OILRIG: Oxidation is loss of electrons and increase in oxidation number. Reduction is gain of
electrons and decrease in oxidation number.
Redox: A reaction where both oxidation and reduction take place.
Oxidation Number: A measure of the number of electrons that an atom uses to bond with
atoms of a different element.
Iodine is not soluble in water since it does not form H bonds with water.
(Reduction also gain of hydrogen and oxidation loss of hydrogen).
Element = 0. Hydrogen = +1. Oxygen = -2. Ion of element = ionic charge.
Oxidising + Reducing Agent:
– Oxidation agent removes electrons from species being oxidised and will contain species that
is reduced. It oxidises another species.
– Reducing agent adds electrons to species being reduced and will contain species that is
oxidised.
– They will both be on LHS and reactants.
– So nitric acid is oxidising agent but only the N atom is reduced using oxidation numbers.
Half Equations to Redox Equation:
–
– Balance the electrons by multiplying both by a factor. Now can cancel electrons on both
sides.
– Put half equations together. Cancel any other species on both sides.
Exact same logic when working backwards from overall redox equation- but remember if
need to divide electrons, or need to add any H+ and H2O. For example if overall equation
has no H+ but oxidation half equation does H+, means reduction half equation must also
have H+ on opposite side.
Constructing and Balancing Redox Equations using Oxidation Numbers: Some equations hard
to balance as an element may appear in multiple compounds in equation. Use this method as
ensures charges are balanced which is hard to do normally when more than two reactants/
products- do with full redox reactions only, including disproportionate reactions.
– Assign oxidation numbers to each element.
– Determine change in oxidation numbers of the two elements from side to side.
– Total increase in oxidation number = total decrease in oxidation number. So multiply the
change by factor to balance totals.
– Use this factor for the balancing number of the two compounds that only contained the
element which changed oxidation number (but not if same element in another compound
which did not change oxidation number).
– Then may need to balance the rest- often the H2O,
H+ and OH-.
– Check charges on both sides of equation balanced.
– Remember to double check whole equation
balances as still may need to make minor
adjustments.
Writing Half Equations:
– Assign oxidation numbers of element and find change.
Counteract the positive.
– Decrease by 5 means need 5e- on LHS.
– Predict any more species and balance all of the equation. H2O, H+ and OH- are likely
products in redox reactions to make sure oxygen and hydrogen’s are balanced on each side.
When polyatomic atom within a compound, assign these oxidation numbers first to match the
charge of the polyatomic atom. Then do the other elements in compound to match overall
compound charge.
Oxidation of iron oxide- questions tells us Cl added as well.
Half equation-
Balancing Equations: Find factor like above to get the same total number of oxygen’s.
If a molecule like SO4
2- stays as it is both sides, then don’t count its oxygens and sulphur in totals.
Treat it as an individual thing need to balance.
First balance the elements that only appear in a few substances. Then play around with H2O, any
acids etc. to balance number of hydrogens and oxygen’s. Keep numbers all odd or all even.
Make sure charges are balanced, alarm bells when see any charges.
Here have to balance both hydrogens and oxygen’s at the same time. Try even AND odd
numbers- trial and error.
Here as well as doing method with oxidation numbers, need to adjust to match the number of
nitrogen, oxygen and hydrogen.