# StudySquare

# Redox equations for SQA National 5 Chemistry

This page covers the following topics:

1. Redox half-equations

2. Redox equations

Reduction is the gain of electrons or the decrease of an oxidation number. Oxidation is the loss of electrons or the increase of the oxidation number. A particle gains electrons when it is reduced and a particle loses electrons when it is oxidised. Redox half-equations provide information about the change in particles and the loss or gain of electrons. Many half-equations can be balanced by adding electrons, water molecules and hydrogen ions.

Fe³⁺ + e⁻ → Fe²⁺ (reduction)

Mg → Mg²⁺ + 2e⁻ (oxidation)

Redox half-equations can be combined to construct full ionic equations by matching the number of electrons and removing duplicate particles from both sides of an equation. If the electrons do not cancel out after adding half-equations, they can be multiplied by suitable integers beforehand. Both sides of a half-equation can also be switched around to ensure that electrons appear on both sides of an ionic equation. In a redox equation a reducing agent is an electron donor, while an oxidising agent is an electron acceptor.

# 1

Identify all particles that are reduced in the half-equations provided.

2Cl⁻ → Cl₂ + 2e⁻

Mn⁵⁺ + 3e⁻ → Mn²⁺

A particle gains electrons when it is reduced. In these half-equations Mn⁵⁺ gains electrons.

Mn⁵⁺

# 2

Which of the following half-equations could be combined together to create an ionic equation that corresponds to the reaction presented in the image?

A) Na → Na⁺ + e⁻

B) Cl₂ + 2e⁻ → 2Cl⁻

C) Br₂ + 2e⁻ → 2Br⁻

In the reaction bromine and chlorine change their oxidation numbers. Redox half-equations can be combined to construct full ionic equations by matching the number of electrons and removing duplicate particles from both sides of an equation. Both sides of a half-equation can also be switched around to ensure that electrons appear on both sides of an ionic equation.

B, C

# 3

Balance the half-equation provided.

MnO₄⁻ + H⁺ + e⁻ → Mn²⁺ + H₂O

MnO₄⁻ + H⁺ + e⁻ → Mn²⁺ + 4H₂O

MnO₄⁻ + 8H⁺ + e⁻ → Mn²⁺ + 4H₂O

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

# 4

Name the process presented by the half-equation provided.

O₂ + 4e⁻ → 2O²⁻

In the half-equation oxygen is gaining electrons. Reduction is the gain of electrons or the decrease of an oxidation number.

reduction

# 5

Combine relevant half-equations from the list below to provide an ionic redox equation between copper metal and lithium ions.

Br₂ + 2e⁻ → 2Br⁻

H₂ → 2H⁺ + 2e⁻

Cu²⁺ + 2e⁻ → Cu

Li → Li⁺ + e⁻

Both sides of a half-equation can also be switched around to ensure that electrons appear on both sides of an ionic equation.

Cu → Cu²⁺ + 2e⁻

Li⁺ + e⁻ → Li

If the electrons do not cancel out after adding half-equations, they can be multiplied by suitable integers beforehand.

Cu → Cu²⁺ + 2e⁻

2Li⁺ + 2e⁻ → 2Li

Redox half-equations can be combined to construct full ionic equations by matching the number of electrons and removing duplicate particles from both sides of an equation.

Cu + 2Li⁺ + 2e⁻ → Cu²⁺ + 2e⁻ + 2Li

Cu + 2Li⁺ → Cu²⁺ + 2Li

Cu + 2Li⁺ → Cu²⁺ + 2Li

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