Oxidation and Reduction

Oxidation is the loss of electrons by a substance, while reduction is the gain of electrons by a substance. These two processes always occur together — when one substance loses electrons (oxidation), another substance must gain them (reduction).

A simple way to remember:

• OIL — Oxidation Is Loss of electrons
• RIG — Reduction Is Gain of electrons

Oxidizing and Reducing Agents

• An oxidizing agent is a substance that accepts electrons from another substance, causing it to be oxidized. Examples: potassium permanganate (KMnO₄), potassium dichromate (K₂Cr₂O₇), hydrogen peroxide (H₂O₂).
• A reducing agent is a substance that donates electrons to another substance, causing it to be reduced. Examples: zinc metal, iron(II) ions, oxalic acid (H₂C₂O₄).

Oxidation Number

The oxidation number is a number assigned to an element in a compound that represents the number of electrons lost or gained. Rules for assigning oxidation numbers:

• Free elements have oxidation number zero (e.g., O₂, Fe, Na)
• Group 1 elements are always +1
• Group 2 elements are always +2
• Fluorine is always -1
• Oxygen is usually -2 (except in peroxides where it is -1)
• Hydrogen is +1 (except in metal hydrides where it is -1)
• The sum of oxidation numbers in a compound equals zero

Identifying Redox Reactions

A reaction is a redox reaction if there is a change in oxidation numbers. Consider this reaction:

$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$

• Zn goes from 0 to +2 (loses 2 electrons → oxidation)
• Cu²⁺ goes from +2 to 0 (gains 2 electrons → reduction)

Definition

Electrolysis is the decomposition of an electrolyte (a conductive solution or molten substance) by passing an electric current through it. This process causes chemical changes at the electrodes.

Electrolytes and Non-electrolytes

Types of Electrolytes

Strong electrolytes ionize completely in solution and conduct electricity well:

• Strong acids: HCl, H₂SO₄, HNO₃
• Strong bases: NaOH, KOH
• Soluble salts: NaCl, KNO₃

Weak electrolytes ionize partially in solution and conduct electricity poorly:

• Weak acids: CH₃COOH, H₂CO₃
• Weak bases: NH₃, NH₄OH
• Water

The Electrolysis Process

During electrolysis:

• The anode is the positive electrode where oxidation occurs (loss of electrons)
• The cathode is the negative electrode where reduction occurs (gain of electrons)
• Positive ions (cations) move toward the cathode
• Negative ions (anions) move toward the anode

Example: Electrolysis of copper(II) sulfate using copper electrodes

• Impure copper is used as the anode
• Pure copper sheet is used as the cathode
• Copper ions (Cu²⁺) from the anode dissolve into solution
• Pure copper deposits onto the cathode
• Impurities fall to the bottom as anode sludge

Faraday's First Law

The mass of a substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

Mathematically: $m = ZIt$

Where:

• $m$ = mass of substance liberated (g)
• $Z$ = electrochemical equivalent (g/C)
• $I$ = current (A)
• $t$ = time (s)

The electrochemical equivalent (Z) is the mass of substance liberated by 1 coulomb of electricity.

Faraday's Second Law

When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their chemical equivalents.

The chemical equivalent of an element is: $\text{Chemical equivalent} = \frac{\text{Relative atomic mass}}{\text{Valency}}$

Or equivalently: $m = \frac{A_r \times Q}{V \times 96500}$

Where:

• $A_r$ = relative atomic mass
• $V$ = valency
• $Q$ = quantity of electricity (It)

Faraday's constant (F) = 96500 C/mol represents the charge of one mole of electrons.

A current of 0.5 A is passed through a solution of copper(II) sulfate for 30 minutes. Calculate the mass of copper deposited. (Ar of Cu = 63.5, valency = 2)

Solution:

Given:

• Current, I = 0.5 A
• Time, t = 30 minutes = 30 × 60 = 1800 s
• Ar of Cu = 63.5
• Valency, V = 2

Step 1: Calculate quantity of electricity: $Q = It = 0.5 \times 1800 = 900 \text{ C}$

Step 2: Calculate mass of copper using the formula: $m = \frac{A_r \times Q}{V \times 96500}$

$m = \frac{63.5 \times 900}{2 \times 96500}$

$m = \frac{57150}{193000} = 0.296 \text{ g}$

The mass of copper deposited is 0.30 g (approximately).

1. Purification of Metals

Impure metals are refined by electrolysis. The impure metal forms the anode, pure metal forms the cathode, and a solution of the metal's salt is the electrolyte.

• Copper purification: Impure copper is refined to 99.99% purity for electrical wiring
• Zinc purification: Zinc is refined electrolytically for galvanizing

2. Extraction of Metals

Very reactive metals cannot be extracted by reduction with carbon. They are extracted by electrolysis of their molten chlorides or oxides:

• Aluminum from bauxite (Al₂O₃) in cryolite
• Sodium from molten NaCl
• Magnesium from molten MgCl₂

3. Electroplating

Electroplating coats a metal object with a thin layer of another metal for protection and decoration:

• The object to be plated is the cathode
• The plating metal is the anode
• A solution of the plating metal's salt is the electrolyte

Common examples: silver plating, chrome plating, nickel plating, and gold plating.

In Tanzania, electroplating is applied in Dar es Salaam and Arusha workshops to restore and decorate metal parts such as motorcycle parts, bicycle handles, and household utensils. When a local mechanic re-chromes a motorcycle exhaust or a jeweler plates cheap metals with silver, they are using the principles of electrolysis learned in this topic. Understanding redox reactions also helps students recognize why iron roofs in coastal areas (like in Zanzibar) corrode faster due to saltwater exposure — a common issue affecting many Tanzanian households and businesses.