ELECTROCHEMISTRY

redox – basic concepts

The term REDOX stands for REDUCTION-OXIDATION.

Oxidation can be defined as gain of oxygen or loss of hydrogen.

Reduction can be defined as loss of oxygen or gain of hydrogen.

The most important definition is given in terms of electrons.

OXIDATION is LOSS of ELECTRONS

REDUCTION is GAIN of ELECTRONS

remember OIL RIG

One way of accounting for electrons is to use OXIDATION NUMBERS

Examples:

Fe²⁺ needs to gain two electrons for it to become neutral iron atom therefore its oxidation numberis +2.

Using oxidation numbers it is possible to decide whether redox has occurred.Increase in oxidation number is oxidation. Decrease in oxidation number is reduction.

We can apply a series of rules to assign an oxidation state to each atom in a substance.

Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom

Oxidation numbers

The oxidation number of an atom can be calculated as the charge that would exist on an individual atom if the bonding were completely ionic.

You can also think about it on “how many electrons each element put for the compound to be formed and if it is more electronegative or not regarding the other elements in the compound.

In simple ions, the oxidation number of the atom is the charge on the ion:

Oxidation Number Rules

1. The oxidation number of an uncombined element is 0.

ELEMENT	OXIDATION NUMBERS
S	S=0
O₂	O=0
Cl₂	Cl=0

2. Certain elements have fixed oxidation numbers.

All group 1 elements are +1.

All group 2 elements are +2.

Hydrogen is always +1 except in hydrides.

Fluorine is always –1.

Oxygen is always –2 except in peroxides, superoxides and when combined with fluorine.

Chlorine is always – 1 except when combined with fluorine and oxygen.

In General, In simple ions, the oxidation number of the atom is the charge on the ion:

IONS	OXIDATION #
Na⁺, K⁺, H⁺	+1.
Mg²⁺, Ca²⁺, Pb²⁺	+2
Cl^–, Br^–, I^–	-1
O^2-, S^2-	-2

3. In molecules or compounds, the sum of the oxidation numbers on the atoms is zero.

MOLECULE	OXIDATION NUMBERS	CALCULATION
SO₃	S=+6; O= -2	+6 + 3(-2) = 0
H₂O₂	H =+1; O= -1	2(+1) + 2(-1) = 0
SCl₂	S=+2; Cl= -1	2 + 2(-1) = 0

4. The sum of oxidation numbers in an ion always adds up to the charge on the ion.

POLYATOMIC ION	OXIDATION NUMBERS	CALCULATION
SO₄^2-	S=+6; O= -2	+6 + 4(-2) = -2
PO₄^3-	P =+5; O= -1	+5) + 4(-2) = -3
ClO^–	Cl=+1; O= -2	+1 +(-2) = -1

Calculating Oxidation Numbers Examples:

redox reactions

When magnesium is placed into a solution of copper sulphate, a reaction occurs which in simple terms is called a ” displacement reaction”

EXAMPLES
CHEMICAL EQUATION	Mg + CuSO₄ → MgSO₄ + Cu
IONIC EQUATION	Mg_(s) + Cu²⁺ → Mg²⁺ + Cu_(s)
ESPECTATOR IONS	SO₄^2- SO₄^2-

The copper in this reaction is taking electrons from the magnesium.

The copper gains electrons – it is REDUCED

The magnesium loses electrons – it is OXIDISED

So this is a REDOX reaction

Whenever one substance gains an electron another substance must lose an electron, so reduction and oxidation always go together.

oxidizing and reducing agents

An oxidizing agent causes another material to become oxidized. In the above example of adding magnesium to copper sulphate, the magnesium is oxidized.

Since the copper ions in the copper sulphate cause this oxidation, they are the oxidizing agent.

In the same way the Mg causes the reduction of copper ions so it is the reducing agent.

Let’s analyze the reaction: Mg_(s) + Cu²⁺ → Mg²⁺ + Cu_(s)

Mg_(s)

reducing
agent

Cu²⁺

oxidizing
agent

→

Mg²⁺

Cu_(s)

In this example the oxidizing agent (copper ions) is reduced and the reducing agent (magnesium) is oxidized.

This always happens with redox reactions:-in a redox reaction the oxidizing agent is reduced and the reducing agent is oxidized.

oxidation number and redox reactions

When a redox reaction occurs an electron transfer takes place and so the oxidation numbers of the substances involved changes.

In the reaction

Mg + CuSO₄ → MgSO₄ + Cu

Mg oxidizes from 0 to +2
Cu reduces from +2 to 0
Sulfur and Oxygen stay unchanged before and after the reaction.
- The sulfate ion is called an SPECTATOR ION

Example:

3NaClO → 2NaCl + NaClO₃

Na does not change the oxidation number (+1 in all compounds) It is the spectator ion.
Oxygen does not change oxidation number (-2) since it is combined in both compounds
Chlorine has the following oxidation numbers:
- +1 in NaOCl
- it reduces to -1 in NaCl
- it oxidizes to +5 in NaClO₃
In this case, Cl oxidizes and reduces at the same time. This process where an element oxidizes and reduces at the same time in a reaction is called DISPROPORTIONATION

half equations – full equations

Reduction and oxidation processes can be represented separately by the use of half equations. Each half equation represents only ONE of the processes, oxidation or reduction.

Simple half-equations

In inorganic chemistry, oxidation and reduction are best defined in terms of electron transfer.

Rule of thumb: you always ADD electrons in the half equations. You never subtract electrons

Oxidation is the loss of electrons. When a species loses electrons it is said to be oxidised.

Na → Na⁺ + e^–(each sodium atom loses one electron)

2I ^– → I₂ + 2e^–(each iodide ion loses one electron, so two in total)

Reduction is the gain of electrons. When a species gains electrons it is said to be reduced.

Cl₂ + 2e^– → 2Cl^– (each chlorine atom gains one electron, so two in total)

Al³⁺ + 3e^– → Al (each aluminum ion gains three electrons)

Processes which show the gain or loss of electrons by a species are known as half-equations. They show simple oxidation or reduction processes.

More complex half-equations

Many oxidation and reduction processes involve complex ions or molecules and the half-equations for these processes are more complex. In such cases, oxidation numbers are a useful tool:

This is the easiest way to balance half-equations:

First: Balance the masses. (Do these steps in order):

- Identify the atom being oxidised or reduced, and make sure there are the same number of that atom on both sides (by balancing).
- Balance O atoms by adding water.
- Balance H atoms by adding H⁺_^.

Second: Balance the charges.

- add the necessary number of electrons to ensure the charge on both sides is the same

Worked Example:

Balance the following half-equation:

H₂SO₄ H₂S

- There is one sulfur on each side, so the S is already balanced
- There are four O atoms on the left and none on the right, so four waters are needed on the right:

H₂SO₄ H₂S + 4H₂O

- there are two H atoms on the left and ten on the right, so eight H ions are needed on the left:

H₂SO₄ + 8H⁺ H₂S + 4H₂O

- The total charge on the left is +8 and on the right is 0. So eight electrons must be added to the left to balance the charge:

H₂SO₄ + 8H⁺ + 8e^– H₂S + 4H₂O

The oxidation number of the S is +6 in the sulfuric acid and -2 in the sulfide. Sulphur will gain 8 electrons which are added to the left of the equation, since it is a reduction process.

Every time we have a redox reaction we actually have a transference of electrons from the element that oxidizes to the element that reduces. so the number of electrons in each half equation has to be the same.

We can obtain the whole redox reaction equation by combining an oxidation and reduction half-equations so the total number of electrons gained is equal to the total number of electrons lost.

Half equations	Type	amount of electrons involved
H₂SO₄ + 8H⁺ + 8e^– H₂S + 4H₂O	reduction	8
2I ^– I₂ + 2e-	oxidation	2

A full equation is written by adding two half equations together.

The process is as follows:

1. Write first half equation
2. Write second half equation
3. Balance in terms of electrons
4. Add equations together

(the oxidation half-equation must be multiplied by 4 to equate the electrons)

Half equations	Type	amount of electrons involved
H₂SO₄ + 8H⁺ + 8e^– H₂S + 4H₂O	reduction	8
8I ^– 4I₂ + 8e-	oxidation	8
overall reaction: – redox H₂SO₄ + 8H⁺ + 8I^– H₂S + 4H₂O + 4I₂

Example 1

1. Chlorine reacts with potassium iodide to form potassium chloride and iodine.

1. Write the half equation for the oxidation of Iodine

2I^– → I2 + 2e ^–

2. Write the half equation for the reduction of fluorine

Cl₂+ 2e ^–→ 2Cl^–

3. Adding the equations together

2I^–+ Cl₂→ I2 +2Cl^–

Example 2

Potassium reacts with fluorine to form potassium fluoride.

1. Write the half equation for the oxidation of potassium

K → K⁺+ e ^–

2. Write the half equation for the reduction of fluorine

F₂+ 2e ^–→ 2F^–

3. To balance for electrons, the first equation must be multiplied by 2

2K → 2K⁺+ 2e ^–

F₂+ 2e ^–→ 2F^–

3. Adding the equations together

2K + F₂ + 2e –→ 2F^–+ 2K⁺+ 2e –

4. Adding the equations together

2K + F₂→ 2F^–+ 2K⁺

Example 3

1. Bromine reacts with iron(II) to form iron(III) and bromide

orm iron(III) and bromideBromine reacts with iron(II) to f

orm iron(III) and bromide

1. Write the half equation for the oxidation of Iron

Fe²⁺ → Fe3+ + e ^–

2. Write the half equation for the reduction of Bromine

Br₂+ 2e ^–→ 2Br^–

3.To balance for electrons, the first equation must be multiplied by 2

2Fe²⁺ → 2Fe3++ 2e ^–

Br₂+ 2e ^–→ 2Br^–

3.Combine both equations

Br₂+ 2Fe²⁺ → 2Br^– + 2Fe3+

more about oxidizing and reducing agents.

The species which is reduced is accepting electrons from the other species and thus causing it to be oxidized. It is thus an oxidizing agent.

H₂SO₄, Al³⁺ and Cl₂ are all oxidizing agents.

The species which is oxidized is donating electrons to another species and thus causing it to be reduced. It is thus a reducing agent.

Na, O^2-, I^– and S₂O₃^2- are all reducing agents.

A redox reaction can thus be described as a transfer of electrons from a reducing agent to an oxidising agent.

Example 1

I₂ + 2S₂O₃^2- → 2I^– + S₄O₆^2-

Half-equations:

I₂ + 2e^– → 2I^– (reduction)

2 S₂O₃^2- → S₄O₆^2- + 2e^– (oxidation)

I₂ is the oxidising agent; S₄O₆^2- is the reducing agent.

Example 1

H₂SO₄ + 8H⁺ + 8I^– → H₂S + 4H₂O + 4I₂

Half-equations:

H₂SO₄ + 8H⁺ + 8e^– → H₂S + 4H₂O (reduction)

2I ^– → I₂ + 2e^– (oxidation)

H₂SO₄ is the oxidising agent, I^– is the reducing agent

more about disproportionation

The simultaneous oxidation and reduction of the same species is known as disproportionation.

This occurs when a substance suffer both oxidation and reduction, and which can therefore behave as both oxidizing agents and reducing agents. H₂O₂and ClO^– are two examples:

Examples

H₂O₂ + 2H⁺ + 2e^{– →} 2H₂O reduction

H₂O₂ → O₂ + 2H⁺ + 2e^– oxidation

ClO^– + 2H⁺ + 2e→ Cl^– reduction

ClO^– →ClO₃^– + 4H⁺ + 4e oxidation

halogens reactivity

The bigger the halide ion, the less attraction from the nucleus because the shielding increases. This reason makes easier for the halide ions to lose electrons as you go down the Group because there is less attraction between the outer electrons and the nucleus.

redox reactions between halide ions and concentrated sulfuric acid

Concentrated sulfuric acid can act both as an acid and as an oxidizing agent.

Two main reactions occur, when sulfuric acid is added to solid sodium halides.

Concentrated sulfuric acid acting as an oxidizing agent

With fluoride or chloride ions

The fluoride and chloride ions aren’t strong enough reducing agents to reduce the sulfuric acid

Steamy fumes of the hydrogen halide – hydrogen fluoride or hydrogen chloride are produced.

NaF + H₂SO₄ → HF ↑ + NaHSO₄

NaCl + H₂SO₄ → HCl ↑ + NaHSO₄

With Bromine and Iodide ions the hydrogen halides

NaBr + H₂SO₄ → HBr ↑ + NaHSO₄

(further oxidation)

In a further oxidation, Bromine changes the oxidation number from -1 to 0

Sulfur in sulfuric acid reduces and changes the oxidation number from +6 to +4

H₂SO₄ + 2HBr → Br₂+ SO₂+ H₂O

NaI+ H₂SO₄ → HI ↑ + NaHSO₄

(further oxidation)

In a further oxidation, Iodine changes the oxidation number from -1 to 0

Sulfur in sulfuric acid reduces and changes the oxidation number from +6 to -2

H₂SO₄ + 8HI → 4I₂+ H₂S + 4H₂O

redox titrations

Redox titrations involve calculations very similar to the ones we carry out during an acid base titration.

Redox titrations can be useful for:

- find the concentration of a solution
- determine the stoichiometry of a redox titration and so, suggest the possible equation for a reaction.

Using potassium permanganate (Potassium Manganate VII) as the oxidant (oxidizing agent)

Things you need to know about these titrations

- Potassium manganate (VII) – also called “permanganate” has a deep purple color.
- It is used as its own indicator
- Most of the time, the solution is placed in the burette.
- As you run the solution, it gets colorless because the MnO₄^– is reduced to Mn²⁺

MnO₄^–_(aq) →Mn²⁺_(aq)

balancing masses and charges we obtain

MnO₄^–_(aq) + 8H⁺_(aq) + 5e^–→ Mn²⁺_(aq) + 4H₂O_(l)

- at the end point the solution turns pink again (with the last drop)
- Most of the time we use a 0.02M solution since the solid is not very soluble
- the reaction is carried in presence of sulfuric acid 1M or more.
- You should read at the top of the meniscus in the burette, since it is very difficult to see the lower meniscus. Because the volume delivered is the difference between your readings, it does not make any difference in the calculations.
- The solution may turn brown if you carry it very fast because of the presence of MnO2. This can be solved by adding more acid or by warming up the solution.

There are several examples of redox titrations using Manganate (VII).

- Iron (II) sulfate → iron (III) sulfate
- Ethanedioic acid → Carbon Dioxide + water
- Hydrogen peroxide → Oxygen gas

In the third example, the H₂O₂ is used as the reducing agent. Since H₂O₂ decomposes with time, we can calculate the concentration of H₂O₂ by titration.

If this is the case, we will have the following reactions:

Half equations	Type	amount of electrons involved
MnO₄^– + 8H⁺ + 5e^– → Mn₂⁺ + 4H₂O	reduction MnO₄^– is the oxidizing agent	5 x 2
H₂O_2(aq) → O_2(g) + 2H⁺_(aq) + 2e^–	oxidation H₂O₂ is the reducing agent	2 x 5
OVERALL REACTION multiplying the first equation by 5 and the 2^nd by 2 we obtain the following reaction: 2MnO₄^–_(aq) + 16H⁺_(aq) + 10e^–+5H₂O_2(aq)_→2Mn₂⁺_(aq) + 8H₂O_(l)+5O_2(g) + 10H⁺_(aq) + 10e^– subtracting the protons from the right from the ones on the left, we have the following overall reaction 2MnO₄^–_(aq) + 6H⁺_(aq) +5H₂O_2(aq)_→2Mn₂⁺_(aq) + 8H₂O_(l)+5O_2(g)

Using Iodine as the oxidant (oxidizing agent)

Things you need to know about these titrations

- The half equation for Thiosulfate:

2S₂O₃^2-_(aq) S₄O₆^2-_(aq)

balancing masses and charges we obtain

2S₂O₃^2-_(aq) S₄O₆^2-_{(aq) +}2e^–

- The reaction between iodine I_2(aq) and 2 thiosulphate ions 2S₂O₃^2-_(aq) is a redox reaction that is useful in chemical analysis.
- In this reaction, aqueous iodine is reduced to iodide ions by aqueous sodium thiosulphate which forms the tetrathionate ion S₄O₆^2-_(aq)
- IMPORTANT: Do not try to calculate the oxidation number of S in both species. Just balance the masses and charges and you will be fine.

2S₂O₃^2-_(aq) + I_2(aq) 2I^–_(aq) + S₄O₆^2-_(aq)

thiosulfate ion + Iodine iodide ions + tetrathionate ion

- The concentration of iodine in a solution can be determined by titration with a solution of thiosulphate of known concentration.
- The above reaction can be used to determine the concentration of a solution of an oxidizing agent that reacts with iodide ions to form iodine.
- The thiosulfate is placed in the burette.
- The iodine solution is initially brown and as the thiosulfate is added it fades to pale yellow.
- Near the ending point, some drops of starch solution are used to show the presence of iodine that appears as almost invisible to our eyes. This forms a deep blue (almost black) solution. (do not add the starch at the beginning. it can form clumps of blue solids that difficult the run of the titration.
- The end point is a sharp change between the blue solution to colorless.
- This titration is not used to calculate the concentration of iodine, but all the oxidizing agents that produce iodine in solution. Some of these oxidizing agents are listed below:

Oxidizing agent	Equation
Cl₂	Cl₂ + 2I^–2Cl^–+ I₂
Br₂	Br₂ + 2I^–2Br^–+ I₂
IO₃^–	IO₃^– + 6H⁺ + 5 I^–3I₂ + 3 H₂O
MnO₄^–	MnO₄^– + 5I^– + 8H⁺ 2½ I₂ + 4H₂O
Cu²⁺	Cu + 2I^– CuI + ½ I₂

ELECTROCHEMISTRY

redox – basic concepts

Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom

Oxidation Number Rules

ELEMENT

OXIDATION NUMBERS

S

S=0

O2

O=0

Cl2

Cl=0

IONS

OXIDATION #

Na+, K+, H+

+1.

Mg2+, Ca2+, Pb2+

+2

Cl–, Br–, I–

-1

O2-, S2-

-2

MOLECULE

OXIDATION NUMBERS

CALCULATION

SO3

S=+6; O= -2

+6 + 3(-2) = 0

H2O2

H =+1; O= -1

2(+1) + 2(-1) = 0

SCl2

S=+2; Cl= -1

2 + 2(-1) = 0

POLYATOMIC ION

OXIDATION NUMBERS

CALCULATION

SO42-

S=+6; O= -2

+6 + 4(-2) = -2

PO43-

P =+5; O= -1

+5) + 4(-2) = -3

ClO–

Cl=+1; O= -2

+1 +(-2) = -1

Calculating Oxidation Numbers Examples:

redox reactions

oxidizing and reducing agents

→

oxidation number and redox reactions

half equations – full equations

Half equations

Type

amount of electrons involved

H2SO4 + 8H+ + 8e– H2S + 4H2O

reduction

8

2I – I2 + 2e-

oxidation

2

Half equations

Type

amount of electrons involved

H2SO4 + 8H+ + 8e– H2S + 4H2O

reduction

8

8I – 4I2 + 8e-

oxidation

8

overall reaction: – redox

H2SO4 + 8H+ + 8I– H2S + 4H2O + 4I2

Example 1

2I– → I2 + 2e –

Cl2 + 2e – → 2Cl–

2I–+ Cl2 → I2 +2Cl–

Example 2

K → K++ e –

F2 + 2e – → 2F–

2K → 2K++ 2e –

O₂

Cl₂

Na⁺, K⁺, H⁺

Mg²⁺, Ca²⁺, Pb²⁺

Cl^–, Br^–, I^–

O^2-, S^2-

SO₃

H₂O₂

SCl₂

SO₄^2-

PO₄^3-

ClO^–

H₂SO₄ + 8H⁺ + 8e^– H₂S + 4H₂O

2I ^– I₂ + 2e-

H₂SO₄ + 8H⁺ + 8e^– H₂S + 4H₂O

8I ^– 4I₂ + 8e-

H₂SO₄ + 8H⁺ + 8I^– H₂S + 4H₂O + 4I₂

2I^– → I2 + 2e ^–

Cl₂+ 2e ^–→ 2Cl^–

2I^–+ Cl₂→ I2 +2Cl^–

K → K⁺+ e ^–

F₂+ 2e ^–→ 2F^–

2K → 2K⁺+ 2e ^–

F₂+ 2e ^–→ 2F^–

2K + F₂ + 2e –→ 2F^–+ 2K⁺+ 2e –

2K + F₂→ 2F^–+ 2K⁺

Fe²⁺ → Fe3+ + e ^–

Br₂+ 2e ^–→ 2Br^–

2Fe²⁺ → 2Fe3++ 2e ^–

Br₂+ 2e ^–→ 2Br^–

Br₂+ 2Fe²⁺ → 2Br^– + 2Fe3+

H₂SO₄, Al³⁺ and Cl₂ are all oxidizing agents.

Na, O^2-, I^– and S₂O₃^2- are all reducing agents.

H₂O₂ + 2H⁺ + 2e^{– →} 2H₂O reduction

H₂O₂ → O₂ + 2H⁺ + 2e^– oxidation

ClO^– + 2H⁺ + 2e→ Cl^– reduction

ClO^– →ClO₃^– + 4H⁺ + 4e oxidation

NaF + H₂SO₄ → HF ↑ + NaHSO₄

NaCl + H₂SO₄ → HCl ↑ + NaHSO₄

NaBr + H₂SO₄ → HBr ↑ + NaHSO₄

H₂SO₄ + 2HBr → Br₂+ SO₂+ H₂O

NaI+ H₂SO₄ → HI ↑ + NaHSO₄

H₂SO₄ + 8HI → 4I₂+ H₂S + 4H₂O

MnO₄^–_(aq) →Mn²⁺_(aq)

MnO₄^–_(aq) + 8H⁺_(aq) + 5e^–→ Mn²⁺_(aq) + 4H₂O_(l)

2S₂O₃^2-_(aq) S₄O₆^2-_(aq)

2S₂O₃^2-_(aq) S₄O₆^2-_{(aq) +}2e^–

2S₂O₃^2-_(aq) + I_2(aq) 2I^–_(aq) + S₄O₆^2-_(aq)