Unit 1- Oxidising and reducing agents

Question 1

Definition of reduction?

Answer 1

Reduction is gain of electrons by a reactant (or loss of oxygen).

Question 2

Definition of oxidation?

Answer 2

Oxidation is loss of electrons by a reactant (or gain of oxygen).

Question 3

Definition of a redox equation?

Answer 3

A Redox Reaction is when an oxidation and a reduction reaction takes place at the same time.

Question 4

Definition of a oxidising agent?

Answer 4

Oxidising agent is a substance that accepts electrons (helps another reactant to be oxidised by being reduced itself)

Question 5

Definition of a reducing agent?

Answer 5

Reducing agent is a substance that donates electrons (helps another reactant to be reduced by being oxidised itself).

Question 6

How do you recognise oxidising and reducing agents in reactions, on the periodic table and by using electronegativities?

Answer 6

Elements with low electronegativities can form ions by losing electrons so act as reducing agents (elements with high electronegativities can act as oxidising agents).

Question 7

How do you identify compounds, ions and molecules that can act as oxidising and reducing agents?

Answer 7

Oxidising agents are at the bottom of the left-hand side column of the electrochemical series:

permanganate (MnO4-)
dichromate (Cr2O72-)
hydrogen peroxide (H2O2)
Reducing agents are at the top right-hand column of the electrochemical series.

Question 8

What are the uses of oxidising agents?

Answer 8

Oxidising agents can be used to kill fungi and bacteria and also as a bleach.

Question 9

How to balance ion-electron equations?

Answer 9

You need to be able to balance ion-electron equations by adding water, Hydrogen ions and electrons eg:

Cr2O72-(aq) → Cr3+(aq)

Step 1
Balance the chromium ions.

Cr2O72-(aq) → 2Cr3+(aq)

Step 2
Balance the oxygen by adding water.

Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)

Step 3
Balance the hydrogen by adding hydrogen ions.

Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l)

Step 4
Balance the charge. In this example there is a charge of 12+ on the LHS and 6+ on the RHS. This means 6 electrons will need to be added to the LHS to give both sides a charge of 6+.

Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

Question 10

How do you combine ion-electron equations to produce redox equations?

Answer 10

Combine the following oxidation and reduction reactions to show the overall redox reaction:

1: Cl2(g) + 2e- → 2Cl-(aq)
2: Fe2+(aq) → Fe3+(aq) + e-

Step 1
Before combining the equations the number of electrons must be equal in each equation. Remember whatever you multiply e- by, everything else in the equation must be multiplied by same value.

1: Cl2(g) + 2e- → 2Cl-(aq)
2: Fe2+(aq) → Fe3+(aq) + e- × 2
2: 2Fe2+(aq) → 2 Fe3+(aq) + 2e-

Step 2
Combine equations.

Cl2(g) + 2e- + 2Fe2+(aq) → Cl-(aq) + 2 Fe3+(aq) + 2e-

Step 3
Cancel out electrons (and anything else that appears on both sides).

Cl2(g) + 2Fe2+(aq) → 2Cl-(aq) + 2Fe3+(aq)