Chapter 23 - Redox and Electrode Potentials

Question 1

Define redox

Answer 1

A reaction where both reduction and oxidation takes place.

Question 2

What is the manganate half reaction?

Answer 2

MnO4-(aq) + 8H+(aq) + 5e- —> Mn2(aq) + 4H2O(l)

Question 3

What is the procedure for a Manganate(VII) titration?

Answer 3

1. Add a standard solution of KMnO4 to the burette and measure initial reading.
2. Using a pipette, add a measured volume of the other solution to a conical flask.
3. Add an excess of dilute sulphuric acid to the flask to provide H+ ions to reduce MnO4- ions. No indicator is needed as reaction is self-indicating.
4. During titration, the manganate(VII) is decolourised as it lands in the conical flask until the end point - where it will turn pink. Record new reading on burette.
5. Repeat titration until concordant results obtained.

Question 4

How is the meniscus read differently when measuring out KMnO4 in manganate(VII) titrations?

Answer 4

Read from the top of the meniscus because potassium permanganate has a deep purple colour so it is hard to see from bottom of meniscus.

Question 5

Which 2 reducing agents can be analysed using manganate(VII) titrations?

Answer 5

Iron (II) ions, Fe2+(aq)

Ethanedioic acid, (COOH)2(aq)

Question 6

What is the iodine/thiosulfate reaction?

Answer 6

2S2O3-(aq) + I2(aq) —> 2I-(aq) + S4O62-(aq)

Question 7

What is the oxidation reaction in an iodine/thiosulfate reaction?

Answer 7

2S2O3-(aq) —> S4O62- + 2e-

Question 8

What is the reduction reaction in an iodine/thiosulfate reaction?

Answer 8

I2(aq) + 2e- —> 2I-(aq)

Question 9

What can iodine/thiosulfate titrations be used to determine?

Answer 9

The ClO- content in household bleach
The Cu2+ content in copper(II) compounds
The Cu content in copper alloys

Question 10

What is the procedure for an iodine/thiosulfate titration?

Answer 10

1. Add a standard solution of sodium thiosulfate, Na2S2O3, to the burette
2. Using a pipette, add solution with oxidising agent to conical flask.
3. Add an excess of potassium iodide. Oxidising agent reacts with iodide ions to produce iodine, which turns the solution yellow-brown
4. Titrate the solution with sodium thiosulfate. During titration, iodine is reduced back to I- ions so brown colour fades and end point is hard to judge.
5. Add starch indicator, which turns solution blue-black. End point is when blue-black colour disappears, as all iodine has been reduced to I- ions.

Question 11

Why is starch indicator added towards the end of the iodine-thiosulfate titration?

Answer 11

Because if added too early, the starch-iodine complex may precipitate out of the solution, preventing some of the iodine reacting with the thiosulfate.

Close to the end-point, the iodine concentration is low enough for the complex not to precipitate out.

Question 12

What is a voltaic cell?

Answer 12

A type of electrochemical cell which converts chemical energy into electrical energy.

Question 13

What property does a voltmeter have in a cell?

Answer 13

High resistance

Question 14

Which way does the current flow in the cell?

Answer 14

From negative electrode to positive electrode

Question 15

How do ions flow between cells?

Answer 15

Salt bridge

Question 16

What is the salt bridge generally made of?

Answer 16

Potassium nitrate

Ammonium nitrate

Question 17

What is the electrode made of in a standard hydrogen cell?

Answer 17

Platinum

Question 18

When 2 half cells are connected, what is oxidised and what is reduced?

Answer 18

More reactive metal is oxidised

Less reactive metal is reduced

Question 19

Define standard electrode potential (E°)

Answer 19

The e.m.f. of a half cell compared with a standard hydrogen half cell. Measured at 298K with solution concentrations of 1 mol dm-3 and a gas pressure of 100kPa

Question 20

What is the electrode potential of a hydrogen half-cell?

Answer 20

0.00V

Question 21

How do you calculate electrode potential of a cell?

Answer 21

E° of the more positive terminal - E° of the more negative terminal.

Question 22

How do we know if a cell is feasable?

Answer 22

If the electrode potential of the cell is positive, the reaction is fesable

Question 23

Which electrode goes on the right when asked about a labelled diagram?

Answer 23

The half-cell with the more +ve E° (electrode potential)
The metal ions that are more easily reduced

Question 24

Which electrode goes on the left when asked about a labelled diagram?

Answer 24

The half-cell with the more -ve E°


The metal ions that are more easily oxidised

Question 25

What does the more positive E° value mean for the position of the equilibrium?

Answer 25

Equilibrium lies further to the RIGHT than the other half-cell
So the metal ions are more easily REDUCED than the other half-cell
So the metal is an OXIDISING agent

Question 26

What does the more negative E° value mean for the position of the equilibrium?

Answer 26

Equilibrium lies further to the LEFT than the other half-cell
So the metal ions are more easily OXIDISED than the other half-cell.
So the metal is an REDUCING agent

Question 27

State 2 reasons why the actual cell potential may be different from the value calculated.

Answer 27

Non- standard conditions


High activation energy, so very slow rate of reaction

Question 28

What’s in a Metal/Metal ion half cell?

Answer 28

Metal rod (e.g. Cu)
Dipped into aqueous solution of metal ion (e.g. Cu2+)

Question 29

What’s in a, ion/ion half-cell?

Answer 29

Contains ions of the same element in different oxidation states (e.g Fe3+ and Fe2+)
Pt electrode

Question 30

State the key difference between a primary cell and a secondary cell.

Answer 30

A primary cell is only used once and is non-rechargeable

A secondary cell is rechargeable

Question 31

State the key characteristics of a fuel cell.

Answer 31

A fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage.


A fuel cell can be run continuously provided that the fuel and oxygen are continually being supplied to the cell.