redox

RedoxPrinciples and application

Course outlineOxidation and reductionRedox reactions

explain oxidation and reduction as an electron transfer process

calculate oxidation numbers

identify and name oxidants and reductants in equations

identify oxidation-reduction reactions using oxidation numbers

Oxidation and reductionRedox reactions

describe, write equations for and interpret observations for:

metal displacement reactions

halogen displacement reactions

write balanced simple redox equations (metal/metal ion, metal/hydrogen ion and halogen/halide ion)

Course outline

OXIDATION

Used to be viewed as

Gaining oxygen 2 Mg + O2 → 2 MgO

Magnesium gains oxygen

Losing hydrogen CH4 + 2 S → C + 2 H2S

Carbon loses hydrogen

REDUCTIONUsed to be viewed as

Losing oxygen FeO + C → Fe + CO

Iron loses oxygen

(Note: carbon also gains oxygen = oxidation)

Gaining hydrogen CH4 + 2 O2 → CO2 + 2 H2O

Oxygen gains hydrogen

(Note: carbon also loses hydrogen and gains oxygen = oxidation)

It was observed that very often when one substance was being oxidised another was being reduced at the same time.

Modern view of oxidation and

reductionOxidation is defined as losing of electrons LEO

Reduction is defined as gaining of electrons GER

Modern view of oxidation and

reductionOxidation Is Loss of electronsReduction Is Gain of electrons

OIL RIG

A redox reaction is an

electron transfer process

REDOX = electron transfer

Reducing agent

Oxidising agent

Loses electrons

Gains electrons

2 Fe (ON=0)

O2 (ON=0)

2 Fe2+O2 –+

4e-

`

4e-4e-

4e-

4e-

4e-

4e-

4e- 4e- 4e- 4e-

REDOX = electron transfer

Reducing agent

Oxidising agent

Loses electrons

Gains electrons

2 Na (ON=0)

Cl2 (ON=0)

2 Na+Cl –+

2e-

`

2e-2e-

2e-

2e-

2e-

2e-

2e- 2e- 2e- 2e-

An oxidising An oxidising agentagent

Oxidises another substance by taking electrons from it

Is also called the oxidant

Gains electrons, and therefore

Becomes reduced

A reducing A reducing agentagent

Reduces another substance by donating electrons to it

Is also called the reductant

Loses electrons, and therefore

Becomes oxidized

A redox equation analysed

Mg + S → Mg2+ + S2–

Mg loses electrons

Is oxidized

Is the reductant

S gains electrons

Is reduced

Is the oxidant

Electron

transfer

0 +20 –2

What is the ON of P in PH3

ONP + 3 ONH = 0 (neutral)

ONP + 3 (+1) = 0

ONP + 3 = 0

ONP = –3

ON of P in H3PO4

3H + P + 4O = 0

3(+1) + P + 4(–2) = 0

(+3) + P + (–8) = 0

P + (–5) = 0

P = +5

ON of P in H3PO3

P = +3

3H + P + 3O = 0

3(+1) + P + 3(–2) = 0

(+3) + P + (–6) = 0

P + (–3) = 0

ON of Cr in CrO3

Cr + 3O = 0

Cr + 3(– 2) = 0

Cr + (– 6) = 0

Cr = + 6

ON of Cr in Cr2O72-

2Cr + 7O = – 2

2Cr + 7(– 2) = – 2

2Cr = 14 – 2

2Cr = + 12

Cr = + 6

ON of Cr in CrO42-

Cr + 4O = – 2

Cr + 4(– 2) = – 2

Cr = 8 – 2

= + 6

ON of Cr in (NH4)2CrO4

2NH4 + Cr + 4O = 0

2(+1) + Cr + 4(–2) = 0

Cr +2 – 8 = 0

Cr = + 6

ON of C in CH4

C + 4H = 0

C + 4(+1) = 0

C = – 4

ON of C in CH2O

C + 2H + O = 0

C + 2(+1) +(– 2) = 0

C + 2 + (– 2) = 0

C = 0

O is more electronegative

than C

C has ‘gained’

these 2

electrons

CeO

e e

e

H

Heee

ee

e e

e

H is less electronegative than

C

How can combined C have an ON of zero?

Carbon has ‘lost’ 2 electrons to Oxygenbut ‘gained’ 2 from Hydrogen

C has ‘lost’ these 2

electrons

ON of C in CH3OH

C + 4H + O = 0

C + 4(+1) + (– 2) = 0

C + 4 + (– 2) = 0

C + 2 = 0

C = – 2

ON of C in C6H6

6C + 6H = 0

6C + 6(+1) = 0

6C + 6 = 0

6C = – 6

C = – 1

ON of Cl in HClO4

H + Cl + 4O = 0

(+1) + Cl + 4(– 2) = 0

(+1) + Cl + (– 8) = 0

Cl + (– 7) = 0

Cl = +7

ON of Se in H2SeO3

2H + Se + 3O = 0

2(+1) + Se + 3(– 2) = 0

(+2) + Se + (– 6) = 0

Se + (– 4) = 0

Se = +4

ON of Se in SeO3

Se + 3O = 0

Se + 3(– 2) = 0

Se + (– 6) = 0

Se = +6

ON of Mn in MnO4 -

Mn + 4O = – 1

Mn + 4(– 2) = – 1

Mn + (– 8) = – 1

Mn = +7

ON of Mn in MnO4 2 –

Mn + 4O = – 2

Mn + 4(– 2) = – 2

Mn + (– 8) = – 2

Mn = +6

ON of Mn in MnO2

Mn + 2O = 0

Mn + 2(– 2) = 0

Mn + (– 4) = 0

Mn = +4

ON of Mn in Mn2O3

2Mn + 3O = 0

2Mn + 3(– 2) = 0

2Mn + (– 6) = 0

2Mn = +6

Mn = +3

In HClO4 the

•ON of Cl is +7

•ON of O is – 2

•ON of H is +1

H + Cl + 4O = 0

+1 + Cl – 8 = 0

Cl = 8 – 1 = +7

With the aid of an electron-dot diagram, show this is consistent with the

electronegativities of the atoms in each bond.EN: O = 3.4 Cl = 2.8 H = 2.2

And also these

2 electrons

As well these 2

electrons

And this 1

electron

e e

e

e e

e

e

e

e e

ClO O

Oe e

O

e e

e e

e e

e

e e

e

e

e

e e

e e e e

e eH

Cl has ‘lost’ these 2

electrons

Cl has ‘lost’ 7

electrons

ON = +7

Each O has

‘gained’ 2

electrons

ON = – 2

H has ‘lost’ 1

electron

ON = +1

In H2SO4 the

•ON of S is +6

•ON of O is – 2

•ON of H is +1With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond.

EN: O = 3.4 S = 2.6 H = 2.2

2H + S + 4O = 0

+2 + S – 8 = 0

S = 8 – 2 = +6

And also these

2 electrons

And this 1

electron

S has ‘lost’ these 2

electrons

S has ‘lost’ 6

electrons

ON = +6

Each O has

‘gained’ 2

electrons

ON = – 2

Each H has

‘lost’ 1

electron

ON = +1

S

e ee eOe e

e e

e eOe e

e e

e e

e eOe ee e

e e

H

e Oe

e e

e e

ee H

As well as this

1 electron

Write the half equation for the conversion of IO3

– to I21. Write the ‘skeleton’ equation including

the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced.

I + 3(–2) = –1

so I = (+5)

IO3– → I2

(+5)(–2) (0)

I is reduced


– to I22. Balance the numbers of atoms of the

element that is being oxidized/reduced

2 IO3– → I2

IO3– → I2


– to I23. Balance the number of oxygen atoms by

adding H2O to the side that needs more O

2 IO3– → I2

IO3– → I2

2 IO3– → I2 + 6 H2O


– to I24. Balance the number of hydrogen atoms by

adding H+ to the side that needs more H

2 IO3– → I2

IO3– → I2

2 IO3– → I2 + 6 H2O

2 IO3– + 12 H+ → I2 + 6 H2O


– to I25. Balance the charge by adding electrons to

the side that has more positive charge

2 IO3– → I2

IO3– → I2

2 IO3– → I2 + 6 H2O

2 IO3– + 12 H+ → I2 + 6 H2O

2 IO3– + 12 H+ + 10e – → I2 + 6 H2O


– to I2 Now amend the equation for this reaction

in BASIC solution

2 IO3– + 12 H+ + 10e – → I2 + 6 H2O

Add 12 OH– to both sides of the half equation.

12 OH– 12 OH–



in BASIC solution

2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH–

The OH– and H+ neutralise



in BASIC solution

2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH–

Cancel out excess H2O

2IO3–+ 6H2O + 10e –→ I2 + 12 OH–

6

Write the half equation for the conversion of Mn2+ to

MnO4–

1. Write the ‘skeleton’ equation including the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced.

Mn + 4(–2) = –1

So Mn = +7

Mn2+ → MnO4–

(+2) (+7)(–2)

Mn is oxidized


MnO4–

2. Balance the numbers of atoms of the element that is being oxidized/reduced

Mn is already balanced

Mn2+ → MnO4–


MnO4–

3. Balance the number of oxygen atoms by adding H2O to the side that needs more O

Mn2+ → MnO4–

Mn2+ + 4 H2O → MnO4–


MnO4–

4. Balance the number of hydrogen atoms by adding H+ to the side that needs more H

Mn2+ + 4 H2O → MnO4–

Mn2+ → MnO4–

Mn2+ + 4 H2O → MnO4– + 8 H+


MnO4–

5. Balance the charge by adding electrons to the side that has more positive charge

Mn2+ + 4 H2O → MnO4–

Mn2+ → MnO4–

Mn2+ + 4 H2O → MnO4– + 8 H+

Mn2+ + 4 H2O → MnO4– + 8 H+ + 5e –


MnO4–

Now amend the equation for this reaction in BASIC solution

Mn2+ + 4 H2O → MnO4– + 8 H+ + 5e –

Add 8 OH– to both sides of the half equation.

8 OH– 8 OH–


MnO4–


Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e –

The OH– and H+ neutralise


MnO4–


Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e –

Cancel out excess H2O

Mn2+ + 8 OH– → MnO4– + 4 H2O + 5e –

4

Write the two half equationsand the net redox equation for

the disproportionation (self-redox) of

Cu+ to Cu2+ and Cu

The two half reactions are:

Cu+ → Cu2+ and Cu+ → Cu

Disproportionation (self-redox) of

Cu+ to Cu2+ and CuThe two half reactions are:

Cu+ → Cu2+ and Cu+ → Cu

These equations need only to be balanced for charge

Cu+ → Cu2+ + e– Cu+ + e– → Cu

Disproportionation (self-redox) of

Cu+ to Cu2+ and CuNow combine the two half equations

Each equation has one electron, so simply add all reactants

and all products

Cu+ → Cu2+ + e–

Cu+ + e– → Cu

2 Cu+ → Cu2+ + Cu

Write the two half equationsand the net redox equation

for the disproportionation (self-redox) of ClO2 to ClO3

–

and ClO2– in basic solution

Some ClO2 molecules are converted to ClO3– and

other ClO2 molecules are converted to ClO2–

Use the rules to write the two half equations for acidic solution, and then for basic solution.

Disproportionation (self-redox) of ClO2 to ClO3

– and ClO2

–


ClO2 → ClO3– and ClO2 → ClO2

– (+4) (+5) (+4) (+3)

Use the rules to complete each half equation


– and ClO2

–

Balance each for O by adding H2O

ClO2 + H2O → ClO3–

ClO2 → ClO2–


– and ClO2

–

Balance each for H by adding H+

ClO2 + H2O → ClO3– + 2H+

ClO2 → ClO2–


– and ClO2

–

Balance charges by adding electrons

ClO2 + H2O → ClO3– + 2H+ + e –

ClO2 + e – → ClO2–

Now add the two half equations

ClO2 + H2O → ClO3– + 2H+ + e –

ClO2 + e – → ClO2–

2ClO2 + H2O → ClO3– + ClO2

– + 2H+


– and ClO2

–

Add 2 OH– to each side


– + 2H+


– and ClO2

–

2 OH– 2 OH–

Add 2 OH– to each side

Then cancel excess H2O


– and ClO2

–


– + 2H2O2 OH–

1

2ClO2 + 2OH– → ClO3– + ClO2

– + H2O

Disproportionation (self-redox) of MnO4

2– to MnO4– and

MnO2


MnO42– → MnO4

–

(+6) (+7)

MnO42– → MnO2

(+6) (+4)

Use the rules to complete each half equation


2– to MnO4– and

MnO2

Balance O by adding H2O

MnO42– → MnO4

–

MnO42– → MnO2

+ 2H2O


2– to MnO4– and

MnO2

Balance H by adding H+

MnO42– → MnO4

–

MnO42– + 4H+ → MnO2

+ 2H2O


2– to MnO4– and

MnO2

Balance charges by adding electrons

MnO42– → MnO4

– + e –

MnO42– + 4H+ + 2e –→ MnO2

+ 2H2O


2– to MnO4– and

MnO2

Electrons lost = Electrons gained

2MnO42– → 2MnO4

– + 2e –

MnO42– + 4H+ + 2e –→ MnO2

+ 2H2O


2– to MnO4– and

MnO2

Add the two half equations

2MnO42– → 2MnO4

– + 2e –

MnO42– + 4H+ + 2e – → MnO2

+ 2H2O

3MnO42– + 4H+ → 2MnO4

– +MnO2 +2H2O

Chemistry ProjectRedox Titration

Content

Definition

Experiment

Discussion

Definition

Redox reaction• A redox reaction is involving transfer of

electrons.

• Oxidation is a process in which a substance loses electrons.

• Reduction is a process in which a substance gains electrons.

• They cannot take place without each other.

Definition

Redox reaction• An oxidizing agent is a substance that

oxidizes others by accepting electrons.

• A reducing agent is a substance that reduces others by donating electrons.

Definition

Titration• Titration is a process of volumetric analysis

that is used to determine the amount of solute in a solution.

• In simple titration, indicator is used to detect the equivalence point of reaction.

Definition

Example

• Iodometric titration

• Titration involving oxidation of iron (Ⅲ)

• Titration involving reduction of iron (Ⅱ)

( In some case,indicators need not to add.

e.g. Titration with potassium manganate (Ⅶ)

or potassium dichromate (Ⅵ) )

Experiment of iodimetric titrition

1) Thoery

A standard iodine solution is titrated with a thiosulphate solution with unknown molarity. The standardized thiosulphate solution can be used to titrate another iodine solution of unknown concentration.


2)• Since iodine is volatile, we cannot prepare

standard solution of iodine directly by accurately

weighing a certain amount of iodine solid &

dissolving the iodine in water .

• Moreover , iodine is only very slightly soluble in

water


• so a standard solution of iodine is first prepare by

dissolving a known of pure potassium iodate(v) solid into an

acidic medium (diluted H2SO4)containing excess iodide.

IO3- (aq) +5I- (aq) + 6H+ (aq) ------> 3 I2 (aq) +3H2O (l)

or [IO3- (aq) +8I- (aq) + 6H+ (aq) ------> 3 I3

- (aq) + 3H2O (l) ]


3) 1ST titration • The standard iodine solution is then used to titrate with a thiosulphate solution of unknown concentration . • Since the colour of iodine in iodide cannot be used to accurately detect the end point starch is used as the indicator. (the change in colour ‘brown->yellow->colourless’ is

very difficult to observe) Starch + iodine blue ‘complex’


• Since starch irreversibly combines with iodine at a high concentration of I2(aq), so that the I2 will not be released from starch at the end point .

• The starch solution should be added at the later stage of the titration (when the solution just turns from brown to pale yellow ). After the addition of starch ,the mixture turns deep blue .

• The end point is shown by the complete decolourisation of the blue colour .


4) 2nd titration • After standardization ,the thiosulphate solution can be used to titrate another solution of unknown concentration of iodine.

This method of titration (using iodine as an oxidizing

agent in the titration ) is known as the iodimetric titration

or iodimetry.


IO3- (aq) +5I- (aq) + 6H+ (aq) ------> 3 I2 (aq) +3H2O (l) …..(1)

or [IO3- (aq) +8I- (aq) + 6H+ (aq) ------> 3 I3

- (aq) + 3H2O (l)]

I2 (aq) + 2S2O32- (aq) -------> S4O6

2- (aq) + 2I- (aq) ……(2)

or [I3- (aq) + 2S2O3

2- (aq) -------> S4O62- (aq) +3I- (aq)]


The concentration of thiosulphate solution can be determined as follow:

no. of moles of I2 = 3 x no. of moles of IO3-

no. of moles of I2 = 1/2 x no. of moles of S2O32-

no. of moles of S2O32- = 6 x no. of moles of IO3

-

Determination of iron(Ⅲ)

1) Excess NaI (aq) is added to an oxidizing agent (e.g. Fe3+) with unknown molarity. 2) The iodine liberated is then determined by the

titration with standard thiosulphate solution . 2Fe3+ (aq) + 2I- (aq) ----> 2Fe2+ (aq) + I2 (aq)

This method of titration (using iodide as reducing agent in the titration )is known as the iodometric titration or iodometery.

Determination of iron (Ⅱ) 1)Standard acidified KMnO4 (aq) is added to

an reducing agent (e.g. Fe2+)

MnO4- (aq) +8H+ (aq) +5Fe2+ (aq) Mn2+ (aq) +4H2O

(l) +5Fe3+ (aq)

2) An indicator is not required as acidified potassium permanganate(Ⅶ) itself serves as an indicator.

Determination of iron (Ⅱ) 3) When manganate (Ⅶ) is added to iron (Ⅱ) in

acidic solution , manganate(Ⅶ) is reduced to manganate(Ⅱ) while iron (Ⅱ) is oxidized to iron (Ⅲ).

4) The colour of the reaction mixture at the equivalence point is yellow (due to the presence of FeFe3+ (aq) )

5) When an extra drop of acidified potassium manganate (Ⅶ)and thus become light purple .

ConclusionReducing agent can be determined by titration with potassium permanganate

Oxidizing agent can be determined by titration with sodium iodide (iodometric titration)

Discussion Error in the idiometric titration A standard solution of iodine must be used immediately because its molarity changes with time because: • iodine is volatile =>I2 can escape from the solution ,

causing the decrease of [I2] with time • iodine can be oxidized by air (promoted by acids, heat

& light )=>[I2]increase with time

Discussion

Error in the titration with KMnO4

• An aqueous solution of KMnO4 is unstable .

4MnO4- + light -------> MnO2 (s) +2 O2 (g) +

MnO22- (aq)

• So the KMnO4 (aq) should be stored in a brown bottle & should be standardized before use.

Making Galvanic Cells

3) The salt bridge allows the cell to continue to function. K+ or NO3

– can move into one of the half cells to balance the charge created by + metal ions forming or leaving solution.

4) No. We did not get the same values in the lab

No bridge Voltage Points to Deposit

Cu – Al 0 V 2.0 V Cu Cu

Cu – Zn 0 V 1.1 V Cu Cu

Al – Zn 0 V 0.9 V Zn Zn

Cu – Zn cell

Zn Zn2+ + 2e– (oxidation - LEO)Cu2+ + 2e– Cu (reduction - GER)

Cu2+ + Zn Cu + Zn2+

Electron flow

Salt bridgeCu (+) Zn (–)

Cu2+ Zn2+

Al – Zn cell

Zn2+ + 2e– Zn (reduction - GER)Al Al3+ + 3e– (oxidation - LEO)

2Al + 3Zn2+ 2Al3+ + 3Zn

Electron flow

Salt bridge Zn (+) Al (–)

Zn2+Al3+

Making Galvanic Cells

AlDepositCuDeposit

AlPoints toCuPoints to

0.530 sec 0.930 sec

0.50 sec 0.90 sec

0No bridge0No bridge

voltageAl-ZnvoltageCu-Zn

3) The salt bridge allows the cell to continue to function. K+ or NO3

– can move into one of the half cells to balance the charge created by + metal ions forming or leaving solution.

4) No. We did not get the same values in the lab

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Mg Mg2+ + 2e– (oxidation - LEO)Fe2+ + 2e– Fe (reduction - GER)

Mg + Fe2+ Mg2+ + Fe

Electron flow e-

Salt bridgeFe (+) Mg (–)

Fe2+ Mg2+

Cu and Al

Al Al3+ + 3e– (oxidation - LEO)Cu2+ + 2e– Cu (reduction - GER)

3Cu2+ + 2Al 3Cu + 2Al3+

Salt bridgeCu (+) Al (–)

Cu2+ Al3+

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Electron flow e-

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redox

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