Acids, Bases & Electrochemistry

Redox & Electrochemistry

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define oxidation and reduction in terms of electron transfer and oxidation numbers.
  • By the end of this lesson students will be able to assign oxidation numbers to elements in compounds and ions, and use them to identify oxidising and reducing agents.
  • By the end of this lesson students will be able to construct balanced half-equations for oxidation and reduction reactions, and combine them to form overall redox equations.
  • By the end of this lesson students will be able to describe the principles and applications of electrolysis, including the roles of the anode, cathode, and electrolyte.
  • By the end of this lesson students will be able to explain the operation of electrochemical (voltaic/galvanic) cells, including the calculation of standard cell potentials (HL).
  • By the end of this lesson students will be able to discuss the causes and prevention of corrosion, with a focus on its electrochemical nature and galvanic corrosion.

Key concepts

Oxidation Numbers

Oxidation numbers (also called oxidation states) are a way of keeping track of electron distribution in compounds. They represent the hypothetical charge an atom would have if all bonds were ionic. Rules for assigning oxidation numbers: 1. The oxidation number of an element in its uncombined state is 0 (e.g., O₂ = 0, Na = 0). 2. The oxidation number of a monatomic ion is equal to its charge (e.g., Na⁺ = +1, Cl⁻ = -1). 3. In compounds, Group 1 metals are +1, Group 2 metals are +2. 4. Fluorine is always -1 in compounds. 5. Oxygen is usually -2 in compounds, except in peroxides (e.g., H₂O₂) where it is -1, and with fluorine (e.g., OF₂) where it is +2. 6. Hydrogen is usually +1 in compounds, except in metal hydrides (e.g., NaH) where it is -1. 7. The sum of the oxidation numbers in a neutral compound is 0. 8. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Oxidation

Oxidation is the loss of electrons by a species, resulting in an increase in its oxidation number. The species that undergoes oxidation is called the reducing agent because it causes another species to be reduced.

Reduction

Reduction is the gain of electrons by a species, resulting in a decrease in its oxidation number. The species that undergoes reduction is called the oxidising agent because it causes another species to be oxidised.

Redox Reaction

A redox reaction is a chemical reaction in which both oxidation and reduction occur simultaneously. Electrons are transferred from one species to another.

Half-equations

Redox reactions can be split into two half-equations: one representing the oxidation process and the other representing the reduction process. These half-equations show the electron transfer explicitly. To balance half-equations in acidic solution: 1. Balance atoms other than oxygen and hydrogen. 2. Balance oxygen atoms by adding H₂O molecules. 3. Balance hydrogen atoms by adding H⁺ ions. 4. Balance the charge by adding electrons (e⁻) to the more positive side.

Electrolysis

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction. It involves passing an electric current through an electrolyte (a molten ionic compound or an ionic solution) to cause chemical decomposition. - **Anode**: The positive electrode where oxidation occurs (anions lose electrons). - **Cathode**: The negative electrode where reduction occurs (cations gain electrons). - **Electrolyte**: The substance containing free ions that conducts electricity. Applications include electroplating, extraction of reactive metals (e.g., aluminium from bauxite), and purification of metals (e.g., copper).

Electrochemical Cells (Galvanic/Voltaic Cells) (HL)

Electrochemical cells are devices that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two half-cells, each containing an electrode immersed in an electrolyte. - **Anode**: The electrode where oxidation occurs (negative terminal). - **Cathode**: The electrode where reduction occurs (positive terminal). - **Salt Bridge**: Connects the two half-cells, allowing ion flow to maintain electrical neutrality. - **External Circuit**: Wires connecting the electrodes, allowing electron flow. Standard electrode potential (E°) is the potential difference of a half-cell compared to the standard hydrogen electrode (SHE, defined as 0.00 V) under standard conditions (298 K, 1 atm, 1 M concentration). The cell potential (E°_cell) is calculated as E°_cell = E°_cathode - E°_anode. The more positive E° value corresponds to the species that will be reduced (cathode).

E°_cell = E°_cathode - E°_anode
Corrosion (Galvanic)

Corrosion is the deterioration of a metal due to its reaction with its environment, typically an electrochemical process. Rusting of iron is a common example of corrosion. - **Mechanism of Rusting**: Iron acts as the anode, oxidising to Fe²⁺ (Fe → Fe²⁺ + 2e⁻). Oxygen and water act as the cathode, reducing to hydroxide ions (O₂ + 2H₂O + 4e⁻ → 4OH⁻). The Fe²⁺ and OH⁻ ions then react to form iron(II) hydroxide, which is further oxidised by oxygen to hydrated iron(III) oxide (Fe₂O₃.nH₂O), which is rust. - **Galvanic Corrosion**: Occurs when two different metals are in electrical contact in the presence of an electrolyte. The more reactive metal (with a more negative standard electrode potential) acts as the anode and corrodes preferentially, protecting the less reactive metal (cathode). - **Prevention**: Methods include barrier protection (painting, oiling, plastic coating), galvanising (coating with zinc, which acts as a sacrificial anode), and cathodic protection (connecting the metal to a more reactive metal or an external power supply).

Key facts to remember

  • 1Oxidation is the loss of electrons (OIL), reduction is the gain of electrons (RIG).
  • 2An increase in oxidation number means oxidation; a decrease means reduction.
  • 3The oxidising agent is reduced; the reducing agent is oxidised.
  • 4In electrolysis, oxidation occurs at the anode (positive electrode) and reduction at the cathode (negative electrode). It requires an external power source.
  • 5In an electrochemical (galvanic) cell, oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive terminal). It generates electricity spontaneously.
  • 6Standard electrode potentials (E°) are measured relative to the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V.
  • 7Corrosion, such as the rusting of iron, is an electrochemical process involving oxidation of the metal and reduction of oxygen.
  • 8Sacrificial protection involves connecting a more reactive metal (e.g., zinc or magnesium) to the metal being protected, causing the more reactive metal to corrode instead.

Worked examples

Example 1

For the reaction: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (in acidic solution): a) Assign oxidation numbers to chromium and iron in all species. b) Identify the oxidising agent and the reducing agent.

Ia) Assigning oxidation numbers:
II For Cr₂O₇²⁻:
III Oxygen is -2. Let Cr be x.
IV 2x + 7(-2) = -2
V 2x - 14 = -2
VI 2x = 12
VII x = +6. So, Cr in Cr₂O₇²⁻ is +6.
VIII For Fe²⁺: The charge is +2. So, Fe in Fe²⁺ is +2.
9 For Cr³⁺: The charge is +3. So, Cr in Cr³⁺ is +3.
10 For Fe³⁺: The charge is +3. So, Fe in Fe³⁺ is +3.
11b) Identifying oxidising and reducing agents:
12 Chromium changes from +6 (in Cr₂O₇²⁻) to +3 (in Cr³⁺). The oxidation number decreases, so Cr₂O₇²⁻ is reduced. Therefore, Cr₂O₇²⁻ is the oxidising agent.
13 Iron changes from +2 (in Fe²⁺) to +3 (in Fe³⁺). The oxidation number increases, so Fe²⁺ is oxidised. Therefore, Fe²⁺ is the reducing agent.

Answer

a) Cr in Cr₂O₇²⁻ = +6, Fe in Fe²⁺ = +2, Cr in Cr³⁺ = +3, Fe in Fe³⁺ = +3. b) Oxidising agent: Cr₂O₇²⁻. Reducing agent: Fe²⁺.

Remember that the oxidising/reducing agent is the *reactant* species that contains the element undergoing the change.

Example 2

Balance the following redox equation in acidic solution: MnO₄⁻ + SO₃²⁻ → Mn²⁺ + SO₄²⁻

I1. Separate into half-equations:
II Reduction: MnO₄⁻ → Mn²⁺
III Oxidation: SO₃²⁻ → SO₄²⁻
IV2. Balance the reduction half-equation (MnO₄⁻ → Mn²⁺):
V a) Balance Mn: Already balanced.
VI b) Balance O: Add 4H₂O to the right: MnO₄⁻ → Mn²⁺ + 4H₂O
VII c) Balance H: Add 8H⁺ to the left: 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
VIII d) Balance charge: Left side charge = 8(+1) + (-1) = +7. Right side charge = +2 + 0 = +2. Add 5e⁻ to the left: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
93. Balance the oxidation half-equation (SO₃²⁻ → SO₄²⁻):
10 a) Balance S: Already balanced.
11 b) Balance O: Add 1H₂O to the left: H₂O + SO₃²⁻ → SO₄²⁻
12 c) Balance H: Add 2H⁺ to the right: H₂O + SO₃²⁻ → SO₄²⁻ + 2H⁺
13 d) Balance charge: Left side charge = 0 + (-2) = -2. Right side charge = -2 + 2(+1) = 0. Add 2e⁻ to the right: H₂O + SO₃²⁻ → SO₄²⁻ + 2H⁺ + 2e⁻
144. Equalise electrons and combine:
15 Multiply the reduction half-equation by 2: 10e⁻ + 16H⁺ + 2MnO₄⁻ → 2Mn²⁺ + 8H₂O
16 Multiply the oxidation half-equation by 5: 5H₂O + 5SO₃²⁻ → 5SO₄²⁻ + 10H⁺ + 10e⁻
17 Add the two balanced half-equations:
18 (10e⁻ + 16H⁺ + 2MnO₄⁻) + (5H₂O + 5SO₃²⁻) → (2Mn²⁺ + 8H₂O) + (5SO₄²⁻ + 10H⁺ + 10e⁻)
195. Simplify by cancelling common species (electrons, H⁺, H₂O):
20 Cancel 10e⁻ from both sides.
21 Cancel 10H⁺ from both sides (16H⁺ - 10H⁺ = 6H⁺ on left).
22 Cancel 5H₂O from both sides (8H₂O - 5H₂O = 3H₂O on right).

Answer

6H⁺ + 2MnO₄⁻ + 5SO₃²⁻ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O

Always double-check that both mass (atoms of each element) and charge are balanced in the final equation.

Example 3

(HL) An electrochemical cell is constructed using a zinc electrode in 1.0 M ZnSO₄ solution and a copper electrode in 1.0 M CuSO₄ solution. Given the standard electrode potentials: Zn²⁺(aq) + 2e⁻ → Zn(s) E° = -0.76 V Cu²⁺(aq) + 2e⁻ → Cu(s) E° = +0.34 V a) Identify the anode and cathode. b) Write the half-reactions occurring at each electrode. c) Calculate the standard cell potential (E°_cell).

Ia) Identify anode and cathode:
II Compare the standard electrode potentials. The species with the more positive (or less negative) E° value will be reduced (cathode).
III E°(Cu²⁺/Cu) = +0.34 V (more positive) → Cu²⁺ is reduced, so copper is the cathode.
IV E°(Zn²⁺/Zn) = -0.76 V (more negative) → Zn is oxidised, so zinc is the anode.
Vb) Write the half-reactions:
VI At the anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
VII At the cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
VIIIc) Calculate the standard cell potential (E°_cell):
9 E°_cell = E°_cathode - E°_anode
10 E°_cell = (+0.34 V) - (-0.76 V)
11 E°_cell = +0.34 V + 0.76 V

Answer

a) Anode: Zinc electrode. Cathode: Copper electrode. b) Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) c) E°_cell = +1.10 V

A positive E°_cell indicates a spontaneous reaction, meaning the cell will generate electricity.

Common mistakes

  • Confusing oxidation (loss of electrons, increase in ON) with reduction (gain of electrons, decrease in ON). A common mnemonic is OIL RIG.
  • Incorrectly assigning oxidation numbers, especially for oxygen and hydrogen in unusual compounds (peroxides, hydrides) or in polyatomic ions.
  • Failing to balance both mass (atoms) and charge (electrons and ions) when constructing half-equations.
  • Mixing up the roles of anode and cathode, or the signs of the electrodes, between electrolysis and electrochemical cells.
  • Forgetting to reverse the sign of the standard electrode potential for the oxidation half-reaction when calculating E°_cell, or using E°_anode - E°_cathode instead of E°_cathode - E°_anode.

Exam tips

  • Practise assigning oxidation numbers for a wide range of compounds and ions until it becomes second nature.
  • Master the systematic steps for balancing half-equations in acidic (and basic, if covered) conditions. Show all steps clearly in your exam answers.
  • Draw and label diagrams for electrolysis cells and electrochemical cells, clearly indicating the electrodes, electrolyte, direction of electron flow, and ion movement.
  • For electrochemical cell calculations (HL), always identify the cathode (reduction, more positive E°) and anode (oxidation, more negative E°) first, then apply the formula E°_cell = E°_cathode - E°_anode.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.