Oxidation and Reduction
# Definition of Oxidation and reduction or Redox reaction:
“Redox (reduction-oxidation) reactions include all chemical reactions in which atoms have
their oxidation state changed; in general, redox reactions involve the transfer of electrons
between species.”
Redox reaction can be a simple process, such as the oxidation of carbon (like coal burning) to
yield carbon dioxide (CO2) or the reduction of carbon by hydrogen to yield methane (CH4), or a
complex process such as the oxidation of glucose (C6H12O6) in the human body through a series
of complex electron transfer processes.
C (coal) + O2(air) → CO2 + H2O + heat
C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) + heat
The term "redox" comes from two concepts involved with electron transfer: reduction and
oxidation. It can be explained in simple terms:


Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom,
or ion.
Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or
ion.
Oxidation (loss of electrons): H2 (Reducing agent) ─ 2e─ → 2H+
Reduction (gain of electrons): F2 (Oxidizing agent) + 2e─→ 2F─
#Oxidation Number:
“The oxidation number of an element indicates the number of electrons lost, gained, or shared as
a result of chemical bonding.”
In other way, it is defined as the charge the atom would have if electrons were not shared but
were transferred completely.
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Example 1. Simply show the oxidation number (O.N.) of each element in these compounds:
A. CaO (s)
B. KNO3(s)
C. NaHSO4 (aq)
D. CaCO3(s)
E. N2(g)
F. H2O (l)
Solution:
A.
C.
B.
D.
E.
F.
Example 2. Determine the oxidation number (O.N.) of each element in these compounds:
A. KMnO4
B. K2Cr2O7
C. [Cr (CN)6]3−
Solution:
A. Let us consider, the Oxidation Number of Mn in the KMnO4 = x.
Then, (+1) + x + (-2) × 4 = 0.
Or, x +1− 8 = 0.
Or, x = +7
B. Let us consider, the Oxidation Number of Cr in K2Cr2O7 = x
Then, 2 (+1) + 2x + 7(− 2) = 0.
Or, + 2 + 2x – 14 = 0.
Or, 2x – 12 = 0.
Or, x = +6
C. Let us consider, the Oxidation Number of Cr in [Cr (CN)6]3− = x
Then, x + 6 (− 1) = − 3. Or, x = +3
Example 3. Identify the oxidizing agent and reducing agent in each of the following:
A. 2H2 (g) + O2(g) → 2H2O (g)
B. Cu (s) + 4HNO3 (aq) → Cu(NO3)2 (aq) + 2NO2(g)+ 2H2O (l)
Solution:
A.
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O2 was reduced (O.N. of Oxygen: 0 → −2); O2 is the oxidizing agent.
H2 was oxidized (O.N. of Hydrogen: 0 → +1); H2 is the reducing agent.
B.
Cu was oxidized (O.N. of Cu: 0→+2); Cu is the reducing agent
HNO3 was reduced (O.N. of N: +5 → +4); HNO3 is the oxidizing agent
# Balancing Redox Equations:
Two methods can be used:
1. Oxidation number method
2. Half-reaction method
Method 1: Oxidation number method
Example 4: Use the oxidation number method to balance the following equations:
A. Al(s) + H2SO4 (aq) →Al2(SO4)3(aq) + H2(g). Or, aluminium reacts with sulpuric
acid to generate aluminium sulfate [Al2(SO4)3].
B. Cu(s) + HNO3 (aq) → Cu (NO3)2 (aq) + NO(g) + H2O(l). Or, Copper (Cu) reacts
with nitric acid (HNO3) to produce copper (II) nitrate [Cu(NO3)2], nitric oxide
[NO] and water.
Solution:
A. Step 1. Let’s find the skeletal and Ionic equation.
Skeletal equation: Al(s) + H2SO4 (aq) → Al2(SO4)3(aq) + H2(g)
Ionic equation:
Al
+
H+
→
Al3+ +
H2
Step 2. Now let’s identify oxidized and reduced species


Al was oxidized (O.N. of Al: 0 → +3) = reducing agent
H2SO4 was reduced (O.N. of H: +1 →0) = Oxidizing agent
Step 3. Let’s multiply by factors to make e− lost equal to e− gained, and use the factors
as coefficients. From the above reaction we can see that, the change in oxidation number
of aluminium is [3− 0=3] 3, so multiply the other reactant with 3. Then the change in
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oxidation number of hydrogen in sulfuric acid is [0−2= −2] 2, so multiply the other
reactant with 2.
Step 4. Complete the balancing by inspection
2Al
+
6H+ →
2Al3+ +
3H2
Step 5. Now adding the spectator ions in above reaction,
Balanced Reaction: 2Al(s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3H2(g)
B. Step 1. Let’s find the skeletal and Ionic equation.
Skeletal equation: Cu(s) + HNO3 (aq) → Cu (NO3)2 (aq) + NO(g) + H2O(l)
Ionic equation:
Cu
+
NO3− →
Cu2+ +
NO
Step 2. Identify oxidized and reduced species


Cu was oxidized (O.N. of Cu: 0 → +2) = Reducing agent
HNO3 was reduced (O.N. of N: +5 →+2) = Oxidizing agent
Step 3. Let’s multiply by factors to make e− lost equal to e− gained, and use the factors
as coefficients. From the above reaction we can see that, the change in oxidation number
of copper is [2− 0=2] 2, so multiply the other reactant with 2. Then the change in
oxidation number of nitrogen in nitric acid is [2−5= −3] 3, so multiply the other reactant
with 3.
Step 4. Complete the balancing by inspection
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3Cu + 2NO3− + 8H+→
3Cu2+ +2NO + 4H2O
Step 5. Now adding the spectator ions in above reaction,
Balanced Reaction: 3Cu(s) + 8HNO3 (aq) → 3Cu (NO3)2 (aq)+ 2NO(g) + 4H2O(l)
Method 2: Half-reaction method
Example 5: Use the Half-reaction method to balance the following reaction:
A. Ferric chloride [FeCl3] reacts with stannous chloride [SnCl2] to produce ferrous
chloride [FeCl2] and Stannic chloride [SnCl4].
B. Ferrous chloride [FeCl2] salts reacts with acid potassium dichromate [K2Cr2O7] to
produce ferric chloride [Fe2Cl3], potassium chloride [KCl], chromium chloride
[CrCl3] and water.
C. Acidic (in H2SO4) potassium permanganate [KMnO4] reacts with oxalic acid
[H2C2O4] to produce manganese sulfate [MnSO4], potassium sulfate [K2SO4], carbon
dioxide and water.
Solution:
A. Here, Ferric chloride [FeCl3] is oxidizing agent and stannous chloride [SnCl2] and is
reducing agent.
(Oxidation) half reaction:
Sn2+ ─ 2e─ → Sn4+ ------------- (1)
(Reduction) half reaction:
Fe3+ + e─ → Fe2+ -------------- (2)
To equalize the no. of electron in both side multiply the equation (2) with 2 and then
adding with equation (1) we found:
(Oxidation) half reaction:
Sn2+ ─ 2e─ → Sn4+
(Reduction) half reaction:
2Fe3+ +2e─ → 2Fe2+
Over-all reaction: Sn2+ +2Fe3+ → 2Fe2+ + Sn4+---------- (3)
Adding the spectator ions in both side of the equation (3),
2FeCl3 + SnCl2 → 2FeCl2 + SnCl4
B. Here, ferrous chloride [FeCl2] is reducing agent and potassium dichromate [K2Cr2O7] is
oxidizing agent.
(Oxidation) half reaction: Fe2+ − e─ → Fe3+ ---------------------------------------- (1)
(Reduction) half reaction: Cr2O72- + 6e─ +14 H+ → 2Cr3+ + 7H2O-------------- (2)
To equalize the no. of electron in both side multiply the equation (1) with 6 and then
adding with equation (2) we found:
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(Oxidation) half reaction: 6Fe2+ − 6e─ → 6Fe3+
(Reduction) half reaction: Cr2O72− + 6e─ +14 H+ → 2Cr3+ + 7H2O
Over- all reaction: 6Fe2 + Cr2O72− + 6e─ +14 H+ → 2Cr3+ + 6Fe3+ + 7H2O---- (3)
Adding the spectator ions in both side of the equation (3),
6FeCl2 + K2Cr2O7 + 14 HCl → 2CrCl3 + 6 FeCl3 + 2KCl + 7 H2O
Practice: Ferrous chloride [FeCl2] salts reacts with acid potassium dichromate [K2Cr2O7]
to produce ferric chloride [Fe2Cl3], potassium chloride [KCl], chromium chloride [CrCl3]
and water.
C. Here, oxalic acid [H2C2O4] is reducing agent and potassium permanganate [KMnO4] is
oxidizing agent.
(Oxidation) half reaction: C2O42+ − 2 e─ → 2CO2 ------------- (1)
(Reduction) half reaction: MnO4─ +5e─ + 8H+ → Mn2+ + 4H2O------------- (2)
To equalize the No. of electron in both side multiply the equation (1) with 5 and equation
(2) with 2 and then adding we got:
(Oxidation) half reaction: 5C2O42+ − 10e─ → 10CO2
(Reduction) half reaction: 2MnO4─ +10e─ + 16H+ → 2Mn3+ + 8H2O
Over-all reaction: 5C2O42+ + 2MnO4─ + 16H+ → 2Mn3+ + 8H2O + 10CO2 ----- (3)
Adding the spectator ions in the both side of the equation (3)
5H2C2O4 + 2KMnO4 + 3H2SO4 → 2MnSO4 + K2SO4+ 8H2O + 10CO2
Practice:
1. Acidic (in H2SO4) potassium permanganate [KMnO4] reacts with sodium oxalate
[Na2C2O4] to produce manganese sulfate [MnSO4], potassium sulfate [K2SO4],
sodium sulfate [Na2SO4] carbon dioxide and water.
Answer: 5Na2C2O4 + 2KMnO4 + 8H2SO4 → 2MnSO4 + K2SO4 + 5Na2SO4 +
8H2O + 10CO2
2. Acidic (in HCl) potassium permanganate [KMnO4] reacts with sodium oxalate
[Na2C2O4] to produce manganese chloride [MnCl2], potassium chloride [KCl],
sodium chloride [NaCl] carbon dioxide and water.
Answer: 5Na2C2O4 + 2KMnO4 + 16HCl → 2MnCl2 + 2KCl + 10NaCl + 8H2O +
10CO2
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# Reagents:
A reagent is a "substance or compound that is added to a system in order to bring about a
chemical reaction, or added to see if a reaction occurs." Although the terms reactant and reagent
are often used interchangeably, a reactant is more specifically a "substance that is consumed in
the course of a chemical reaction". Solvents, although they are involved in the reaction, are
usually not referred to as reactants. Similarly, catalysts are not consumed by the reaction, so are
not described as reactants.
In organic chemistry, reagents are compounds or mixtures, usually composed of inorganic or
small organic molecules, that are used to effect a transformation on an organic substrate.
Examples of organic reagents include the Collins reagent, Fenton's reagent, and Grignard
reagent. There are also analytical reagents which are used to confirm the presence of another
substance. Examples of these are Fehling's reagent, Millon's reagent and Tollens' reagent.
In another use of the term, when purchasing or preparing chemicals, reagentgrade describes chemical substances of sufficient purity for use in chemical analysis, chemical
reactions or physical testing. Purity standards for reagents are set by organizations such
as ASTM (American Society for Testing and Materials ) International or the American Chemical
Society. For instance, reagent-quality water must have very low levels of impurities like
sodium and chloride ions, silica, and bacteria, as well as a very high electrical resistivity.
# Chemical grade and subtle differences between them:
Technical grade :
These products are suitable for non-critical tasks in the laboratory such as rinsing, dissolving or
are used as raw materials in production tasks.
Synthesis reagents :
Those reagents those are suitable for organic synthesis and preparative applications.
Extra Pure grade :
This quality is associated to products that are suitable for qualitative and semi-quantitative
grades. These chemicals suitable for general laboratory works, and in most cases, meet Most
Pharmacopoeia (BP, USP etc.) Standards.
Pharmacopoeia Grade :
Products and solutions meeting the purity requirements of chemical products admitted in the
pharmaceutical sector (pharmacopoeia requirement such as NF, BP, USP, Ph Eur, DAB, DAC,
JP, Ph Franc..)
For Analytical Purpose :
This quality is suitable for most analytical application, research and quality control.
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Guaranteed Reagent (GR) :
For laboratory purposes. The grade is equivalent to Analytical grade (A.R.) Reagent grade
(R.G.).
Electronic/ VLSI/ ULSI/ SLSIgrade :
Reagents used for semiconductor industry, cations, anions and particles value guarantee to ppm
to ppb level.
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