1. Starter What is the full electron configuration for Calcium?

2. Redox equations To understand what redox means
To be able to write ionic half equations
To be able to confidently balance ionic equations

3. e- Redox – what? Oxidation and reduction happen at the same time
There is no net gain or loss of electrons. e-
You can’t just create them or destroy them!

4. Example: the thermit reaction
Fe2O Al  Fe Al2O3
Fe Al Fe Al3+
Fe e-  Fe Reduction
Al  Al e- Oxidation
Ionic half-equations

5. Some rules to help Species Symbol Charge
An element in its standard state
O2, Br2, Mg, Al
Hydrogen in a compound H+ +1
Oxygen in a compound
O2- -2

6. Remember: OIL - Oxidation Is Loss of electrons
RIG - Reduction Is Gain of electrons
And often... (as a quick and simple way to tell):
Oxidation is gain in oxygen or loss of hydrogen
Reduction is loss of oxygen or gain of hydrogen

7. +7 +6 +5 +4 +3 +2 +1 -1 -2 -3 -4 -5 -6 -7
Oxidation
Reduction
Oxidation and reduction can be seen as movement up or down a scale of oxidation states
Oxidation state

8. Redox in presence of acid
Example: a past exam question
a) Identify, as oxidation or reduction, the formation of NO2 from NO3- in the presence of H+ and deduce the half-equation for the reaction.
NO3-  NO2
Reduction

9. Beautiful! Steps to take NO3-  NO2 +5 +4 NO3- + e-  NO2
Write a balanced equation for the species
Work out “before and after” oxidation states
Balance oxidation states with electrons
NO3-  NO2
NO3- + e-  NO2
If all the charges don’t balance, add H+ ions to one of the sides to balance them
NO3- + e- + 2H+  NO2
If the equation still doesn’t balance, add enough water to one side so it balances
NO3- + e- + 2H+  NO2 + H2O
Beautiful!

10. Practical
Add a small amount of Fe (III) ions to a solution of iodide ions.
Describe what you see.
Use the following two equations to produce a balanced redox equation
I e-  2I-
Fe e-  Fe2+

11. Give these a go Balance the half equations Fe2+  Fe3+ I2  I¯
Na  Na+
Fe2+  Fe3+
I2  I¯
C2O  CO2
H2O  O2
H2O  H2O
NO  NO
NO  NO3-
SO  SO2

12. All good? Fe2+  Fe3+ + e- I2 + 2e-  2I¯ C2O42-  2CO2 + 2e-
Na  Na e-
Fe2+  Fe e-
I e-  2I¯
C2O  CO e-
H2O  O H e-
H2O H e-  2H2O
NO H e-  NO H2O
NO H2O  NO H+ + e-
SO H e  SO H2O

13. Practical 2
Add a small amount of chlorine water to a solution of bromide ions.
Describe what you see.
Use the following two equations to produce a balanced redox equation
Br e-  2Br-
2Cl e-  Cl2

14. Combining half equations
Just a mashing together of two half-equations!
... followed by some satisfying cancelling-out.

15. Back to our exam question
b) Deduce the overall equation for the reaction of copper with NO3- in acidic conditions to give Cu2+.
Reduction
NO3- + e- + 2H+  NO2 + H2O
Oxidation
Cu  Cu2+ + 2e-
Now what?

16. Steps to take to combine equations
Multiply the equations so that the number of electrons in each is the same
Add the two equations and cancel out the electrons on either side of the equation
If necessary, cancel out any other species which appear on both sides of the equation
Cu  Cu2+ + 2e-
2NO3- + 2e- + 4H+  2NO2 + 2H2O
X 2
Cu + 2NO3- + 2e- + 4H+  Cu2+ + 2e- + 2NO2 + 2H2O
Cu + 2NO3- + 4H+  Cu2+ + 2NO2 + 2H2O

17. Give these a go 6Fe2+ + ClO3- + 6H+  Fe3+ + Cl- + 3H2O
Fe2+ ions are oxidised to Fe3+ ions by ClO3- ions in acidic conditions. The ClO3- ions are reduced to Cl- ions. Write the overall reaction.
Write an overall reaction for MnO4- reducing H2O2 to O2 and creating Mn2+ .
Fe2+  Fe3+ + e- ClO3- + 6e- + 6H+  Cl- + 3H2O
6Fe2+ + ClO3- + 6H+  Fe3+ + Cl- + 3H2O
2MnO4¯ + 5H2O2 + 6H+  2Mn O H2O

18. Electron is on the wrong side! Electrons should be cancelled out
Complete the half-equation
Na  Na+
Na  Na++ e-
Pb4+  Pb2+
Pb e-  Pb2+
H2  H+
H2  2H e-
Cr2O72-  Cr3+
Cr2O e-  2Cr3+
What’s wrong with this equation?
Ce3+ + e-  Ce4+
Electron is on the wrong side!
Mg + 2H+ + e-  Mg+ + H2 + e-
Should be Mg2+
Electrons should be cancelled out