Redox

Question 1

Define oxidation

Answer 1

Loss of electrons or increase in oxidation number

Question 2

Define reduction

Answer 2

Gain of electrons or decrease in oxidation number

Question 3

What is an oxidising agent

Answer 3

1. A reagent that oxidises (takes electrons from) another species.
2. It contains the species that is reduced.

Question 4

What is a reducing agent

Answer 4

1. A reagent which reduces (adds electrons to) another species
2. Contains the species that is oxidised

Question 5

Write the steps for writing a redox equations from half-equations

Answer 5

1. Write out the half equations
2. Balance the electrons
3. Add the half equations and cancel the electrons
4. Cancel any species that are on both sides of the eqaution

Question 6

Some ideas of how to predict products of redox reactions

Answer 6

1. In aqueous redox reactions H2O is often formed.
2. H+ and OH- ions are also likely products
3. Check that both sides are balanced by charge

Question 7

Write the half-equations for the following reactions and state whether oxidation or reduction:

1. Manganate (VII) ions, MNO4- reacting in acidic conditions to form manganese(II) ions
2. VO3- reacting in acidic conditions to produce Vanadium (III) ions
3. Chromium (III) ions reacting in alkaline solution to produce chromate (VI) ions

Answer 7

1. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O = Reduction
2. VO3- + 6H+ + 2e- →V3+ +3H2O = Reduction
3. Cr3+ + 8OH- →CrO4 2- + 3e- + 4H2O = Oxidation

Question 8

Describe the procedure of a manganate (VII) titration

Answer 8

1. A standard solution of potassium manganate is added to the burette
2. Using a pipette, add a measured volume of the solution being analysed to the conical flask.
3. An excess of dilute sulfuric acid is also added to provide the H+ ions required for the reduction of MnO4- ions.- no indicator needed
4. During the titration the manganate solution reacts and is decolourised and the endpoint is judged by the first permanent pink colour, indicating when there is an excess of MnO4- ions present.
5. Repeat until you obtain concordant titres.
6. The manganate titrations can be used for the analysis of many different reducing agents e.g Fe2+ and COOH2

Question 9

Describe how you would analyse the purity of an iron (II) compound

Answer 9

1. Do a manganate titration with impure iron compound e.g. FeSO4.7H2O -this will be in a conical flask and have the KMnO4 added
2. Calculate the meant titre- volume of KMnO4- and from this calculate the mols of MnO4- reacted
3. Then determine the mols of Fe2+ that reacted using the equation and molar ratios.
4. Scale up to find the amount of Fe2+ in the original solution
5. Find mass of FeSO4.7H2O in the impure sample
6. Find percentage purity= mass of FeSO4.7H2O/mass of impure sample* 100

Question 10

Experiments involving iodine/thiosulfate titrations

Answer 10

Haven’t been told to learn so leaving for now

Question 11

What is a half-cell

Answer 11

1. A half-cell contains the chemical species present in a redox half-equation.
2. A voltaic cell can be made by connecting together two different half-cells, which then allow electrons to flow.
3. A voltaic cell is a type of electrochemical cell which converts chemical energy into electrical energy - takes place in modern cells and batteries that power devices such as mobile phones.
4. In the cell the chemicals in the two half-cells must be kept apart- if allowed to mix, the electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy.

Question 12

Describe what a metal/metal ion half-cell looks like

Answer 12

1. A metal rod dipped into a solution of its aqueous metal ion. A vertical line is used for the phase boundary between the aqueous solution and the metal e.g Zn2+(aq)|Zn(s)
2. At the phase boundary an equilibrium will be set up
3. Convention is that the equilibrium is written so that the forward reactions shows reduction and the reverse shows oxidation.
4. In an isolated half-cell there is no net transfer of electrons either into or out of the metal.

Question 13

Describe what an ion/ion half-cell looks like

Answer 13

1. Contains ions of the same element in different oxidaiton states e.g Fe2+ and Fe3+
   Fe3+ (aq) + e- ↔ Fe2+ (aq)
2. There is no metal to transport electrons in this half-cell so and inert metal electrode made out of platinum is used.

Question 14

How do you know which electrode has a greater tendency to gain or lose electrons in a cell with two metal/metal ion half-cells.

Answer 14

1. The more reactive metal releases electrons more readily and is oxidised
2. The electrode with the more reactive metal loses electrons and is oxidised- negative electrode
3. The electrode with the less reactive metal gains electrons and is reduced- positive electrode.

Question 15

Define standard electrode potential

Answer 15

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

Question 16

Describe the standard hydrogen half-cell

Answer 16

1. Beaker containing 1 mol dm-3 of H+(aq)
2. Glass tube with holes to allow bubbles of H2 (g) to escape
3. H2(g) Going into the glass tube at the top
4. Platinum electrode.
5. The standard electrode potential of a standard hydrogen electrode is exactly 0V

Question 17

Describe how to measure a standard electrode potential

Answer 17

1. Connect the half-cell to a standard hydrogen electrode
2. The two electrodes are connected by a wire to allow a controlled flow of electrons
3. The two solutions are connected with a salt bridge which allows ions to flow. The salt bridge typically contains a concentrated solution of an electrolyte that does not react with either solution e.g. a strip of filter paper soaked in aqueous potassium nitrate
4. A voltmeter is connected in the middle of the wires.

Question 18

Why is potassium nitrate used as a salt bridge over potassium iodide and chloride and sulfate for an Ag+ cell

Answer 18

1. All the other chemicals will react with Ag+ to produce precipitates.

Question 19

Suggest why, when measuring a standard electrode potential, it is necessary to arrange things such that no current is allowed to flow

Answer 19

1. Once current starts to flow, solution concentrations will have changed, so conditions will no longer be standard.

Question 20

Describe what the standard electrode potential values mean

Answer 20

1. The more negative the E value- the greater the tendency to lose electrons and undergo oxidation and the less the tendency to gain electrons and undergo reduction.
2. The more positive the E value- the greater the tendency to gain electrons and undergo reduction and the less the tendency to lose electrons and undergo oxidation.
3. Metals tend to have negative E values and lose electrons
4. Non-metals tend to have positive E values and gain electrons.

Question 21

Describe how to set up cells to measure standard cell potentials Ecells

Answer 21

1. Prepare two standard half-cells using standard conditions
2. Connect the metal electrodes of the half-cells to a voltmeter using wires.
3. Prepare a salt bridge by soaking a strip of filter paper in KNO3
4. Connect the two solutions of half-cells with a salt bridge
5. Record the standard potential from the voltmeter
6. The cell potential is the potentials difference between the two half-cells

Question 22

Write the equation to find the standard cell potential from the standard electrode potentials

Answer 22

1. Ecell= E(reduction)-E(oxidation)

2. Measured in Volts- remember to have capital letter

Question 23

Suggest two reasons why, even when half-cells are correctly connected together, the predicted result may not take place.

Answer 23

1. The activation energy may be very high, so the rate of reaction is too slow to be observed
2. The actual conditions (often concentration) used may not be standard, so the predictions are no longer valid (the half reactions concerned are reversible equilibria so their position and hence the electrode potential will change.)
3. Need to be aqueous equilibria but many reactions take place that are not aqueous.

Question 24

How should you answer questions about reactions feasibility

Answer 24

1. Talk about the E values and talk about them as being more positive or more negative than each other.
2. Use this to say whether species will oxidise or reduce each other.
3. If the Ecell value is negative then the reaction is not feasible
4. The more positive E value species will oxidise the more negative species.

Question 25

Explain whether Fe2+ can oxidise magnesium metal and write the equation.
Fe2+ 2e-↔Fe E= -0.44
Mg2+ + 2e- ↔ Mg E= -2.37

Answer 25

1. Fe2+ has a more positive E value than Mg2+/Mg
2. Therefore Fe2+ can oxidise Mg
3. Fe2+ + Mg → Mg2+ + Fe

Question 26

Manganate(VII) ions in acidic solution are often used as an oxidising agent. Sulfuric acid is frequently used. Explain why it is not a good idea to use hydrochloric acid in such solutions.

Answer 26

1. Cl2/Cl- has a more negative E value than MnO4- and H+/ Mn2+
2. Therefore MnO4- ions in acid solution can oxidise the Cl- ions to Cl2 gas, which is toxic
3. Also, MnO4- will have been used to oxidise the Cl- ions rather than what it is supposed to be oxidising