Starter What is the full electron configuration for Calcium?

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

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!

Example: the thermit reaction Fe2O3 + Al  Fe + Al2O3 Fe3+ Al0 Fe0 Al3+ Fe3+ + 3e-  Fe Reduction Al  Al3+ + 3e- Oxidation Ionic half-equations

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

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 +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

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 +5 +4

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 +5 +4 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!

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 I2 + 2e-  2I- Fe3+ + e-  Fe2+

Give these a go Balance the half equations Fe2+  Fe3+ I2  I¯ Na  Na+ Fe2+  Fe3+ I2  I¯ C2O42-  CO2 H2O2  O2 H2O2  H2O NO3-  NO NO2  NO3- SO42-  SO2

All good? Fe2+  Fe3+ + e- I2 + 2e-  2I¯ C2O42-  2CO2 + 2e- Na  Na+ + e- Fe2+  Fe3+ + e- I2 + 2e-  2I¯ C2O42-  2CO2 + 2e- H2O2  O2 + 2H+ + 2e- H2O2 + 2H+ + 2e-  2H2O NO3- + 4H+ + 3e-  NO + 2H2O NO2 + H2O  NO3- + 2H+ + e- SO42- + 4H+ + 2e  SO2 + 2H2O

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 Br2 + 2e-  2Br- 2Cl- + 2e-  Cl2

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

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?

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

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+  2Mn2+ + 5O2 + 8H2O

Electron is on the wrong side! Electrons should be cancelled out Complete the half-equation Na  Na+ Na  Na++ e- Pb4+  Pb2+ Pb4+ + 2e-  Pb2+ H2  H+ H2  2H+ + 2e- Cr2O72-  Cr3+ Cr2O72- + 6e-  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