Redox Poerpoint 2017.ppt

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Oxidation and Reduction

Definitions of oxidation and reduction Definitions of oxidation and reduction

Oxidation numbers Oxidation numbers

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Oxidation - reduction

 _{Oxidation is loss of electrons}_{Oxidation is loss of electrons}  _{Reduction is gain of electrons}_{Reduction is gain of electrons}

 Oxidation is always accompanied by reduction_{Oxidation is always accompanied by reduction}

• The total number of electrons is kept constant_{The total number of electrons is kept constant}

 _{Oxidizing agents oxidize and are}_{Oxidizing agents oxidize and are} themselves reduced

themselves reduced

 _{Reducing agents reduce and are}_{Reducing agents reduce and are} themselves oxidized

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Oxidation 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 oxidation would possess negative oxidation numbers.

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 compound

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Predicting 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 charge

the 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 charge

charge

 F has oxidation number –1_{F has oxidation number –1}

 H has oxidn no. +1; except in metal hydrides where it is –1_{H has oxidn no. +1; except in metal hydrides where it is –1}

 Oxygen is Oxygen is usually –2. Except:usually –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 KO_{O is –1/2 in superoxides KO}₂₂

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Position of element in periodic table

determines oxidation number

 Group 1 is +1_{Group 1 is +1}

 Group 2 is +2_{Group 2 is +2}

 Group 13 is +3 (some rare exceptions)_{Group 13 is +3 (some rare exceptions)}

 Group 15 are –3 in compounds with metals, H or with _{Group 15 are –3 in compounds with metals, H or with}

NH

NH4+4+. Exceptions are in compounds to the right; in _{. Exceptions are in compounds to the right; in} which case use rules 3 and 4.

which case use rules 3 and 4.

 Group 16 below O are –2 in binary compounds with _{Group 16 below O are –2 in binary compounds with}

metals, H or NH

metals, H or NH4+4+. When they are combined with O or _{. When they are combined with O or} with a lighter halogen, use rules 3 and 4.

with a lighter halogen, use rules 3 and 4.

 Group 17 elements are –1 in binary compounds with _{Group 17 elements are –1 in binary compounds with}

metals, H or NH

metals, H or NH4+4+ or with a heavier halogen. When _{or with a heavier halogen. When}

combined with O or a lighter halogen, use rules 3 and 4.

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Identifying reagents

 _{Those elements that tend to give up}_{Those elements that tend to give up}

electrons (metals) are typically categorized

as reducing agents and those that tend to

accept electrons (nonmetals) are referred to

as oxidizing agents.

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Iron can reduce Cu

2+2+

to Cu

 _{The iron nail reduces the Cu}_{The iron nail reduces the Cu}2+2+ ions and _{ions and}

becomes coated with metallic Cu. At the

same time, the intensity of the blue color

diminishes due to loss of Cu

diminishes due to loss of Cu2+2+ ions from _{ions from}

solution.

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Any element can be both an oxidizer and

reducer depending on relative positions

in the activity series

 _{Fe reduced Cu}_{Fe reduced Cu}2+2+, but Cu can reduce Ag_{, but Cu can reduce Ag}+ +

(lower activity

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More active metals are strongly

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Predicting results of displacement

reactions

 _{In this reaction the element metal A}_{In this reaction the element metal A}

displaces 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 series

activity series

 _{Displacement reaction exercises}_{Displacement reaction exercises}

)

(

)

(

)

(

)

(

s

BX

aq

B

s

AX

aq

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Nuggets of redox processes

 Where there is oxidation there is _{Where there is oxidation there is}alwaysalways

reduction reduction

Oxidizing agent

Oxidizing agent Reducing agentReducing agent

Is

Is itselfitself reduced reduced Is Is itselfitself oxidized oxidized

Gains

Gains electrons electrons LosesLoses electrons electrons

Causes oxidation

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Identify redox by change in oxidation

numbers: follow the flow of electrons

 Reducing agent increases its oxidation number_{Reducing agent increases its oxidation number}

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In reaction with metals, nonmetals

are always oxidizers

 _{Reactions of elements are always redox}_{Reactions of elements are always redox}

 _{The nonmetal gains electrons, becomes a}_{The nonmetal gains electrons, becomes a} negative ion

negative ion

 _{The metal loses electrons, becomes a}_{The metal loses electrons, becomes a} positive ion

positive ion

 _{Identification is harder when there are no}_{Identification is harder when there are no}

elements involved: oxidation numbers must

be used

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Balancing redox equations:

systematic methods

 _{Oxidation number method – tracking}_{Oxidation number method – tracking} changes in the oxidation numbers

changes in the oxidation numbers

 _{Half-reaction method – tracking changes in}_{Half-reaction method – tracking changes in} the flow of electrons

the flow of electrons

 _{Same principles, different emphasis}_{Same principles, different emphasis}

 _{Use is a matter of choice, but familiarity with}_{Use is a matter of choice, but familiarity with} both is important

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Oxidation 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 zero

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Six habits of the redox equation

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Walking through the steps

 STEP 1: write the unbalanced net ionic equation_{STEP 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 Mn

bromide ion to give Mn2+2+ ion and bromine_{ion and bromine}

) (

)

( 2 ₂

4 aq Br aq Mn aq Br aq

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STEP 2: Balance the equation for

elements 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 equation

equation

*

*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 later and hydronium ions later

)

(

)

(

)

(

2

)

(

2 ₂

4

aq

Br

aq

Mn

aq

Br

aq

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STEP 3: Assign oxidation numbers

 Use the rules of oxidation numbers_{Use 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} charge

charge

 Oxide is -2_{Oxide is -2}

)

(

)

(

)

(

2

)

(

2 ₂

4

aq

Br

aq

Mn

aq

Br

aq

MnO











0 +2

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STEP 4: Identify oxidized and

reduced

 Mn is reduced from +7 to +2 _{Mn is reduced from +7 to +2}

 Net gain of 5 electrons_{Net gain of 5 electrons}

 Br is oxidized from -1 to 0_{Br is oxidized from -1 to 0}

 Net loss of 1 electron_{Net loss of 1 electron}

)

(

)

(

)

(

2

)

(

2 ₂

4

aq

Br

aq

Mn

aq

Br

aq

MnO











0 +2

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STEP 5: Balance the oxidized and

reduced species

 _{For every Mn reduced (decrease in}_{For every Mn reduced (decrease in} oxidation number of 5), need five Br

oxidation number of 5), need five Br-

-oxidized (increase in oxidation number of 1)

oxidized (increase in oxidation number of 1)

 _{Equation becomes}_{Equation becomes}

 _{Redox is now complete but material balance}_{Redox is now complete but material balance} is not

is not

) (

2 )

( 2 ₂

4

5

2

5

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STEP 6: Material balance with H

₂₂

O

and H

++

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

and add H+ + (the reaction is in acid solution) to _{(the reaction is in acid solution) to} the other side

the other side

 In basic solution we add OHIn basic solution we add OH-- and H_{and H} 2

2O instead O instead

of H

of H₂₂O and H_{O and H}++ respectively_respectively

 Test equation for both atoms and charges_{Test equation for both atoms and charges}

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

( 2 ₂ ₂

4 2 5

2MnO aq  Br aq  H aq  Mn  aq  Br aq  H O l

) ( ) ( ) ( 2 )

( 2 ₂

4

5

2

5

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The 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 reduction

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Six habits of the redox equation

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STEP 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)

) (

)

( 3 ₂

2 7

2O aq Cl aq Cr aq Cl aq

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STEP 2: identify the oxidized and

reduced and write the half reactions

 _{Oxidation half-reaction}_{Oxidation half-reaction}

 _{Reduction half-reaction}_{Reduction half-reaction}

)

(

)

(

aq

Cl

₂

aq

Cl





) (

)

( 3

2 7

2O aq Cr aq

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STEP 3: Balance the half reactions

 _Oxidation_Oxidation

 _Reduction_Reduction

)

(

)

(

2

Cl



aq



Cl

₂

aq

) (

2 )

( 3

2 7

2O aq Cr aq

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STEP 4: Material balance

 _{As with the oxidation number method, add}_{As with the oxidation number method, add} H

H₂₂O 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)

 _{Oxidation reaction – unchanged}_{Oxidation reaction – unchanged}

 _{Reduction reaction}_{Reduction reaction}

)

(

)

(

2

Cl



aq



Cl

₂

aq

) ( 7 ) ( 2 ) ( ) (

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STEP 5: Balance half-reactions for

charge 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-reaction

sides of each half-reaction

 

aq



Cl

aq



e

Cl

(

)

(

)

2

₂ ) ( 7 ) ( 2 6 ) ( ) (

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STEP 5 cont: Multiply by factors to

balance total electrons

 _{Overall change in electrons must be zero}_{Overall change in electrons must be zero}  _{Multiply the oxidation half reaction by 3}_{Multiply the oxidation half reaction by 3}



Cl



(

aq

)



Cl

(

aq

)



2

e





3

₂

) ( 7

) (

2 6

) (

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STEP 6: Add half reactions and

eliminate common items

+

=

Atoms and charges balance

 

aq



Cl

aq



e

Cl

(

)

3

(

)

6

₂ ) ( 7 ) ( 2 6 ) ( ) (

14H  aq  Cr₂O₇2 aq  e  Cr3 aq  H₂O l

) ( 7 ) ( 3 ) ( 2 ) ( 6 ) ( ) (

14 2 3 ₂ ₂

7

2O aq Cl aq Cr aq Cl aq H O l

Cr aq

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Redox 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 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

oxidized and reduced states

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Using 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 MnO

determine the concentration of the MnO₄₄-

- MnO_MnO₄₄- - has an intense purple colour, whereas Mn_{has an intense purple colour, whereas Mn}2+2+

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Strategy

 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 H

purple 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}

MnO

MnO₄₄-

- The volume of solution yields the _{The volume of solution yields the}

concentration