Analía Bellizzi – Chemistry Classes

Ronald Reagan High School

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

O2-, S2-

    -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

          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

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

SO42-                                       SO42-                       

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  H2

    • 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

Cl+ 2e – → 2Cl

3. Adding the equations together

2I+ Cl I +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

F+ 2e – → 2F

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

                2K  → 2K++ 2e

F+ 2e – → 2F

3. Adding the equations together

2K  +  F  +  2e →  2F–  +  2K+  +  2e 

4. Adding the equations together

2K + F→ 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

Br+ 2e – → 2Br

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

2Fe2+   → 2Fe3++ 2e

Br+ 2e – → 2Br

3.Combine both equations

Br+ 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, O2-, I and S2O32- 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 + 2S2O32- → 2I + S4O62-

Half-equations:    

I2 + 2e → 2I (reduction)         

2 S2O32- →  S4O62- + 2e (oxidation) 

I2 is the oxidising agent; S4O62- 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. H2O2 and ClO are two examples:

Examples

H2O2 + 2H+ + 2e–  → 2H2O reduction

H2O2 O2 + 2H+ + 2e oxidation    

 

ClO + 2H+ + 2e→ Cl reduction

ClOClO3 + 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 + H2+ 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

 H2O2(aq)   →  O2(g) + 2H+(aq) + 2e

   

oxidation H2O2

is the reducing agent

2 x 5

OVERALL REACTION

multiplying the first equation by 5 and the 2nd by 2 we obtain the following reaction:

2MnO4(aq) + 16H+(aq) + 10e+5H2O2(aq)   →   2Mn2+(aq) + 8H2O(l)+5O2(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) +5H2O2(aq)  →  2Mn2+(aq) + 8H2O(l)+5O2(g) 

Using Iodine as the oxidant (oxidizing agent)

Things you need to know about these titrations

    •  The half equation for Thiosulfate: 

2S2O32-(aq)        S4O62-(aq)

balancing masses and charges we obtain

  2S2O32-(aq)        S4O62-(aq) + 2e

    • The reaction between iodine I2(aq)  and 2 thiosulphate ions 2S2O32-(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  S4O62-(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. 

2S2O32-(aq)   +  I2(aq)    2I(aq)   +   S4O62-(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            
Cl2Cl2 + 2I– 2Cl– + I2
Br2Br + 2I– 2Br– + I2
IO3 IO3 + 6H+ + 5 I3I2 + 3 H2O   
MnO4MnO4 + 5I + 8H+ 2½ I2 + 4H2O
Cu2+Cu + 2I CuI + ½ I2