6. Oxidation and Reduction
🎯 Learning outcomes
By the end of this chapter, Students should be able to:
- 6.1 Define oxidation and reduction in terms of the electronic concept.
- 6.2 Define oxidation number and explain the rules of assigning oxidation numbers.
- 6.3 Calculate oxidation numbers of elements in compounds and ions.
- 6.4 Explain redox processes in terms of changes in oxidation number.
- 6.5 Use oxidation number change to identify oxidizing and reducing agents.
- 6.6 Balance the given redox reaction by oxidation number change or the half-equation method.
- 6.7 Explain the qualitative and quantitative aspects of Faraday’s laws of electrolysis.
1The Classical Concept of Oxidation and Reduction▾
Addition of oxygen or electronegative element / removal of hydrogen or electropositive element — OXIDATION
2 Na + Cl₂ → 2 NaCl
Removal of oxygen or electronegative element / addition of hydrogen or electropositive element — REDUCTION
Electronic interpretation of oxidation and reduction
LOE — oxidation; GOE — reduction
2Oxidation Numbers and Rules for Assigning Them▾
Solution: the residual charge that an atom appears to bear when all other atoms are removed from the chemical species as their respective ions
Rules to assign oxidation numbers:
- Free state – any element – 0
- O – generally −2 (e.g., Na₂O), exceptions: peroxide −1 (e.g., Na₂O₂) and superoxide −1/2 (e.g., NaO₂); OF₂ (+2, fluorine= −1)
- H – generally +1, except metal hydrides −1
- Alkali metals +1 (Li, Na, K, Rb, Cs, Fr)
- Alkaline earth metals +2 (Be, Mg, Ca, Sr, Ba, Ra)
- Cl – only in binary compounds with metals −1.
- Total ON of neutral molecule = 0, total ON of ion = its charge
- More electronegative elements are always assigned a negative charge.
a) N₂O₂
2x = 4
x = +2
k) NH₄NO₃
In NH₄⁺,
x + 4(+1) = +1 or, x = −3
l) NH₄NO₃
m) Cr₂(SO₄)₃
Solution:
+4 −2 +4 −2 +1 −2
SO₂ + NO₂ + H₂O
In this equation, the ON of S has increased; an increase in the ON is called oxidation, hence S has been oxidized.
The one that gets oxidised by reducing others is called a reducing agent. Hence, S is the reducing agent.
Similarly, ON of N in HNO₃ has decreased; the decrease in ON is called reduction, hence N or HNO₃ has been reduced. The one that gets reduced by oxidising another is called an oxidising agent. Hence, N or HNO₃ is the oxidising agent.
Or,
Sulphur (S) has been oxidized into SO₂ (or S⁺⁴).
Nitrogen or N⁺⁵ (or HNO₃) has been reduced into N⁺⁴ or NO₂.
HNO₃ or N is an oxidising agent (oxidant), as it oxidises S into SO₂.
S is the reducing agent (reductant), as it reduces HNO₃ into NO₂.
3Balancing Redox Reactions by the Oxidation Number Method▾
Step 1: writing oxidation number of each atom,
Step 2: showing the change in ON per atom — Mn decreases in ON by 5 per atom; C increases in ON by 1 per atom.
Step 3: Calculating the changes in ON per molecule on the reactant side,
Increase in ON of C per molecule = 1 × 2 = 2 → C₂H₂O₄
Step 4: Doing criss-cross multiplication on the reactant side,
Step 5: Balancing the oxidized and reduced atoms correspondingly on the product side,
Step 6: Balancing remaining atoms by the hit and trial method (first metals, then other non-metals, and finally oxygen and hydrogen)
2 KMnO₄ + 5 C₂H₂O₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂
4Balancing Redox Reactions by the Ion-Electron (Half-Reaction) Method▾
Solution: Changing into ionic form;
Balancing oxidation half-reaction,
Since the Zn (oxidized atom) is already balanced,
Zn + 2 OH⁻ → ZnO₂⁻² (balancing oxygen)
Zn + 2 OH⁻ → ZnO₂⁻² + 2 H⁺ (balancing hydrogen)
Zn + 2 OH⁻ → ZnO₂⁻² + 2 H⁺ + 2e⁻ (balancing charge)
The reduction half equation is,
NO₃⁻ → NH₃ + 3 H₂O (balancing oxygen)
NO₃⁻ + 9 H⁺ → NH₃ + 3 H₂O (balancing hydrogen)
NO₃⁻ + 9 H⁺ + 8 e⁻ → NH₃ + 3 H₂O (balancing charge)
Adding the balanced oxidation half and reduction half-reactions in such a way that electrons cancel out.
NO₃⁻ + 9 H⁺ + 8 e⁻ → NH₃ + 3 H₂O
─────────────────────────────────────
4 Zn + 8 OH⁻ + NO₃⁻ + H⁺ → 4 ZnO₂⁻² + NH₃ + 3 H₂O
4 Zn + 7 OH⁻ + NO₃⁻ + H₂O → 4 ZnO₂⁻² + NH₃ + 3 H₂O
4 Zn + 7 OH⁻ + NO₃⁻ → 4 ZnO₂⁻² + NH₃ + 2 H₂O
4 Zn + NaNO₃ + 7 NaOH → 4 Na₂ZnO₂ + NH₃ + 2 H₂O
The above is the balanced chemical equation.
5Disproportionation Reactions▾
Disproportionation can be thought of as a ‘self-reduction / oxidation’ reaction.
When chlorine gas is passed into cold, dilute aqueous sodium hydroxide, the following reaction takes place and two salts are formed:
6Electrolysis▾
Conductors and non-conductors
The substances that conduct electricity are called conductors. e.g.:
- Metals – Cu, Ag, Fe, Au
- Salts – NaCl, CdS, PbS etc.
- Acids – HCl, H₂SO₄, HNO₃
- Alkali – NaOH, KOH, Ca(OH)₂
The substances that cannot conduct electricity are called non-conductors. e.g.: sugar, glucose, P, S, etc.
Classification of conductors
Metallic conductors: The conductors through which conduction of electricity takes place by the migration of electrons under the influence of an applied potential are called metallic conductors. E.g.: metals, alloys, graphite, etc.
Electrolytic conductors: The conductors through which the conduction of electricity takes place by the migration of ions towards oppositely charged electrodes due to the occurrence of chemical changes at the surface of electrodes are called electrolytic conductors. E.g.: a solution of weak and strong electrolytes, molten salt, etc.
Electrolysis
The process of decomposition of an electrolyte by passing an electric current through its aqueous solution or fused (molten) state is called electrolysis.
- The electrochemical cell in which the electrical energy is used to bring about a chemical reaction is called an electrolytic cell.
- In this cell, two metallic electrodes are dipped into the solution of a suitable electrolyte.
- Then, the electrodes are connected to the external source of electricity, such as a battery. (DC current is required)
- The electrode that is connected to the positive terminal of the battery is called the anode, and the electrode that is connected to the negative terminal of the battery is called the cathode.
- When electricity is passed, chemical reactions take place inside the cell. i.e. Cations move towards the cathode and anions move towards the anode.
- For example, let us consider the electrolysis of molten NaCl.
Reactions involved:
At anode: Cl⁻ − 1e⁻ → Cl (oxidation) Cl + Cl → Cl₂
At cathode: Na⁺ + 1e⁻ → Na (reduction)
Electrolysis finds many applications in
- purification of precious metals like: Cu, Ag
- manufacture of Ca, Na, Al
- manufacture of NaOH, Cl₂, F₂
7Faraday’s Laws of Electrolysis▾
First Law
The amount of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through it.
Let, W gram of substance be deposited or liberated at the electrode by passing Q amount of electricity (charge), then according to the first law.
or, w = zQ
or, w = zIt (since I = Q/t; or Q = It)
where z is a proportionality constant called electrochemical equivalent (ECE). I is current electricity, t is time.
when, I = 1 ampere and t = 1 second, W = z
Hence, ECE (z) is defined as the mass of substance deposited or liberated when a current of 1 A is passed through the electrolytic cell for one second.
In other words, it is the mass of substance deposited or liberated on passing 1 coulomb charge.
Faraday (F)
It has been found that the same amount of electrons (charge) is required to deposit 1 gram equivalent of different substances, which is nearly equal to 96500 coulomb.
One faraday can also be defined as the amount of charge carried by 1 mole of electrons.
Relation between ECE and CE (Chemical equivalent)
According to Faraday’s first law of electrolysis,
or, z = w / Q
We know that 1 F (96500 Coulomb) of electricity deposits one equivalent of a substance. Therefore, when w = E; Q = 1 F or 96500 coulomb
Hence,
where E is the equivalent weight or chemical equivalent.
The equivalent weight of an element is defined as the parts by weight of that element that combine with or displace 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, or 35.5 parts by weight of chlorine.
The equivalent weight of an element can be calculated by dividing its atomic weight by its valency, i.e.,
Second Law
When the same quantity of charge (electricity) is passed through different electrolytic solutions, the amount of different substances deposited or liberated at respective electrodes is directly proportional to their equivalent weights (or chemical equivalents).
Let W grams of substance be deposited or liberated by a certain quantity of electricity, and E is the equivalent weight of the substance. Then,
or W = constant × E
or, W/E = constant ……………(i)
If the same quantity of electricity deposits W₁, W₂,… grams of different substances with equivalent weights E₁, E₂,…, respectively. From equation (i),