1.6 redox

Question 1

Dynamic equilibrium

Answer 1

As the reactants get used up, the forward reaction slows down - and as more
product is formed, the reverse reaction speeds up. After a while, the forward
reaction will be going at exactly the same rate as the backward reaction.
The concentration of reactants and products won’t be changing any more. This is called a dynamic equilibrium. A dynamic equilibrium can only happen in a closed system (this just means nothing can get in or out) which is at a constant temperature.

Question 2

Le Chatelier’s principle

Answer 2

If a reaction at equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium will move to counteract the change.

This means you’ll end up with different amounts of reactants and products at equilibrium.

Question 3

Using Le Chatelier’s principle - Changing concentration

Answer 3

If you increase the concentration of a reactant, the equilibrium tries to get rid of the extra reactant. It does this by making more product. So the equilibrium’s shifted to the right. If you increase the concentration of the product, the equilibrium tries to remove the extra product. This makes the reverse reaction go faster - so the equilibrium shifts to the left. Decreasing the concentrations has the opposite effect.

Question 4

Using Le Chatelier’s principle - Changing pressure

Answer 4

Changing the pressure only affects equilibria involving gases. Increasing the pressure shifts the equilibrium to the side with fewer gas molecules. This reduces the pressure. Decreasing the pressure shifts the equilibrium to the side with more gas molecules. This raises the pressure again.

Question 5

Using Le Chatelier’s principle - Changing temperature

Answer 5

Increasing the temperature means adding heat. The equilibrium shifts in the endothermic (positive AH) direction to absorb this heat. Decreasing the temperature removes heat. The equilibrium shifts in the exothermic (negative AH) direction to try to replace the heat. If the forward reaction’s endothermic, the reverse reaction will be exothermic, and vice versa.

Question 6

Following equilibrium reactions

Answer 6

There are some reactions you can do in the lab to follow equilibrium shifts.
e.e.

Question 7

Following equilibrium reactions - conc eg

Answer 7

Question 8

Following equilibrium reactions - temp eg

Answer 8

Question 9

Compromise conditions in industry - Ethanol production

Answer 9

Companies have to think about how much it costs to run a reaction and how much money they can make from it. This means they have a few factors to think about when they’re choosing the best conditions for a reaction.

Ethanol can be made via a reversible reaction between ethene and steam:

C2H4(g) + H20(g) = C2H5OH(g)

AH = -46 kJ mol-1

The industrial conditions for the reaction are:
* a pressure of 60-70 atmospheres,
* a temperature of 300 C,
* a phosphoric acid catalyst.

Because it’s an exothermic reaction, lower temperatures favour the forward reaction. This means that at lower temperatures more ethene and steam are converted to ethanol - you get a better yield. But lower temperatures mean a slower rate of reaction. You’d be daft to try to get a really high yield of ethanol if it’s going to take you 10 years. So the 300 C is a compromise between maximum yield and a faster reaction.

Higher pressure favours the forward reaction, since it moves the reaction to the side with fewer molecules of gas, so a pressure of 60-70 atmospheres is used. Increasing the pressure also increases the rate of reaction. Cranking up the pressure as high as you can sounds like a great idea, but high pressures are expensive to produce. You need stronger pipes and containers to withstand high pressure. So the 60-70 atmospheres is a compromise between maximum yield and minimum expense.

Only a small proportion of the ethene reacts each time the gases pass through the reactor. To save money and raw materials, the unreacted ethene is separated from the ethanol and recycled back into the reactor. Thanks to this, around 95% of the ethene is eventually converted to ethanol.

Question 10

Compromise conditions in industry - Methanol production

Answer 10

Methanol is also made industrially in a reversible reaction:
2H2(g) + CO(g) = CH3OH(g)
ΔΗ = -90 kJ mol-1
Just like ethanol production, the conditions are a compromise between
keeping costs low and yield high. The conditions for this reaction are:
* a pressure of 50-100 atmospheres,
* a temperature of 250°C,
* a catalyst of a mixture of copper, zinc oxide and aluminium oxide.

Question 11

What are redox reactions?

Answer 11

Oxidation is loss of electrons. Reduction is a gain in electrons - OIL RIG
Reduction and oxidation happen simultaneously - hence the term “redox”
reaction.

Question 12

oxidising / reducing agent

Answer 12

An oxidising agent accepts electrons and gets reduced.
A reducing agent donates electrons and gets oxidised

Question 13

Oxidation states

Answer 13

The oxidation state / oxidation number of an element tells you the total number of electrons it has donated or accepted. There are lots of rules for working out oxidation states.

Question 14

Oxidation states rules - elements

Answer 14

Uncombined elements have an oxidation state of 0.
Elements just bonded to identical atoms also have an oxidation state of 0.
The oxidation state of a simple monatomic ion is the same as its charge.

Question 15

Oxidation states rules - compounds

Answer 15

In compounds or compound ions, each of the constituent atoms has an oxidation state of its own and the sum of the oxidation states equals the overall oxidation state which is equal to the overall charge on the ion.

Within a compound ion, the most electronegative element has a negative oxidation state (equal to its ionic charge). Other elements have more positive oxidation states.

Question 16

Oxidation states rules - oxygen & hydrogen

Answer 16

Combined oxygen is nearly always -2, except in peroxides, where it’s -1 (see Figure 6). Combined hydrogen is +1, except in metal hydrides where it is -1 (see Figure 7) and H2, where it’s 0.

Question 17

Finding oxidation states e.g. Find the oxidation state of Zn in Zn(OH)2.

Answer 17

Question 18

Finding oxidation states of elements with multiple

Answer 18

If an element can have multiple oxidation states (or isn’t in its ‘normal’ oxidation state) its oxidation state is sometimes shown using Roman numerals,
e.g. (I) = +1, (II) = +2, (III) = +3 and so on. The Roman numerals are written
after the name of the element they correspond to.

Examples
* In iron(II) sulfate, iron has an oxidation state of +2. Formula = FeSO4
* In iron(III) sulfate, iron has an oxidation state of +3. Formula = Fe2(SO4)3

This is particularly useful when looking at -ate ions. lons with names ending in -ate (e.g. sulfate, nitrate, carbonate) contain oxygen, as well as another element. For example, sulfates contain sulfur and oxygen, nitrates contain nitrogen and oxygen … and so on. But sometimes the ‘other’ element in the ion can exist with different oxidation states, and so form different ‘-ate ions’. You can use the systematic name to work out the formula of the ion.

Examples
* In sulfate(VI) ions the sulfur has oxidation state +6. This is the SO 2- ion.
* In sulfate(IV) ions, the sulfur has oxidation state +4. This is the SO32- ion.
* In nitrate(III), nitrogen has an oxidation state of +3. This is the NO,- ion.

Question 19

Half-equations and redox equations

Answer 19

lonic half-equations show oxidation or reduction (see Figure 1). The electrons
are shown in a half-equation so that the charges balance.

Question 20

Magnesium burns in oxygen to form magnesium oxide:
2Mg + O2-> 2MgO
Write half-equations for the oxidation and reduction reactions
that are part of this process.

Answer 20

Question 21

complicated Half-equations and redox equations

Answer 21