YEAR 4 CHEM PPA 1 NOTES
Qualitative Analysis
General Tips: Be logical and systematic.
1. Things you really have to know
1.1. Tests for Anions (Refer to QA Notes)
1.2. Tests for Cations (Refer to QA Notes)
1.3. Tests for Gases (Refer to QA Notes)
1.4. Action of Heat on Solids
You only need to know metal carbonate = metal oxide + carbon dioxide. :)
2. How to tackle QA questions
2.1. Look for Keywords
✓ Colour of solutions, precipitates, solids
e.g. If it’s blue you know it’s something to do with Cu.
✓ Take note of what reagents are added and know what test it is for
e.g. If you add in acidified silver nitrate to Solution X, you know that you are testing for halides!
✓ Descriptions of Gases
e.g. Turns litmus paper blue/red, bleaching, pungent, poisonous gases (brown = nitrogen dioxide)
2.2. Work backwards and forwards!
✓ Remember all your reactions!
Redox, Displacement, Precipitation, Acid-Metal, Acid-Base, Acid-Carbonate
✓ Remember solubility of compounds (Refer to QA Notes)
✓ TRICKY STUFF (good to know)
Cu, Ag and Au only react with concentrated acids.
This information might help you for QA! (You will see an example in the Sample Questions)
3. Sample Questions
---When aqueous sodium hydroxide is added to P, a solution, a blue precipitate Q is formed.
When the solution of P is added to aqueous sodium hydroxide followed by powdered aluminum, it
releases a gas R upon gentle heating. Gas R turns damp red litmus paper blue.
When precipitate Q is added to a dilute acid, S, a blue solution T is produced. When solution T is added to
acidified barium nitrate solution, a white precipitate V is formed.
Identify P, Q, R, S, T, V.
Tackling the question! You have to be very systematic. Look for keywords (bolded in this case).
STEP 1: From the first line, we see that a blue ppt is formed from added NaOH. This is a test for a cation, and
you know that if it’s blue, Cu is present, and the ppt is Cu(OH)2! :D so, Q is Cu(OH)2, and P contains Cu2+.
STEP 2: Now we move to the second line, and we see that another test is conducted NaOH, and powdered
aluminum. When you add these two together you know this is a test for the nitrate NO3- ion. (powered
aluminium is part of Devarda’s alloy!) :D So, you know that if it’s positive, ammonia gas is produced which
turns damp red litmus paper blue. Hence, from this you can gather that P contains NO3-, and gas R is
ammonia! From Step 1, you know that P contains Cu2+, so P is Cu(NO3)2, copper (II) nitrate.
STEP 3: We now look at the third line. We see that a test for SO4- is done (addition of acidified barium nitrate
solution) with a positive result. Hence, solution T contains the sulphate ion, and the only way for solution T to get
that is via the addition of sulphuric acid to Q. Hence, S is sulphuric acid, T is CuSO4 (cos it’s blue! meaning
Cu is there) and V is BaSO4 (insoluble).
Page 1 of 6
YEAR 4 CHEM PPA 1 NOTES
---You are given 2 identical coins in an experiment. The coins are made of an alloy known to be a mixture of 2
metals.
(i) One coin was completely dissolved in concentrated nitric acid to form a blue solution.
(ii) The second coin was added to an excess of dilute sulphuric acid. A reddish-brown solid and a
colourless solution was formed.
(iii) The mixture from Step (ii) was filtered and aqueous ammonia was added to the filtrate. A white
precipitate was formed which dissolved in excess aqueous ammonia.
Identify the 2 metals used in the coin.
Look for keywords (bolded in this case).
STEP 1: Tune your mind to note that the coins are made of an ALLOY of only 2 METALS. Then look at the
first line --> This is clearly a metal-acid reaction, producing a salt plus water plus hydrogen gas. This means that
BOTH metals in the coin have reacted with the acid. The solution turns blue, meaning copper is present.
Hence, one of the metals in the alloy is copper.
STEP 2. This is the tricky confusing part! What is this reddish-brown solid of which the question speaks?! BUT
look at what I bolded - dilute sulphuric acid. Remember that COPPER only reacts with CONCENTRATED
acid, hence out of the two metals, only one of them reacts, and only copper is left as the REDDISHBROWN SOLID. As it dissolves to form a colourless solution, transition metal ions are NOT present, meaning
the other metal can be Zn, Al, Pb or Ca. :)
STEP 3: This is a cation test on the filtrate containing the other metal, either Zn, Al, Pb or Ca. The results of the
aq ammonia test shows that it is clearly zinc! Hence, the coin is made up of zinc and copper.
Be systematic I cannot stress this enough!! Look for clues and UNDERLINE things that can give a clue to
the identity of the substances. Remember that everything in QA should make sense once you got the right
chemicals :) Just calm down and think through it :)
----------------------------------------------------------------------------------------------------------------------------------
Redox Reactions
1. Basic Definition of Oxidation and Reduction.
1.1.Gain/Loss of Oxygen
Oxidation = Gain oxygen, Reduction = Lose oxygen
Oxidation
Reduction
O
Gain
Loss
H
Loss
Gain
Electrons
Loss
Gain
Ox. State
Increase
Decrease
1.2. Gain/Loss of Hydrogen
Oxidation = Lose hydrogen, Reduction = Gain hydrogen
1.3. Gain/Loss of Electrons (Acronym: OIL RIG)
Oxidation = Loss of Electrons, Reduction = Gain of Electrons
1.4. Increase/Decrease Oxidation State
Oxidation = Increase Oxidation state, Reduction = Decrease Oxidation state
In proving if the reaction is a redox reaction, we usually explain via oxidation state.
2. General Half-Equations (e.g. for an ion X)
2.1. For oxidation: X = X+ + e
2.2. For reduction: X+ + e- = X
Example: Displacement Reaction (2KI + Cl2 ---> 2KCl + I2)
Half-Equations:
(1) 2I- ---> I2 + 2e- (Oxidation = Lose electrons)
(2) Cl2 + 2e- ---> 2Cl- (Reduction = Gain electrons)
3. Oxidizing and Reducing Agents
3.1. Oxidizing Agents:
They cause other substances to be oxidized, while they themselves get reduced.
Hence, the other substance will be a reducing agent! It’s always opposite.
Page 2 of 6
YEAR 4 CHEM PPA 1 NOTES
Examples of oxidizing agents are:
Acidified potassium dichromate (VI)
K2Cr2O7 turns from orange/yellow to green if a reducing agent is present i.e. when it is reduced.
3.2. Reducing Agents:
They cause other substances to be reduced, while they themselves get oxidized.
Aqueous iron (II) ion
Solution turns from green (Fe2+) to brown (Fe3+).
(Sulphur dioxide is a reducing agent too!)
Acidified potassium manganate (VII)
KMnO4 turns from purple to colourless if a reducing agent is present i.e. when it is reduced.
Examples of reducing agents are:
Potassium iodide
Solution turns from colourless to brown.
ALSO NOTE WEIRD
STUFF which go through
disproportionation, which
means it
SIMULTANEOUSLY gets
both reduced and
oxidized!! e.g. H2O2
(hydrogen peroxide)
*** TIP: LOOK OUT FOR THESE OXIDIZING/REDUCING AGENTS! they make your life so much easier so
you don’t waste your time anyhow go and calculate oxidation state to determine what is reduced/oxidized when
these are present :)
4. Question Mix
-Identify the oxidizing and reducing agents in the following equations. Also, describe what you would
expect to observe for each of the reactions.
Basically, if they give you such identification questions there is a standard form you have to follow. You either say
that the substance is an oxidizing/reducing agent as it causes others to get oxidized/reduced, OR the
substance itself gets oxidized/reduced. You then go on to explain why the substance was oxidized/reduced
based on changes in oxidation state. See examples below!
EXAMPLE 1: Fe + H2SO4 ----> FeSO4 + H2
Iron is an reducing agent as it causes concentrated sulphuric acid to be reduced as the oxidation state
of hydrogen decreases from +1 in H2SO4 to 0 in hydrogen gas. OR
Iron is an reducing agent as it is oxidized, the oxidation state of Fe increases from 0 in iron to +2 in iron (II)
sulphate.
NOTE: The agent is always the WHOLE substance, and spell out the whole substance name when you FIRST
mention it. Meaning you can’t say that “Fe is a reducing agent because....”
And so, for oxidizing agent:
Concentrated sulphuric acid is an oxidizing agent as it causes iron to be oxidized, the oxidation state of
Fe increases from 0 in iron to +2 in iron (II) sulphate. OR
Concentrated sulphuric acid is an oxidizing agent as it is reduced, the oxidation state of hydrogen
decreases from +1 in H2SO4 to 0 in hydrogen gas.
Observation: Effervescence will be observed as H2 gas is given off.
Basically effervescence will be observed whenever a gas is the product of the reaction (examples are oxygen,
hydrogen and chlorine gas)
EXAMPLE 2: 2KBr + Cl2 ----> 2KCl + Br2
What you would observe: The solution turns brown
If you see substances like Br2, I2 produced, the solution will change colour! For bromine it will turn brown, for I2
the solution will turn brown too.
Page 3 of 6
YEAR 4 CHEM PPA 1 NOTES
EXAMPLE 3: Cu + AgNO3 ----> Cu(NO3)2 + 2Ag
Explain how the above shows a redox reaction.
Ok so, when they ask you to show a REDOX REACTION and NOT IDENTIFY oxidizing and reducing agents
then make sure your answer isn’t phrased like that!! So it should be phrased like this:
This is a redox reaction as copper is oxidized, with its oxidation state increases from 0 in Cu metal to
+2 in copper (II) nitrate; and silver is reduced as its oxidation state decreases from +1 in silver nitrate
to 0 in Ag metal.
^ No mention of oxidizing/reducing agent ok!!
--A coin made up of an alloy of zinc and copper was submerged in a solution of silver nitrate. Over a period of
time, the solution turned pale blue and a layer of grey solid was formed on the coin. Write a series of half
equations to represent the redox reactions happening. Hence, or otherwise, explain why the solution turned
blue. (Past year paper question!)
If the solution turns pale blue, it means that Cu2+ ions must somehow be in the solution right? It means that
thus, Cu has been oxidized to Cu2+, so that’s one half equation you have: Cu (s) ----> Cu2+(aq) + 2eNext you see that a layer of grey solid is formed, and in this case, it has to be silver! So there’s your second halfequation: Ag+(aq) + e- ----> Ag (s)
So you might be asking -- what happens to the zinc! yes, the zinc will react accordingly as well, and you should
write that equation down too: Zn (s) ---> Zn2+ (aq) + 2eHowever, you see that the above reaction does nothing to the colour! As Zn2+ is not a transition metal cation, in
solution form it is colourless. Hence, the solution turns blue as copper metal is being oxidized to Cu2+ and
discharged into the solution, thus turning the solution blue. :)
----
Electrolysis
Because I am lazy, and I will just look at questions haha! :D Below are just some quickies you just need to bear
in mind for electrochem :) I assume the basic concept y’all should know so shan’t waste time haha.
1. Types of Cells
The principle of electrolysis forms 2 types of cells - electrolytic cells (means, you SUPPLY electricity to this cell to
make it work) and electrochemical cells (PRODUCES electricity by means of a spontaneous redox reaction).
Electrochemical Cell
Electrolytic - you need a battery, and your electrodes can be inert.
In an electrolytic cell, the ANODE is POSITIVELY CHARGED (attracts anions!). The CATHODE is
NEGATIVELY CHARGED (attracts cations!)
Page 4 of 6
YEAR 4 CHEM PPA 1 NOTES
Electrochemical - you don’t need a battery cos’ your aim is to make one, your electrodes CANNOT
be inert. There’s also a salt bridge: Function --> Close the circuit, maintain electron/ion flow and and
balance the charges on both sides. (usually made up on potassium sulphate/sodium sulfate)
In an electrochemical cell, the ANODE is the NEGATIVE TERMINAL, where the MORE REACTIVE METAL is
placed. The CATHODE is the POSITIVE TERMINAL where the LESS REACTIVE METAL is placed. Yes,
OPPOSITE from electrolytic cell!
THIS IS BECAUSE (I got this off a website as I couldn’t explain it properly myself haha)
The anode of an electrolytic cell is positive (cathode is negative), since the anode attracts anions from the
solution. However, the anode of a electrochemical cell is negatively charged, since the spontaneous oxidation
at the anode is the source of the cell's electrons or negative charge. (can? quite clear hehe :>)
The voltage (how much current) produced by the electrochemical cell is dependent on reactivity! So if the
distance between two metals in the reactivity series is very large, more current is produced. E.g. if you put Mg
and Ag as your electrodes, WHAM loads of electricity. But if you put Zn and Fe for example, your voltage is
much less. So bottom line: memorize reactivity series. (Find a way!!)
2. Important terms
2.1. Electrodes
INERT electrodes are made of carbon/platinum. Inert meaning they don’t participate in the reactions.
METAL electrodes change in SIZE and WEIGHT during electrolysis:
Anode: The anode will corrode, decreasing in size and becoming lighter. OXIDATION HERE.
Cathode: A layer of metal will be coated (EXERCISE CAUTION, not true in some cases) on the cathode
and the cathode will become heavier and increase in size. REDUCTION HERE.
HOW TO REMEMBER: CROA. Cathode Reduction, Oxidation at Anode.
3. Factors affecting what is discharged during electrolysis
2.1. Type of Electrodes Are they inert (carbon or platinum) or reactive metals? They will affect not just what is discharged, but what you
will observe.
2.2. Concentration of electrolyte
Is it concentrated or diluted?
The ion that is more concentrated will be selectively discharged. It affects negative ions only!!! and this is
considered before we bring in the electrochemical series.
2.3. State of electrolyte
Is it molten or in solution? If it is molten, then you don’t have to consider the hydroxide and H+ ions :)
2.4. Reactivity/electrochemical series
Which cation is the one that is easiest to discharge! The one that has a lower position on the reactivity/
electrochemical series will be the one that is discharged.
**GENERAL TIP: If you are confused, write out ALL the ions present and work from there :) ALSO NOTE
THINGS LIKE THE PRODUCT PRODUCED FROM THE REACTION! and HOW MUCH is produced like the
gases, what is the ratio --> you can find out by writing the overall equation :) And what the electrolyte turns to is
also important, it is also a product of the reaction.
4. Stuff you die die must memorize
the very interesting question 3
fluorine question
Page 5 of 6
YEAR 4 CHEM PPA 1 NOTES
5. Sample questions (that are, weird hahaha)
A student conducted an experiment by connecting a 12V battery to a piece of filter paper soaked with dilute
sodium sulfate solution and universal indicator. A diagram of the experiment is shown below.
(a) After 10 minutes, the student removed the filter paper from the
circuit. What would be the colour of the filter paper at electrode A and
electrode B?
Ans: Red patch at electrode A and purple/blue patch at electrode
B. (remember your universal indicator colours hehe they are here:)
Basically what you are doing here is carrying out the electrolysis of
water, cos sodium and sulfate are super hard to discharge in solution (safe to say they will NEVER
discharge in solution.) Hence you are dealing with OH- and H+, which affect pH!
At the electrode A, oxidation of OH- ions occurs! Hence, the region around electrode A turned acidic (red)
because the discharge of OH- ions leaves an excess of H+ ions which are acidic.
Conversely, at electrode B, reduction of H+ ions occurs! The region around electrode B turned basic
because the discharge of H+ ions leaves an excess of OH- ions which are basic.
(b) Another student used distilled water instead of sodium sulfate solution and did not observe any colour
changes. Why didn’t the experiment work with water?
Water does not dissociate completely to form many mobile ions (very important principle to note!!!).
Hence, it does conduct electricity very well, and electrolysis cannot occur, thus there are no colour changes.
---An experiment was set up as follows:
Explain the following observations.
(i) After a short while, a cloud of white precipitate is
observed around the anode.
(ii) The current recorded decreases significantly as the
experiment progresses.
(i) At the anode, some of the Ag is oxidized to form
Ag+ which combines with Cl- ions in solution to
form a white ppt of AgCl! So aha, your solubility stuff
comes into play here, so don’t throw away QA ok :)
(ii) With less ions present to carry electrical charge, there is less current. (Cos all the Ag+ and Cl-is being
combined into AgCl, which is insoluble, and more H+ is also being discharged).
GENERAL TIPS:
Be confident!! You already know everything, so just apply. :))) Electrolysis just take note of the chemicals involved
la can one, can one :) JIAYOU JIAYOU JIAYOU you have chemistry with chemistry!!! :D
Page 6 of 6