H22O and HO and H++ respectively respectively

Test equation for both atoms and chargesTest equation for both atoms and charges

)(8)()()(16)(10)( 222

4 522 lOHaqBraqMnaqHaqBraqMnO

)()()(2)( 22

4 5252 aqBraqMnaqBraqMnO

The Half-Reaction methodThe Half-Reaction method

Any redox process can be written as the Any redox process can be written as the sum of two half reactions: one for the sum of two half reactions: one for the oxidation and one for the reductionoxidation and one for the reduction

Six habits of the redox equation Six habits of the redox equation balancerbalancer

STEP 1: the unbalanced equationSTEP 1: the unbalanced equation

Dichromate ion reacts with chloride ion to Dichromate ion reacts with chloride ion to produce chlorine and chromium (III)produce chlorine and chromium (III)

)()()()( 232

72 aqClaqCraqClaqOCr

STEP 2: identify the oxidized and STEP 2: identify the oxidized and reduced and write the half reactionsreduced and write the half reactions

Oxidation half-reactionOxidation half-reaction

Reduction half-reactionReduction half-reaction

)()( 2 aqClaqCl

)()( 3272 aqCraqOCr

STEP 3: Balance the half reactionsSTEP 3: Balance the half reactions

OxidationOxidation

ReductionReduction

)()(2 2 aqClaqCl

)(2)( 3272 aqCraqOCr

STEP 4: Material balanceSTEP 4: Material balance

As with the oxidation number method, add As with the oxidation number method, add HH22O to the side lacking O and add HO to the side lacking O and add H++ to the to the

other side (for reactions in acid solution)other side (for reactions in acid solution) Oxidation reaction – unchangedOxidation reaction – unchanged

Reduction reactionReduction reaction

)()(2 2 aqClaqCl

)(7)(2)()(14 232

72 lOHaqCraqOCraqH

STEP 5: Balance half-reactions for STEP 5: Balance half-reactions for charge by addition of electronscharge by addition of electrons

No explicit calculation of oxidation numbers No explicit calculation of oxidation numbers is required; we balance the is required; we balance the chargescharges on both on both sides of each half-reactionsides of each half-reaction

eaqClaqCl 2)()(2 2

)(7)(26)()(14 232

72 lOHaqCreaqOCraqH

STEP 5 cont: Multiply by factors to STEP 5 cont: Multiply by factors to balance total electronsbalance total electrons

Overall change in electrons must be zeroOverall change in electrons must be zero Multiply the oxidation half reaction by 3Multiply the oxidation half reaction by 3

eaqClaqCl 2)()(3 2

)(7)(26)()(14 232

72 lOHaqCreaqOCraqH

STEP 6: Add half reactions and STEP 6: Add half reactions and eliminate common itemseliminate common items

++

==

Atoms and charges balanceAtoms and charges balance

eaqClaqCl 6)(3)(6 2

)(7)(26)()(14 232

72 lOHaqCreaqOCraqH

)(7)(3)(2)(6)()(14 2232

72 lOHaqClaqCraqClaqOCraqH

Redox TitrationsRedox Titrations

Acid-base titration is used to determine an Acid-base titration is used to determine an unknown concentration (either acid or base)unknown concentration (either acid or base)

The endpoint is manifested in a color The endpoint is manifested in a color change (of an indicator) or by measuring pHchange (of an indicator) or by measuring pH

In redox titrations, the concentration of one In redox titrations, the concentration of one of the reagents can be measured, provided of the reagents can be measured, provided there is a sharp distinction between the there is a sharp distinction between the oxidized and reduced statesoxidized and reduced states

Using the roadmapUsing the roadmap

Oxalic acid is oxidized by MnOOxalic acid is oxidized by MnO44--

A known quantity of oxalic acid is used to A known quantity of oxalic acid is used to determine the concentration of the MnOdetermine the concentration of the MnO44

--

MnOMnO44- - has an intense purple colour, whereas Mnhas an intense purple colour, whereas Mn2+2+

is almost colourlessis almost colourless

StrategyStrategy

A known amount of HA known amount of H22CC22OO44 is used is used

A solution of KMnOA solution of KMnO44 is titrated till the first is titrated till the first

purple colour – the endpoint. All the Hpurple colour – the endpoint. All the H22CC22OO44 is oxidized.is oxidized.

The equation gives the number of moles of The equation gives the number of moles of MnOMnO44

--

The volume of solution yields the The volume of solution yields the concentrationconcentration