1. (including electrolysis) ALWAYS BRINGS YOU BACK HERE
Working Out Formulae
& Balancing Equations
(including electrolysis)
Valency
Writing Equations
Using Valency
Balancing Equations
Electrolysis half equations
END
ALWAYS BRINGS YOU BACK HERE

2. VALENCY tells you how many BONDS an atom can form
You work the valency out like this…
Work out how many OUTER ELECTRONS the atom has
(That’s it’s GROUP NUMBER in the Periodic Table)
If it has 4 or less then THAT’S THE VALENCY
If it has 5 or more then VALENCY = 8 – OUTER ELECTRONS
Element
Group
Outer Electrons
Valency
Sodium 1
Chlorine 7
8 – 7 = 1
Magnesium 2
Nitrogen 5
8 – 5 = 3
Carbon 4
Oxygen 6
8 – 6 = 2

3. Here’s another method that works for ions…
Sodium forms Na+ ion
It has valency 1
For IONIC BONDS, the valency is just the size of the charge on the ion
Calcium forms Ca2+ ion
It has valency 2
Oxygen forms O2- ion
It has valency 2
Chlorine forms Cl- ion
It has valency 1
Some transition metals can form more than one ion, so they have more than one valency.
Iron forms Fe2+ and Fe3+
So it can have valency 2 or 3
You can use the same idea for ions made of more than one atom
Ammonium ion NH4+
has valency 1
Nitrate ion NO3-
has valency 1
Sulphate ion SO42-
has valency 2

4. USING VALENCIES TO WORK OUT FORMULAE
Write down the two elements (or ions) in the compound
Write down the valency of each as a “small number” next to the other one. Put brackets round any complicated ions, like sulphate, nitrate etc.
Cancel if necessary, and remove any 1s
Sodium sulphate Na SO4
Valency of sodium: 1
Valency of sulphate: 2
So Na2(SO4)1
Remove 1s: Na2SO4
Iron(II) chloride Fe Cl
Valency of iron : 2
Valency of chloride : 1
So Fe1 Cl2
Remove 1s: Fe Cl2
Magnesium nitrate Mg NO3
Valency of magnesium : 2
Valency of nitrate : 1
So Mg1 (NO3)2
Remove 1s: Mg (NO3)2
Calcium oxide: Ca O
Valency of calcium: 2
Valency of oxygen: 2
So Ca2O2
Cancel: CaO
DO NOT “MULTIPLY OUT THE BRACKETS” !
Write (NO3)2 not N2O6

5. WRITING EQUATIONS Step 1: Write a WORD EQUATION Reactants Products
The chemicals you start with
What you make in the reaction
Reactants
Products
Step 2: Write down the FORMULA of each of the chemicals in your word equation
Step 3: Put in the STATE SYMBOLS
(s) for solid, (l) for liquid, (aq) for solution, (g) for gas
Eg: reaction of calcium carbonate with hydrochloric acid:
calcium chloride
+ water +
carbon
dioxide
calcium
carbonate
hydrochloric
acid +
The PRODUCTS
CaCO HCl
The REACTANTS
CaCl H2O CO2
(s)
(aq)
(aq)
(l)
(g)
Make sure you get the formulae right!
No marks for balancing if the formulae are wrong!
Now on to the harder bit – balancing the equation!
Calcium chloride is soluble, so as there is water present, it is a solution
Hydrochloric acid, like other acids, is always a solution
Calcium carbonate is an insoluble solid
Water is a liquid
(not a solution!)
Carbon dioxide is a gas
It forms in bubbles

6. Balancing Equations 2Na + 2H2O  2NaOH + H2 WHAT DOES IT MEAN?
Balancing an equation means making sure the numbers of each type of atom are the same on each side
Let’s look at this balanced equation:
2Na + 2H2O  2NaOH + H2
Atom
Number on Left
Number on Right Na
Have 2Na, so 2
1 in each NaOH
2NaOH. So 2 H
2 in each water
2 waters
2  2 = 4
2 from 2NaOH
+ 2 from H2 makes 4 O
1 in each water
2 waters. So 2
1 in each NaOH
2NaOH. So 2
They all match. So it’s balanced!

7. How to balance an equation
Step 1: Write down the unbalanced formula equation.
Step 2: Work out how many of each atom there are on each side. (in your head, if it’s easy)
Step 3: Look for any atoms where there aren’t the same number on each side
Step 4: Choose the “unbalanced atom” that’s in the smallest number of different formulae.
Step 5: Balance it by putting a number IN FRONT of one of the formulae (don’t change the actual formula!)
Step 6: Recalculate numbers of atoms – and repeat if needed!
NaOH + H2SO4 Na2SO4 + H2O 2 2
Left Right
Na 1 2
O 5 5
H 3 2
S 1 1 2
There are different numbers of O and H
Na is only in one chemical each side
There are different numbers of Na and H 6 6
H is in fewer different formulae. 4 4
We can balance H by putting 2 in front of H2O
We can balance them by putting 2 in front of NaOH
THEY ALL MATCH! IT’S BALANCED!
Now recalculate the numbers of atoms…
Now recalculate the numbers of atoms…

8.   Other examples… 2 2 2 CaCO3 + HCl  CaCl2 + H2O + CO2
balanced
Left Right
Ca 1 1 C
O 3 3
H 1 2
Cl 1 2 2
Let’s balance the hydrogens:
We can do this by putting 2 in front of HCl
CaCO3 + HCl  CaCl2 + H2O + CO2 2 2
Eureka ! That has automatically balanced the chlorines too.
Left Right
Cl 2 1 K
I 1 2 2
Unfortunately, this has also unbalanced the potassium.
However putting 2 in front of KI balances both K and I 2
Let’s balance the chlorines :
We can do this by putting 2 in front of KCl
Cl2 + KI  KCl + I2 2 2 2 2 
balanced

9. THEY ALL MATCH! IT’S BALANCED!
Another example…
Al2O3 + HCl  AlCl3 + H2O 6 2 3
Left Right
Al 2 1
O 3 1
H 1 2
Cl 1 3 2
Now let’s balance the hydrogens:
We can do this by putting 6 in front of HCl
Now let’s balance the oxygens:
We can do this by putting 3 in front of H2O 3
None of the atoms are balanced!
They all occur in just two chemicals
Choose one to balance… 6 6 6 6
Recalculate…
Recalculate…
THEY ALL MATCH! IT’S BALANCED!
We can balance Al by putting 2 in front of AlCl3
Recalculate…

10. THEY ALL MATCH! IT’S BALANCED!
An awkward one! 2
Al + Cl2  AlCl3 3 2
Left Right
Al 1 1
Cl 2 3 2 2
Now we must balance the aluminiums
We can do this by putting 2 in front of Al
Chlorines aren’t balanced. 6 6
But how can we do the balancing?
We haven’t got “nice” numbers!
Recalculate…
THEY ALL MATCH! IT’S BALANCED!
This is like finding the “lowest common denominator” in fractions.
We have two chlorines on one side, and three on the other.
We find the smallest number two and three go into – that’s six.
So we need to aim for six chlorines on each side
If you like maths, you could try balancing ones like this using fractions instead. You’d need to use 1½
To do that, we put 3 in front of Cl2 and 2 in front of AlCl3
Recalculate…

11. Here are Cu2+ ions moving to the negative electrode.
ELECTROLYSIS
Here are Cu2+ ions moving to the negative electrode.
Positive ions in the solution are attracted to negative electrode – opposite charges attract.
When electrons are gained by a positive ion, the name of the chemical change is REDUCTION.
REDUCTION IS THE GAIN OF ELECTRONS.
THE COPPER ION HAS BEEN REDUCED
2e- + Cu2+  Cu
The electrode is negative because it has too many electrons
As they get close, the ions gain electrons from the electrode and the Cu2+ is neutralised.
e- go to ion Cu
This makes copper the element, which covers the electrode.
e- go to ion.
TWO ELECTRONS FROM THE CATHODE
A NEUTRAL ATOM OF THE ELEMENT COPPER.
ARE ADDED TO THE COPPER ION

12. When electrons are lost by a negative ion, the name of the chemical change is OXIDATION.
OXIDATION IS THE LOSS OF ELECTRONS
THE CHLORIDE ION HAS BEEN OXIDISED
e- go to cell
Cl-
This is what happens at the positive electrode when chloride ions, Cl- are present in the electrolyte
Here are negative chloride ions
attracted towards the positive
electrode. Opposite charges attract.
This electrode is positive because some electrons have been removed by the cell.
As they get close, each Cl- ion loses an electron which goes onto the electrode.
The ion becomes electrically neutral
the ion
loses an e-
Cl2
We have made chlorine the element. The neutral atoms join in pairs to make chlorine molecules, Cl2 which bubble off as a gas.
2Cl- - 2e  Cl2
TWO CHLORIDE IONS, EACH WITH AN EXTRA ELECTRON
THE TWO ELECTRONS LOST BY THE IONS GO TO THE ELECTRODE
A NEUTRAL CHLORINE MOLECULE

13. 2 3 2 2 2 2 2 4 2 Cu2+ + e  Cu Al3+ + e  Al Ag+ + e  Ag 1
Half equations for reduction at the negative electrode.
Here are different ions that might be in a solution.
You need to be able to balance the half equations.
the ion GAINS electrons from the electrode
Cu2+ + e  Cu 2
Al3+ + e  Al 3
click for solution
Ag+ + e  Ag 1
Pb2+ + e  Pb 2
In these cases, metallic elements would appear at the negative electrode. 2
H+ + e  H2
click for solution
H+ is present in all acids and hydrogen gas
is evolved from the electrode.
Half equations for oxidation at the positive electrode.
Here electrons are lost by the ion.
When gaseous elements are produced, they bond together in pairs to make a molecule.
The balancing needs to include this. 2
Cl e  Cl2 2
Br e  Br2 2 4
O e  O2
Cu - e  Cu2+ 2
In a special case, a positive copper electrode dissolves in a solution of copper sulphate. Electrons are lost by the copper metal.

14. The end