REDOX REACTIONS

A.k.a. A study on why things

explode

But I Know Every Rock and Tree and Creature…

• Has a life, has a spirit, has a name.

Oxidation of Food: What a Waste!

• Fruits and Vegetables oxidised when left in open air– Solution: Seal in plastic wrap– More radical: Add lemon juice to the cut fruit

Oxidation of…

• Oxidation of nutrients causes increased activity of cells, leading to aging skin– Solution: Beauty products?

People!

What will I learn?

• What is a redox reaction?• What is oxidation and reduction? (and how to

identify them)• What are oxidising and reducing agents?• What is oxidation state?• What is oxidation and reduction in terms of

oxidation state?• How to assign the oxidation state of an atom?• How to use oxidation state to determine which

species is oxidised / reduced

What is a redox reaction?

• Redox – reduction + oxidation

• Both processes occur simultaneously

• Hence, one species is oxidised, another is reduced

• So, what is oxidation, and what is reduction?

• 3 different versions of the definition:

Redox

gain of electronsloss of electrons

gain in hydrogenloss of hydrogen

loss of oxygengain in oxygen

ReductionOxidation

Oxidation and Reduction

• In terms of Oxygen:– Oxidation: Gain of oxygen in a species

• E.g. Mg is oxidized to MgO

– Reduction: Loss of oxygen in a species• E.g. H2O is reduced to H2

– Note: It’s the gain or loss of O, not O2-

Oxidation and Reduction

• In terms of Hydrogen:– Oxidation: Loss of hydrogen in a species

• E.g. H2O is oxidised to O2

– Reduction: Gain of hydrogen in a species• E.g. O2 is reduced to H2O2

– Note: It’s the gain or loss of H, not H+

Oxidation and Reduction

• In terms of Electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain):– Oxidation: Loss of electrons in a species

• E.g. Mg is oxidized to MgO (Mg from 12 electrons to 10 electrons in Mg2+)

– Reduction: Gain of electrons in a species• E.g. O2 is reduced to H2O2 (O from 8 electrons to 9

electrons per O in O22-)

Oxidising and Reducing agent

• An oxidising agent is a chemical species that causes the other reactant in a redox reaction to be oxidised, and it is always reduced in the process.

• A reducing agent is a chemical species that causes the other reactant in a redox reaction to be reduced, and it is always oxidised in the process.

Oxidising and Reducing Agents

• Remember:– An oxidising agent is itself REDUCED when it

oxidises something– A reducing agent is itself OXIDISED when it

reduces something

2Mg + O2 → 2MgO

– Mg is oxidised, and thus is the reducing agent

– O2 is reduced, and thus is the oxidising agent

List of common Oxidising and Reducing Agents

• Realise something?– H2O2 is both an oxidising and a reducing agent!

– If a stronger oxidising agent is present, H2O2 is reducing

Oxidation and Reduction in ‘A’ Level

• In terms of Oxidation States:– Oxidation: Gain in oxidation state in a species

• E.g. Mg is oxidized to MgO (Mg from 0 to +2 in Mg2+)

– Reduction: Gain of electrons in a species• E.g. O2 is reduced to H2O2 (O from 0 to -1 in O2

2-)

• Note: Oxidation states are always written in +x or –x, never just x or x- (e.g. Oxidation State of Mg in MgO is +2, not 2 or -2)

Common Oxidation StatesChemical species Oxidation state and remarks

Any element eg Fe, O2 , S8zero

Oxygen in any compound -2 except in peroxides example H2O2 or Na2O2 then oxygen atom has

oxidation state of -1 or in F2O , then oxygen atom has oxidation state

of +2

Fluorine in any compound -1 being most electronegative

Hydrogen in any compound +1 except in metal hydrides example NaH then hydrogen atom has

oxidation state of -1 as metals have a greater tendency to lose

electrons

Chlorine, bromine, iodine -ve oxidation state if bonded to less electronegative element eg

NaCl; then Cl = -1.

+ve oxidation state if bonded to more electronegative element eg

ClO- , then Cl = +1; ClO3- , then Cl = +5

Determine the oxidation state of…

1) H in H2O

2) N in NH4+

3) S in S2O32-

4) Cr in Cr2O72-

Determine the oxidation state of…

* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom

1) H in H2O

Let the oxidation state of H be x.

Thus, in H2O, 2x + (-2) = 0

x = 1

Determine the oxidation state of…

* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom

2) N in NH4+

Let the oxidation state of N be x.

Thus, in NH4+, x + 4(+1) = +1

x = -3

Determine the oxidation state of…

* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom

3) S in S2O32-

Let the oxidation state of S be x.

Thus, in S2O32-, 2x + 3(-2) = -2

x = +2

Determine the oxidation state of…

* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom

4) Cr in Cr2O72-

Let the oxidation state of Cr be x.

Thus, in Cr2O72-, 2x + 7(-2) = -2

x = +6

Example 1

Let the oxidation state of Mn be x.

Thus, in MnO4-, x + 4(-2) = -1

x = +7• Manganese is reduced from oxidation state of

+7 in MnO4- to +2 in Mn2+, while iron is oxidised

from oxidation state of +2 in Fe2+ to +3 in Fe3+.

OHFeMnHFeMnO 2322

4 4585

+7 +2 +2 +3

A Special Redox: Disproportionation

• Definition: A disproportionation reaction is a redox reaction in which one species is simultaneously oxidised and reduced.

• Chlorine is simultaneously reduced from oxidation state of 0 in Cl2 to -1 in Cl-, and oxidised from oxidation state of 0 in Cl2 to +1 in ClO-.

OHClClOOHCl 22 2

0 +1 -1

Disproportionation Reaction

• Example:Is this a disproportionation reaction?

• This is NOT a disproportionation reaction• Disproportionation requires that the same atom is

both oxidised and reduced simultaneously. • In this case, different atoms (of nitrogen) are

oxidised and reduced.

-3 +5 +1

OHONNONH 2234 2

Non-redox reactions

• The oxidation states of the elements remained unchanged in the following reactions:

• Neutralisation reactions:

OHNaClHClNaOH 2

OHCuSOSOHCuO 2442

Non-redox reactions

• The oxidation states of the elements remained unchanged in the following reactions:

• Precipitation reactions:

)(42)(2)()(4 )(2 aqsaqaq SONaOHCuNaOHCuSO

)(3)(2)(23)( 2)(2 aqsaqaq KNOPbINOPbKI

Non-redox reactions

• The oxidation states of the elements remained unchanged in the following reactions:

• Complex formation:

2

)(43)(3)(2 4

aqaqaq NHCuNHCu

Tetraammine copper(II) complex(deep blue solution)

ligand

Non-redox reactions

• The oxidation states of the elements remained unchanged in the following reactions:

• Another reaction:

HCrOOHOCr 22 242

272

Interlude: More real-life redox!

Balancing redox reactions

• To make calculations in redox titrations, you need a balanced equation

Balancing redox reactions

• Example:Try to balance the following reaction by trial and error.

• Possible answer:

OHOMnHOHMnO 222

224

OHOMnHOHMnO 222

224 222

OHOMnHOHMnO 222

224 632442

Balancing redox reactions

• Example:Try to balance the following reaction by trial and error.

• Actual answer:

• Note: You might not even be told at the beginning that H+ is reactant, H2O is product.

OHOMnHOHMnO 222

224

OHOMnHOHMnO 222

224 852652

The half-equation method

• Write down the given reactants and products of the reaction• Identify the atoms in the given species that are undergoing oxidation

/ reduction and construct the unbalanced oxidation / reduction half-equations

• Balance both the half-equations using the following steps:– Balance the “odd” atoms (“odd” atoms refer to atoms other than

oxygen and hydrogen)– Balance oxygen atoms by adding H2O molecules– Balance hydrogen atoms by adding H+ ions– Balance charges by adding electrons

• Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.

Oh yeah, the blue bottle…• “Angry bottle”• The blue colour is methylene blue in the oxidised form• Before the bottle was shaken, the methylene blue is in

the reduced state (colourless)• After the bottle was shaken, the oxygen in the air

oxidised the methylene blue to the oxidised state (blue)• After a while, the glucose (reducing sugar) in the solution

reduces the methylene blue back to the reduced state (colourless)

))((

)())(( 2

blueoxidisedluemethyleneb

aqOcolourlessreducedluemethyleneb

End of Lecture 1 of Redox

“I have never let my schooling interfere with my education.” Mark Twain

The half-equation method

• Example: Balance the following reaction:

22

224 OMnOHMnO

The half-equation method

• Step 1: Write down the given reactants and products of the reaction

22

224 OMnOHMnO

The half-equation method

• Step 2: Identify the atoms in the given species that are undergoing oxidation / reduction and write the unbalanced oxidation / reduction half-equations

• Reduction half-equation:

• Oxidation half-equation:

22

224 OMnOHMnO

Reduced Oxidised 2

4 MnMnO

222 OOH

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance the atoms undergoing oxidation / reduction

• Reduction half-equation: 2

4 MnMnO

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance oxygen atoms by adding H2O molecules

• Reduction half-equation: 2

4 MnMnO

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance oxygen atoms by adding H2O molecules

• Reduction half-equation:

OHMnMnO 22

4 4

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance hydrogen atoms by adding H+ ions

• Reduction half-equation:

OHMnMnO 22

4 4

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance hydrogen atoms by adding H+ ions

• Reduction half-equation:

OHMnHMnO 22

4 48

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance charges by adding electrons

• Reduction half-equation:

OHMnHMnO 22

4 48

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance charges by adding electrons

• Reduction half-equation:

OHMneHMnO 22

4 458

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

222 OOH

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance the atoms undergoing oxidation / reduction

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

222 OOH

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance oxygen atoms by adding H2O molecules

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

222 OOH

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance hydrogen atoms by adding H+ ions

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

222 OOH

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance hydrogen atoms by adding H+ ions

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

HOOH 2222

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance charges by adding electrons

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

HOOH 2222

The half-equation method

• Step 3: Balance both the half-equations using the following steps:

– Balance charges by adding electrons

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

eHOOH 22222

The half-equation method

• Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

eHOOH 22222

The half-equation method

• Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

eHOOH 22222

The half-equation method

• Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 458

eHOOH 22222

x 2

x 5

The half-equation method

• Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 8210162

eHOOH 22222

x 2

x 5

The half-equation method

• Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 8210162

eHOOH 101055 222

x 2

x 5

The half-equation method

• Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.

• Reduction half-equation:

• Oxidation half-equation:

OHMneHMnO 22

4 8210162

eHOOH 101055 222

x 2

x 5

The half-equation method

• Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.

OHMneHMnO 22

4 8210162

eHOOH 101055 222 +6H+

The half-equation method

• Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.

• Balanced Equation:

OHMneHMnO 22

4 8210162

eHOOH 101055 222 +6H+

The half-equation method

• Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation.

• Balanced Equation:

OHOMnHOHMnO 222

224 852652