Electrode potential

Question 1

Oxidation agent

Answer 1

Takes electrons from the species being oxidised. The oxidising agent contains the species which is being reduced

Question 2

Reducing agent

Answer 2

Adds electrons to the species being reduced, contains the species which is being oxidised

Question 3

Writing a redox equation from half equations

Answer 3

You have to balance the electrons so that there is the same number each side. You can also cancel species which are on both sides of the equation. For the full redox equation there should not be any electrons

Question 4

Writing a redox equation from oxidation numbers

Answer 4

You work out how much the substances are being oxidised or reduced by writing down the change in oxidation numbers. You then balance the species which have changes oxidation numbers, so that the increase in oxidation number is the same as the decrease in value. You then balance any of the remaining atoms

Question 5

Writing a half equation

Answer 5

The change in oxidation number signifies the addition or deletion of electrons, for example if the oxidation number goes down by 5 then 5 electrons were added to the original species

Question 6

Redox titrations using Fe+2 / MnO4-

Answer 6

A standard solution of KMnO4- is added to a burette. Using a pipette measure out 25 cm of iron sulphate, an excess of dilute sulphuric acid is also added. The end point of the titration is when there is a pink colour which indicates an excess of potassium manganate. You repeat titration till you get concordant results which are within 0.1 cm of each other.

Question 7

How can you use titration to analyse the percentage purity of an Iron compound

Answer 7

You calculate the number of moles of MnO4- that reacted. You determine the number of moles of Iron which react with one mole of MnO4-, which is 5 you then times this by the moles of MnO4-. You then scale it up for the 150 soloution and times by ten. You then devide the expected mass of the impure substance by tits actual mass, by multiplying the mr by the moles

Question 8

X is titrated against Y

Answer 8

Y is in the burette, X is in the flask

Question 9

How do you determine how many H2O molecules are in a hydrated substance using titration

Answer 9

You determine the moles of the hydrated substance which reacted. You then divide the mass of this substance by its moles to find the mr. You then takeaway the mr of the unhydrated substance. The leftover mr is the mr of the water molecules so should be divided by 18 to find the water crystaline

Question 10

Iodine / thiosulfate redox titrations

Answer 10

Add a standard solution of Na2S2O3 to a burette. Prepare a solution of the oxidising agent agent and use a pippette to transfer 25 cm to a burette. Then add an excess of potassium iodide. During titration the yellow brown colour will begin to fade, once it has become a pale straw colour you should add a starch indicator. This forms a blue black colour which disappears when the titration ends. This allows for a more clear end point

Question 11

Why is the Iodine / thiosulphate titration useful

Answer 11

Can be used to obtain information about different oxidising agents, for example ClO- and Cu+2

Question 12

How many of moles of chlorate react with iodine which then reacts with thiosulphate in order to work out the concentration of bleach

Answer 12

1 mole of ClO- produces 1 mole of I2 which reacts with 2 moles of S2O3-2. So once you have worked out the moles of thiosulphate you divide it by 2 to calculate the moles of ClO- which reacted. In order to find out the concentration you would then work out the number of moles per dm-3

Question 13

Analysis of brass which contains copper by titrating it with thiosulphate

Answer 13

2 moles of Cu+2 reacts with 1 mole of iodine which reacts with 2 moles of thiosulphate. So 1 mole of Cu+2 is equivalent to 1 mole of S2O3-2. You then find the moles of Cu+2 and times it by its mr to find its mass. You then divide mass of copper by mass of brass to find the percentage composition of copper in brass. The other metal in brass is zinc.

Question 14

Procedure of analysing brass

Answer 14

The sample of brass is reacted with nitric acid to form a solution of Cu+2 and Zn+2 ions. The solution is then neutralised. Excess KI(aq) is added, then Cu+2 ions react with the I- ions forming I2. The iodine is then titrated with sodium thiosulphate.

Question 15

Standard electrode potential

Answer 15

The tendency for a substance to be reduced and gain electrons

Question 16

How standard electrode potential measured

Answer 16

Using a half cell which contains H2 and a solution containing H+ ions which contains a platinum electrode. This is connected to another half cell containing the substance which is to be measured and its ions. The two electrodes are connected by a wire to allow the continued flow of electrons. The two soloutions are connected by a salt bridge which allows the ions to flow.

Question 17

Example of salt bridge

Answer 17

Filter paper soaked in aqueous potassium nitrate KNO3 (aq)

Question 18

Standard conditions used in a half cell

Answer 18

Solutions have a concentration of 1 mol dm-3. The temperature is 298k, the pressure is 100 kPa

Question 19

Standard electrode potential of a hydrogen electrode

Answer 19

0V rether a substance has a positive or negative electrode potential shows its relative tendency to gain electrons when compared to the hydrogen half cell

Question 20

Half cell

Answer 20

Contains the chemical species present in a redox half-equation

Question 21

Why use a high resistance volt meter

Answer 21

The voltmeter needs to be of very high resistance to stop the current from flowing in the circuit. In this state it is possible to measure the maximum possible potential difference (E).

Question 22

Advantages of fuel cells

Answer 22

Less pollution and less CO2. (Pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2). Greater efficiency

Question 23

Disadvantages of fuel cells

Answer 23

Its not very safe to store and transport hydrogen. Toxic substances are used in production.

Question 24

Metal / metal ion half cell

Answer 24

A metal rod is dipped into a solution of its aqueous metal ion. For example Cu+2(aq) and Cu(s)

Question 25

An ion/ion half cell

Answer 25

Consists of the same element in different oxidation states. For example Fe+2 and Fe+3. As there is no solid metal to transport the electrodes a platinum electrode is used.

Question 26

The more negative the E value

Answer 26

The greater the tendency of the substance to loose electrons and undergo oxidation, the reaction is more likely to be on the left.

Question 27

The more positive the E value

Answer 27

The greater the tendency of a substance to gain electrons and be reduced, the reaction is more on the right

Question 28

E cell

Answer 28

The difference between the electrode potentials of the two cells

Question 29

Feasability of redox equations

Answer 29

If the E value of the reduced substance take away the E value of the oxidised substance is a positive number then the reaction is feasible. The reduced substance must have a high value then the oxidised substance. So for example Cr+3 will not react with a substance with a less positive E value

Question 30

Limitations of predictions using E values

Answer 30

The reaction may have a large activation energy, resulting in a slow rate. The concentration of the products or the reactant may not be at 1 mol dm-3. For example if the concentration of Zn+2 is greater then the equilibrium will shift to the right, removing electrons from the system and making the electrode potential less negative. The actual conditions used in the reaction may be different from the standard conditions. Standard electrode potentials only apply to aqueous equilibria, many reactions take place that are not aqueous.

Question 31

Primary cells

Answer 31

Are not rechargeable and are designed to be used only once. The electrical energy is produced by oxidation and reduction at the electrodes. However, the reaction can not be reversed and the batteries have to be thrown out once the chemicals are used up. Used in wall clocks.

Question 32

Secondary cells

Answer 32

They are rechargable, the cell reaction can be reversed during recharging to regenerate the chemicals. Used in car batteries and laptops.

Question 33

Lithium batteries

Answer 33

Secondary cells which have a very high energy density meaning that the battery can be small. The material is also flexible. Also means that the batteries do not have to be thrown out every time they are used. However, there are limitations as they become unstable at high temperatures and can catch on fire. Care must be taken with recycling as lithium is a very reactive metal, its also toxic

Question 34

Fuel cells

Answer 34

They use the energy from the reaction of a fuel with oxygen to create a voltage. The fuel does not have to be recharges, the most common fuel is hydrogen. They produce no CO2. The redox system is therefore between oxygen and hydrogen.