3.2 Redox Reactions

Question 1

What is an oxidising agent?

Answer 1

The oxidising agent gains electrons from the species being oxidised.
The oxidising agent itself is reduced.

Question 2

What is a reducing agent?

Answer 2

The reducing agent donates electrons to the species being reduced.
The reducing agent itself is oxidised.

Question 3

What is a half equation?

Answer 3

An equation that tells you what is happening at one of the electrodes during an electrochemical reaction.

Question 4

Construct the half equation for the reduction of Cr2O7^2- ions to Cr3+ ions.

Answer 4

Cr2O7^2- + 14H+ + 6e- —> 2Cr3+ + 7H2O

Question 5

Construct the half equation for the reduction of MnO4^- to Mn2+ ions

Answer 5

MnO4^- + 8H+ + 5e- —> Mn2+ + 4H2O

Question 6

Construct the half equation for the oxidation of S2O3^2- —> S4O6^2- + 2e-

Answer 6

2S2O3^2- —> S4O6^2- + 2e-

Question 7

How do you combine two balanced half equations?
Zn(s) —> Zn2+(aq) + 2e-
2Fe3+ (aq) + 2e- —> 2Fe2+ (aq)

Answer 7

Combine the equations into one and cancel any common species that appear on both sides of the equation:
Zn(s) + 2Fe3+ (aq) + [2e-] —> Zn2+(aq) + [2e-] + 2Fe2+(aq)

Overall redox equation:
Zn(s) + 2Fe3+(aq) —> Zn2+(aq) + 2Fe2+(aq)

Question 8

What is a redox titration?

Answer 8

A titration of a reducing agent by an oxidising agent (or vice versa).

Question 9

What equipment is required to carry out a titration?

Answer 9

• Burette
• Pipette and pipette filter
• Conical flask
• Funnel
• White tile
• Clamp and stand

Question 10

How do you carry out a titration?

Answer 10

• Once the pipette has been used to measure one reactant into the conical flask, fill the burette with the other reactant. Record the burette initial volume.
• Add a few drops of indicator to the conical flask.
• Open the burette tap and allow the reactant to flow into the conical flask, swirling it to mix the contents.
• Close the burette tap once the colour change occurs.
• Record final burette volume.
• Repeat until you get concordant results, then calculate a mean titre.

Question 11

What is the ionic equation for the reaction between acidified Fe2+ and potassium manganate(VII)?

Answer 11

5Fe2+ + MnO4^- +8H+ —> 5Fe3+ + Mn2+ + 4H2O

Question 12

Describe the procedure of titrating potassium manganate(VII) with Fe2+ ions in an unknown solution

Answer 12

• Add the potassium manganate(VII) solution into the burette.
• The unknown solution containing Fe2+ ions is acidified with dilute H2SO4 and measured into a conical flask.
• Record the initial volume in the burette.
• Add the potassium manganate(VII) solution slowly into the flask. As it reacts, it decolourises. When there is the first trace of a permanent pale pink solution, close the burette tap.
• Record the final volume of the burette.

Question 13

What is potassium manganate(VII)?

Answer 13

Strong oxidising agent

Question 14

Explain what happens during the redox reaction between Cu2+ ions and I- ions

Answer 14

In this reaction, I- isn’t a strong enough reducing agent to completely reduce the Cu2+ ions, so they are only reduced to Cu+ ions.

Question 15

Give the chemical equation for the reaction between Cu2+ ions and I- ions

Answer 15

2Cu2+ + 4I- —> I2 + 2CuI

Question 16

What happens when iodide is titrated with thiosulfate ions?

Answer 16

Iodine is reduced by the thiosulfate reducing agent.
The thiosulfate ions are oxidised by the iodine oxidised agent.

Question 17

What is the chemical equation for the redox reaction that takes place between iodine and thiosulfate ions?
I2 + 2e- —> 2I-
2S2O3^2- —> S4O6^2- + 2e-

Answer 17

Cancel out the electrons and combine the two half equation:
I2 + 2S2O3^2- —> 2I- + S4O6^2-