chemistry – form 4 page 21 ms. r. buttigieg 8.1 the … 4 chem chapter 2.pdf · chemistry –...

Chemistry – Form 4 Page 21 Ms. R. Buttigieg

8.1 The Ionic Theory. What you should Learn: � Writing formulae for cations and anions � Writing ionic formulae for ionic substances � Representing the change from atoms/molecules to ions, and vice/versa by means of ionic half

equations.

Ionic compounds are compounds containing metal atoms and non-metal atoms. Sodium

chloride and copper(II) sulphate are both ionic compounds.

Working out the formula

Although ions are charged particles, ionic compounds are neutral overall. You need to

have equal numbers of positive charges and negative charges in an ionic compound. The

diagrams show how you can work out the formulae for sodium chloride and iron(III) oxide

from the charges carried by their constituent ions

• Always write the symbol for the positive ion first and do not write any charges in

the final formula. So, NaCl is correct but ClNa and NaCl- are wrong.

• Compound ions, such as OH-, contain more than one element. If you need two or

more compound ions in your formulae, put brackets around them.

Ionic Equations

If we consider the reaction of sodium hydroxide and hydrochloric acid:

NaOH (aq) + HCl (aq) � NaCl (aq) + H2O (l)

These are actually made of ions (except water), so this equation can be written:

Na+ OH- (aq) + H+ Cl- (aq) � Na+ Cl- (aq) + H2O (l)

The simplest equation you can have is: OH- + H+ �� H2O

An equation shows

change, so anything which

appears on both sides

can be removed.

Chemistry – Form 4 2 Ms. R. Buttigieg

Consider the reaction between silver nitrate and sodium chloride. Normally, this is written as,

AgNO3(aq) + NaCl(aq) � NaNO3(aq) + AgCl(s)

Now let’s separate the equation out and into its ions.

Ag+(aq) + NO3-(aq) + Na

+(aq) + C l

-(aq) � Na+(aq) + NO3

-(aq) + AgCl(s)

You can see that the ions NO3-(aq) and Na

+ appear on both sides of the equation. They have

not changed and so have not reacted at all. Sometimes these are called spectator ions. So, the ionic equation reduces the chemical equation to just showing the ions that are actually

participating in the reaction.

In this case the ionic equation is

Ag+(aq) + C l-(aq) � AgCl(s)

The positive ions come from atoms that have lost electrons and the negative ions form

atoms that have gained electrons.

Make sure there is equal charge on each side.

In terms of ionic equations, you need to make sure that the charges on the ions

balance. There are easy rules to follow:

All group 1 ions = +

All group 2 ions = 2+

All group 3 ions = 3+

All group 5 ions = 3-

All group 6 ions = 2-

All group 7 ions = -

Don’t forget that group 4 and 0 don’t form ions and that the transition metals may form

more than one type of ion.

Once you know what charge the ion carries, the rest is easy.

For example to make calcium chloride you need calcium ions and chlorine ions.

Calcium is in group 2 so is Ca2+

Chlorine is in group 7 so is Cl-

The number of charges need to balance (cancel each other out) so one Ca2+ needs

2 Cl-. Therefore calcium chloride would be CaCl2.


Writing ionic Equations

An ionic equation tells you what is really happening. To write an ionic equation:

1. First write the equation

2. Then write the equation again: the ionic compounds write them in ion form,

except when the ionic compound is a precipitate.

3. Cross out the “spectator ions”: i.e. the ions which are common on both sides

of the equation. These ions, didn’t take part in the reaction since they

remained the same.

4. Write the ions, molecules and atoms which remain. This is the ionic equation.

2 FeCl2 (aq) + Cl2 (aq) � 2 FeCl3 (aq)

______________________________________________________________________________

______________________________________________________________________________

Some Ionic Equations you should know Metal Hydroxide/Alkali with an Acid H+ (aq) + OH— (aq) � H2O (l)

(also known as neutralization reaction)

Metal Oxide with an Acid 2H+ (aq) + O- -(aq) � H2O (l)

Metal with an Acid Zn (s) + 2H+ (aq) � Zn++ (aq) + H2 (g)

Displacement Reactions

Metal displacing less reactive metal Zn (s) + Cu++ (aq) � Zn++ (aq) + Cu (s)

Metal displacing hydrogen from an acid Mg (s) + 2H+ (aq) � Mg++ (aq) + H2 (g)

Halogen displacing less reactive halogen Cl2 (g) + 2Br- (aq) � 2Cl- (aq) + Br2 (g)

Carbonate with an Acid 2H+ (aq) + CO3- -(s) � H2O (l) + CO2 (g)

Precipitation Reaction Ag+ (aq) + Cl- (aq) � AgCl (s)

Write ionic equations for the following reactions (from JL 2003) (6 marks)

a) NaCl (aq) + AgNO3 (aq) � NaNO3 (aq) + AgCl (s)

_____________________________________________________________________________________

_____________________________________________________________________________________

b) MgO (s) + 2 HNO3 (aq) � Mg(NO3)2 (aq) + H2O (l)

_____________________________________________________________________________________

_____________________________________________________________________________________

See C4U pg. 151, 305 GCSE Chem pg. 99


c) CaCO3 (s) + 2 HCl (aq) � CaCl2 (aq) + CO2 (g) + H2O (l)

_____________________________________________________________________________________

_____________________________________________________________________________________

d) (i) K2CO3 + 2HCl → 2KCl + H2O + CO2 (i.e. an acid on a carbonate)

_____________________________________________________________

_____________________________________________________________

(ii) NaOH + HNO3 → NaNO3 + H2O (i.e. neutralisation)

_____________________________________________________________

_____________________________________________________ (4 marks)

Electron-half-equations

What is an electron-half-equation?

When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is:

You can split the ionic equation into two parts, and look at it from the point of view of the magnesium and of the copper(II) ions separately. This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them.

These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing!

The reaction between chlorine and iron(II) ions (a metal and non-metal)

Chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions.

You start by writing down what you know for each of the half-reactions. In the chlorine case, you know that chlorine (as molecules) turns into chloride ions:

The first thing to do is to balance the atoms that you have got as far as you possibly can:

Now you have to add things to the half-equation in order to make it balance completely.


All you are allowed to add are:

• electrons • water • hydrogen ions (unless the reaction is being done under alkaline conditions - in which

case, you can add hydroxide ions instead)

In the chlorine case, all that is wrong with the existing equation that we've produced so far is that the charges don't balance. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. That's easily put right by adding two electrons to the left-hand side. The final version of the half-reaction is:

Now you repeat this for the iron(II) ions. You know (or are told) that they are oxidised to iron(III) ions. Write this down:

The atoms balance, but the charges don't. There are 3 positive charges on the right-hand side, but only 2 on the left. You need to reduce the number of positive charges on the right-hand side. That's easily done by adding an electron to that side:

Combining the half-reactions to make the ionic equation for the reaction What we've got at the moment is this:

It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Allow for that, and then add the two half-equations together.

But don't stop there!! Check that everything balances - atoms and charges.


Consider the following list of metals. Zn, Mg, Cu, Ag, Fe, Al.

(a) (i) Rewrite this list of metals, starting with the most reactive.

____________________________________________________ (2 marks)

(ii) Suggest the two metals between which hydrogen could be placed.

______________________ and _______________________ (1 mark)

(b) Complete the following statements by inserting the name of one of the metals given above.

(i) ______________________ cannot displace magnesium but it displaces all the other metals

form solutions of their salts.

(ii) ______________________ can displace silver but cannot displace the other metals. (2)

(c) Some metals, for example calcium, can also displace hydrogen from dilute hydrochloric acid.

The equation for the reaction is. Ca + 2HCl � CaCl2+ H2

(i) Write an ionic equation for this reaction (omitting spectator ions).

___________________________________________________ (2 marks)

(ii) Write the ionic half equations for this reaction.

___________________________________________________ (2 marks)

(iii) Give a reason why this displacement reaction is also a redox reaction.

___________________________________________________ (1 mark)

2 Mg (s) + O2 (g) -------> 2 MgO

1) What is happening to magnesium in this reaction? Answer: Magnesium is losing electrons to become a positive ion . This is the process of oxidation.

The reaction we can write for this process is called the oxidation half - reaction.

Mg (s) --------> Mg 2+ + 2 e - ( electrons )

Since two magnesium atoms are reacting a total of 4 electrons are lost by the magnesium

2) What is happening to the Oxygen in this reaction? Answer: Oxygen is gaining electrons to become a negative ion. This is the process of reduction.

The reaction we write for this process is called the reduction half- reaction.

O (g) ----------> O 2- + 2e –Since oxygen is a diatomic molecule two oxygen atoms are undergoing reduction and a total of 4 electrons are gained 1

1 Note: The total number of electrons lost in oxidation = the total number gained in reduction. This is true of all redox reactions


This question is about oxidation and reduction. Say which substances are oxidized and which are reduced in each equation. 1. C(s) + H2O (g) � CO(g) + H2(g) _____________________________________________

2. C(s) + PbO (s) � CO(g) + Pb(s) _____________________________________________

3. Ca(s) + Cl2 (g) � CaCl2(s) ___________________________________________________

4. 2Ag+(aq) + Mg (s) � 2Ag+(s) + Mg2+(aq) ______________________________________

5. Cl2(g) + 2KBr (aq) � 2KCl(aq) + Br2(g) _______________________________________

For the same equations state the oxidizing agent and the reducing agent. 1. C(s) + H2O (g) � CO(g) + H2(g) _____________________________________________

2. C(s) + PbO (s) � CO(g) + Pb(s) _____________________________________________

3. Ca(s) + Cl2 (g) � CaCl2(s) ___________________________________________________

4. 2Ag+(aq) + Mg (s) � 2Ag+(s) + Mg2+(aq) ______________________________________

5. Cl2(g) + 2KBr (aq) � 2KCl(aq) + Br2(g) _______________________________________

Oxidation Number: The charge an atom has, or appears to have, when the electrons of the compound are counted in accordance with a set of rules.

1. Free elements are assigned an oxidation state of zero. 2. The sum of the oxidation states of all that atoms in a species must be equal to

the net charge on the species. 3. The alkali metals (Li, Na, K, Rb, and Cs) in compounds are always assigned

an oxidation state of +1. 4. Fluorine in compounds is always assigned an oxidation state of -1. 5. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and Cd in

compounds are always assigned an oxidation state of +2. 6. Hydrogen in compounds is assigned an oxidation state of +1. 7. Oxygen in compounds is assigned an oxidation state of -2. 8. Halogen in compounds is assigned an oxidation state of -1.

Application of these rules to a compound must occur in the order given.


What is the oxidation number of . . .

1) N in NO3¯ _________________________________________________________

2) C in CO32¯ _________________________________________________________

3) Cr in CrO42¯ _________________________________________________________

4) Cr in Cr2O72¯ _________________________________________________________

5) Fe in Fe2O3 _________________________________________________________

6) Pb in PbOH+ _________________________________________________________

7) V in VO2+ _________________________________________________________

8) V in VO2+ _________________________________________________________

9) Mn in MnO4¯ _________________________________________________________

10) Mn in MnO42¯ _________________________________________________________

Reactivity Series and Reactions - SUMMARY

Metals Reaction with cold water

Reaction with steam

Reaction with acid

Reduction of Heated Metal oxide by hydrogen

Potassium Sodium Lithium Calcium

These react

Metal (s) + Water (l) �

Metal Hydroxide (aq) +

Hydrogen (g)

React with steam

Metal (s) + Steam

(g) � Metal

hydroxide (aq) +

Hydrogen (g)

Magnesium Reacts very slowly. Form Hydroxide and

Hydrogen

Aluminium Zinc

Are not reduced

Iron

Lead

React.

Metal (s) + Acid (aq) � Salt (aq) + Hydrogen (g)

Aluminium is slow to react as it is protected by an

aluminium oxide layer.

These all displace hydrogen from acid as above it.

Hydrogen Copper

Reduced. Hydrogen is passed over the

heated oxide. Metal oxide +

Hydrogen �� Metal (s) + Steam

The reaction with iron is reversible

Silver

Do not react

Do not react

Do not react

Silver oxide is reduced to silver

even by heat alone


8.2 Electrolysis

� Electricity is a flow of charge. � In fact current is the amount of charge flowing per second. � In metals it is the electrons which flow. In solutions/melts of ionic compounds, it is the ions

which carry the current. � In electrolysis we use a direct current (d.c.) supply. Because if the polarity changes many times

a seconds, the ions are neither attracted towards on electrode or the other. � In a d.c. supply, electron flow is from the negative to the positive. � Conventional current flow is from the positive to the negative.

Metals and graphite are good conductors, because they have free electrons.

Ionic compounds when molten or dissolved in water also conduct electricity because they have free ions. Moreover they are decomposed in the process.

� So they are called electrolytes. � The chemical change brought about by electricity is called electrolysis. � The positive electrode is called anode. � The negative electrode is the cathode.

Positive ions like Al3+ and Cu2+ move to the negative electrode (cathode), and negative ions like Cl- and O2- move towards the positive electrode (anode).

HW: Chemistry for You pg. 116 no 1, pg. 117 no. 5, 6

At the cathode the half equation: Cu2+ (aq) + 2e � Cu (s) At the anode the half equation is: 2Cl- (aq) � Cl2 (g) + 2e The overall equation is: CuCl2 (aq) � Cu (s) + Cl2 (g)


Complete the following:

The positive electrode is called the __________________ and the negative electrode is called the

__________________. The chemical change brought about by electricity is called

__________________. __________________ are substances which conduct electricity and are

decomposed in the process. Only __________________ compounds when

__________________ or ____________________________________ conduct electricity and

are decomposed in the process. The ion which is attracted to the cathode is called the

__________________ and the ion which is attracted to the anode is called the

__________________.

Electrolysis of Molten Lead Bromide

The Lead (II) ion, Pb++ is attracted towards the cathode (-). There it is changed to a lead atom.

Pb++ (l) + 2e � Pb (l) Half equation of what happens at the cathode

The Bromide ion, Br- is attracted towards the anode (+).

2Br- (l) - 2e � Br2 (g) Half equation of what happens at the anode

The overall equation is:

PbBr2 (l) � Pb (l) + Br2 (g)


Electrolysis of Potassium Chloride.

Potassium chloride must be heated until it is molten before it will conduct electricity.

Electrolysis separates the molten ionic compound into its elements.

The reactions at each electrode are called half equations.

2K+ + 2e- 2K (potassium metal at the (-)cathode). 2Cl- - 2e- Cl2 (chlorine gas at the (+)anode).

Potassium ions gain electrons (reduction) to form potassium atoms. Chloride ions lose electrons (oxidation) to form chlorine atoms. The chlorine atoms combine to form molecules of chlorine gas.

The overall reaction is

2K+Cl-(l) 2K(s) + Cl2(g)

Label the diagram:

Write half equations for what is happening At the cathode and anode: At anode ______________________ At cathode _______________________ Overall equation ____________________________________

Molten zinc chloride


Electrolysis of Ionic Compounds Dissolved in Water Aqueous solutions of ionic compounds also conduct electricity because they have

free ions.

Some molecules of water break up into hydrogen ions, H+ and hydroxide ions, OH- .

So sodium chloride solution contains these four ions: Na+, Cl-, H+ and OH-. The sodium ion and hydrogen ion migrate to the cathode but only one ion is discharged. The hydroxide ion and chloride ion migrated to the anode but only one is discharged.

a) If we are electrolyzing concentrated sodium chloride solution, and the chloride ion and hydroxide ion are attracted to the anode, it is the chloride ion, which is discharged, because it is more concentrated.

b) The hydrogen ion is preferred over the sodium ion at the cathode, because the ion of the less reactive element is discharged.

� An atom of a reactive element easily becomes an ion. � So an ion of a reactive element is very difficult for it to become an atom. � An atom of an unreactive element becomes an ion with difficulty. � So an ion of an unreactive element becomes an atom with ease. � So the copper(II) ion and silver ion are discharged instead of the hydrogen ion.

The less stable ion is discharged. The more stable ion remains an ion. So the hydroxide ion is discharged instead of a nitrate or a sulphate ion. c) Sometimes the electrode dictates what happens. So a copper anode in copper (II) sulphate solution is eroded and neither the sulphate nor the hydroxide ion are discharged. A mercury cathode prefers the sodium ion over the hydrogen ion.

Electrolysing Acidified Water We acidify water with sulphuric acid, so that the products are

the elements of water: hydrogen and oxygen in the proportion

of 2:l. We do not use hydrochloric acid, because the product would be chlorine instead of oxygen. We call it the electrolysis of water because the products are the elements of water: hydrogen and oxygen in the proportion of 2:l. We use a Hoffman voltmeter to collect the gases given off. At the cathode (-): 4H'(aq) + 4e � 2H2(g) At the anode (+): 4 OH- (aq) - 4e � 2H2O (l) + O2 (g) The acid becomes more concentrated.


Electrolysing Copper (II) Sulphate Solution Using Carbon Electrodes. Ions present: Cu++, SO4

2-, H+ and OH- At the cathode, the copper(II)ion is discharged since it is the ion of the less reactive element and orange brown copper is formed. At the anode, the hydroxide ion is discharged because it is less stable and oxygen gas is given off. At the cathode (-): Cu++ (aq) + 2e � Cu (s) At the anode (+): 4 OH- (aq) - 4e � 2H2O (l) + O2 (g) The blue solution fades in colour. The solution becomes acidic since the hydroxide ions are decreasing and so there are more hydrogen ions present.

Electrolysing Copper (II) Sulphate Solution Using a Copper Anode. Ions present: Cu++, SO4

2-, H+ and OH- At the cathode, the copper(II)ion is discharged since it is the ion of the less reactive element. At the anode, neither the sulphate nor the hydroxide ion is discharged. But the copper atoms become copper ions and the anode is eroded. At the cathode (-): Cu++ (aq) + 2e � Cu (s) At the anode (+): Cu (s) - 2e � Cu++ (aq)

The solution remains blue because the number of copper ions is decreasing at the same rate as it is increasing. So the amount of copper ions remains the same.

This set-up, i.e. having a copper anode in copper (II) sulphate solution is used in industry in electroplating and in the purification of copper.

Electroplating This means covering a metal object with another metal using electricity

To electroplate with copper,

the metal object is the cathode;

we use copper(II)sulphate solution and a copper anode.

Purification of Copper Copper, which is extracted, is not pure enough to be used for electrical wiring. Very pure copper is needed to make electrical wires so that it has very low resistance.

� To purify copper, copper(II)sulphate solution is used,

the impure copper is the anode

� and a thin strip of pure copper is placed as the cathode. � The anode is eroded and becomes copper(II)ions.

� The impurities fall to the bottom.

� Pure copper collects at the cathode.


Purification of copper

When copper is made from sulphide ores by the first method above, it is impure. The blister copper is first treated to remove any remaining sulphur (trapped as bubbles of sulphur dioxide in the copper - hence "blister copper") and then cast into anodes for refining using electrolysis.

Electrolytic refining The purification uses an electrolyte of copper(II) sulphate solution, impure copper anodes, and strips of high purity copper for the cathodes. The diagram shows a very simplified view of a cell.

At the cathode, copper(II) ions are deposited as copper.

At the anode, copper goes into solution as copper(II) ions.

For every copper ion that is deposited at the cathode, in principle another one goes into solution at the anode. The concentration of the solution should stay the same. All that happens is that there is a transfer of copper from the anode to the cathode. The cathode gets bigger as more and more pure copper is deposited; the anode gradually disappears. In practice, it isn't quite as simple as that because of the impurities involved. What happens to the impurities? Any metal in the impure anode which is below copper in the electrochemical series (reactivity series) doesn't go into solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with any unreactive material left over from the ore. The anode sludge will contain valuable metals such as silver and gold. Metals above copper in the electrochemical series (like zinc) will form ions at the anode and go into solution. However, they won't get discharged at the cathode provided their concentration doesn't get too high. The concentration of ions like zinc will increase with time, and the concentration of the copper(II) ions in the solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion formed. (See the next note if you aren't sure about this.) The copper(II) sulphate solution has to be continuously purified to make up for this.


Industrial Production of Sodium Hydroxide

� Electrolysis of Brine (concentrated sodium chloride solution) is done. � Hydrogen is produced at the cathode and chlorine at the anode. � Sodium hydroxide solution is left.

The cell (the place where the electrolysis takes place) is constructed so that the chlorine and sodium hydroxide solution do not mix. Otherwise they would react.

Calculating the Amount of Substance Produced.

1 mole of electrons carry 96, 500 coulombs of charge and is equal to 1 faraday (F).

Current, I = Charge, Q

Time,t

e.g. What is the mass of

aluminium produced when 20,000

amperes pass for 10 hours?

Q = I x t = 20,000 x (10 x 60 x 60) = 7.2 x 108 C

Al +++ + 3e �� Al 3 moles of electrons produce 1 mole of aluminium � Convert moles of electrons to coulombs, � and mole of aluminium to grams, � because you know the coulombs that passed � and you need to know the mass of aluminium

3 x 96500 C produce 27g

289500 C produce 27g

7.2 X 108 C produce ? 7.2 X 108 x 27 = 67150g = 67.15 kg. 289500

So much electricity for so little aluminium produced!

That's why aluminium is expensive.

Quantity Symbol SI unit

Current I Ampere (A)

Charge Q Coulomb (C)

Time t second


3. The diagram below shows the electrolysis of concentrated sodium chloride solution

in the laboratory.

If a few drops of universal indicator are added to the solution before electrolysis starts, the

indicator is green.

As electrolysis happens, the indicator turns blue around the cathode while around the anode first

it turns red, then colourless.

(a) (i) Name the gases Y and Z.

___________________________ _________________________(2 marks)

(ii) Explain, in terms of the ions present in solution, what causes the indicator to go blue at

the cathode.

_________________________________________________________

_________________________________________________________(2 marks)

(iii) Suggest a reason for the colour changes at the anode.

_________________________________________________________(1 mark)

(b) Name a compound produced on an industrial scale by the electrolysis of

sodium chloride solution. ______________________________________________________

(1 mark)

carbon electrodes

bubbles of gas Y

bubbles of gas Z

sodium chloride solution


Cobalt has been used for electroplating in a similar way to copper. Complete the labelling

of the diagram below, which could be used to electroplate a spoon with cobalt.

Rewrite the following equations as an ionic equation, (omitting spectator ions).

(a) (i) Ca + 2HCl → CaCl2 + H2

_______________________________________________________

(ii) Cl2 + 2NaBr → 2NaCl + Br2

________________________________________________________(4 marks)

(b) Explain, in terms of electrons, why

(i) calcium is oxidised in equation (a)(i) above.

_____________________________________________________

(ii) chlorine is reduced in equation (a)(ii) above

_____________________________________________________ (2 marks)

Complete this table:

(4 marks)

chemical large-scale use

copper

manufacture of bleaches

nitric acid

construction of aircraft

The object to be plated is

made the _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ containing ions

of the metal being used for plating.

The metal being used for

plating is made the

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .


It is important to keep in mind that when metals are extracted from their ores, the process involved is a reduction process.

Extraction of Aluminium – Chemistry for You pg. 107

Aluminium ore is called bauxite. Bauxite contains aluminium oxide, water, iron oxide and other impurities.

The purified dry ore, called alumina, is aluminium oxide - Al2O3.

At the cathode Al+++ (l) + 3e � Al (l) At the anode 2O-- (l) - 4e � O2 (g)

Work out Chemistry for You pg. 138 number 18, pg. 140 number 23, 24

In the electrolysis of bauxite Al2O3, aluminium is the main product. Calculate the mass of aluminium (R.A.M. = 27) produced if a current of 25A is passed for 30 minutes.

Calculating the volume of gas liberated during electrolysis. (Chem 4U pg. 354,355) Moles of Gas = volume of gas (cm3) At r.t. and pressure 1 mole of any gas takes a volume of 24l 24 000 What volume does 8g of oxygen gas occupy at room temperature and pressure?

Electrolysis of the alumina/cryolitesolution gives aluminium at the cathode and oxygen at the anode.

The overall reaction is

aluminium oxide aluminium + oxygen. 2Al2O3(l) 4Al(l) + 3O2(g)

This process is very expensive as a lot of electricity is needed.


Answer the following: A pure specimen of molten lead (II) bromide was electrolysed in a suitable apparatus using inert electrodes. a. Give the formulae of the ions present in the liquid. (i) _____________ (ii) _____________

b. What changes would you expect to SEE at each of the electrodes?

Negative electrode (cathode) ___________________________________________________

_____________________________________________________________________________

Positive electrode (anode) ___________________________________________________

_____________________________________________________________________________ c. Write ionic equations to show the changes taking place at each of the electrodes:

Negative electrode (cathode) ___________________________________________________

Positive electrode (anode) ___________________________________________________

d. In such an experiment, a current of 0.2 A was passed thourhg the molten lead (II) bromide

for four minutes.

i. What quantity of electricity is passed?

ii. What would be the mass of the product at the negative electrode?

iii. Assuming all the product liberated at the positive electrode to be in the form of a gas, what volume would it occupy at room temperature and pressure?

(Don’t forget to write the half equations for (ii) and (iii))


8.3 Reactivity Series

This was already discussed in previous sections (make sure its clear) Producing electrical energy from a chemical reaction - limited to the simple cell.

The Fuel Cell – Chem 4U pg. 125 A scientist noted when electrolyzing sulphuric acid that when he switched off the current, the cell itself began producing electricity. So he concluded that if you have a cell, and you feed it oxygen and hydrogen, these are changed to water and produce electricity. What happens in the fuel-cell is the reverse of what happens when you electrolyze water (acidified with sulphuric acid). In electrolysis of water: electrolysis breaks up water into hydrogen and oxygen. In the fuel cell, hydrogen and oxygen are turned to water and produce electricity. At anode: 2H2 (g) � 4H+ (aq) + 4e At cathode: 2H2O (l) + O2 (g) + 4e � 4OH- (aq)

A simple cell - One of the first practical batteries is called the 'Daniel cell' - GCSE pg, 192 imp

o It uses a half-cell of copper dipped in copper(II) sulphate, o and in electrical contact with a 2nd half-cell of zinc dipped in zinc sulphate solution. o The zinc is the more reactive, and is the negative electrode, releasing electrons

because � on it zinc atoms lose electrons to form zinc ions, Zn(s) �� Zn

2+(aq) + 2e

- o The less reactive metal copper, is the positive electrode, and gains electrons from

the negative electrode through the external wire connection and here .. � the copper(II) ions are reduced to copper atoms, Cu2+(aq) + 2e

- �� Cu(s) o Overall the reactions is: Zn(s) + CuSO4(aq) + ZnSO4(aq) + Cu(s)

� or ionically: Zn(s) + Cu2+(aq) + Zn

2+(aq) + Cu(s)

o The overall reaction is therefore the same as displacement reaction, and it is a

redox reaction involving electron transfer and the movement of the electrons

through the external wire to the bulb or voltmeter etc. forms the working electric

current.

Work out GCSE Chemistry pg. 193 no. 1,2. With the help of diagrams, explain the basic steps in the purification of copper. (see GCSE chemistry pg. 88, 89) Work out GCSE Chemistry pg. 92 no. 1-4


Limestone, Quicklime and slaked lime Limestone, chalk and marble are calcium carbonate (CaCO3) � Quicklime is calcium oxide. � Slaked lime is calcium hydroxide � Limewater is calcium hydroxide solution. Calcium hydroxide is slightly soluble in water. a) Limestone itself is used as a building material, to make iron and steel, to neutralize acid soil and in fertilizers. b) Limestone is used to make: i) Calcium oxide by heating calcium carbonate strongly in a lime kiln (pg. 129 C4U) CaCO3 (s) � CaO (s) + CO2 (g) ii) Quicklime than absorbs water to form calcium hydroxide. CaO (s) + H2O (l) � Ca(OH)2 (s) Slaked lime is the cheapest industrial alkali. It is used to make sodium hydroxide, bleaching powder and mortar. It is also used to control soil acidity. Morta r is a mixture of calcium hydroxide, sand and water. Mortar sets by drying out, then hardens by reacting with carbon dioxide. Limewater is a dilute alkaline solution of calcium hydroxide. Carbon dioxide turns limewater milky: carbon dioxide, an acidic oxide, reacts with calcium hydroxide, an alkali to form a salt, calcium carbonate, which is a white solid, insoluble in water.

Ca(OH)2 (s) + CO2 (g) � CaCO3 (s) + H2O (l) If we continue to pass carbon dioxide through the milky limewater, the milkiness disappears. This happens as calcium carbonate changes into soluble calcium hydrogencarbonate.

CaCO3 (s) + H2O (l) + CO2 (g) � Ca(HCO3)2 (aq)

In the laboratory, limestone is changed to quicklime by heating strongly for a long time on a strong Bunsen flame. The appearance doesn’t change.

CaCO3 (s) � CaO (s) + CO2 (g)

When water is added drop by drop onto the quicklime this puffs up and then crumbles to calcium hydroxide powder. A hissing sound is heard, because some of the water is changed to steam. Because the reaction between quicklime and water is exothermic and changes some of the water into steam.

CaO (s) + H2O (g) � Ca(OH)2 (s)

See Chemistry for You pg. 127-134. Work out pg. 134 numbers 1, 3, 4.


Work out the following for the Holidays: GCSE Chemistry pg 94 number 1, 2 (see GCSE chemistry pg. 88, 89), 5, 8a-d. Chemistry for You pg. 117 numbers 3, 4, 8. pg. 141 numbers 26, 27. 1. Underline the correct answer:

1. Aluminium is extracted using (electrolysis, carbon monoxide in the blast furnace) because it is a fairly (reactive/unreactive) metal.

2. Iron is extracted using (electrolysis, carbon monoxide in the blast furnace) because it is a fairly (reactive/unreactive) metal.

3. Aluminium is (more/less) abundant in the earth’s crust than iron. 4. Aluminium is (more/less) expensive than iron due to the extraction process. 2. Complete the following table: Electrolyte Cathode type Anode type Product at

cathode Product at anode

Molten lead (II) bromide

Graphite Graphite

Sodium chloride solution

Graphite Graphite

Sulfuric acid solution

Platinum Platinum

Copper (II) sulphate solution

Platinum Platinum

Copper (II) sulphate solution

Copper Copper

3. Work out the oxidation number of the underlined element, then state whether it has been oxidized or reduced.

PbO2 + 4HCl � PbCl2 + 2H2O + Cl2

Oxidation number __________ __________ The element has been _________________.

ZnO + C � Zn + CO

Oxidation number __________ _________ The element has been _________________.

H2S + Br2 � 2HBr + S

Oxidation number __________ _________ The element has been _________________. Give 2 other ways in which one can determine if an element has been oxidized or reduced.

______________________________________________________________________________

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chemistry – form 4 page 21 ms. r. buttigieg 8.1 the … 4 chem chapter 2.pdf · chemistry –...

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