Redox Flashcards
What happens to the oxidation state of an element in an equation when it is oxidised?
The oxidation state increases
What is an oxidising agent?
a chemical species, or reactant, which increases the oxidation number of the other species or substance
What happens to the oxidation state of an element in an equation when it is reduced?
it’s oxidation state decreases
What is a reducing agent?
a chemical species, or reactant, which decreases the oxidation number of the other species or substance
What must you do to determine how a particular reactant has behaved?
to deduce the oxidation states within the chemical equation
What is the oxidation state of an ion?
the number of electrons that must either be added or removed from that ion in prefer to give a neutral atom
What do we imagine in terms of oxidation numbers in non-ionic compounds?
that the more electronegative atoms takes the electrons
What are the rules of oxidation numbers?
-uncombined element= 0
-all atoms in a molecule add up to 0
-simple ion= charge of the ion
-molecular ion- adds up to charge of ion
-g1-3 metals+ +1, +2, +3
-fluorine atom+ -1
-hydrogen atom= +1 (some exceptions)
-oxygen atom- -2
-any other can then be calculated
How can an overall redox equation for a reaction be constructed?
by combining the half equations showing
What does a half equation involve?
either oxidation or reduction, never both.
There must be electrons on one side (for reduction, the left and for oxidation, the right).
How do we combine 2 half equations?
To combine the two equations the number of electrons must be the same in each half equation and they must be on opposite sides of the equations.
To do this we multiply the silver half equation by 2.
We can then add the two equations together: this is now a fully balanced redox equation.
What do half equations often involve?
oxyanions
(e.g. PO43-)
what are oxyanions?
these are negative ions which contain both oxygen and another element
What is the formula or for these oxyanions: Thiosulfate, ethanedioate
Manganate (VII), Iodate (V), Sulfate (IV) Sulfate (VI), dichromate (VI), Nitrate (V)
Thiosulfate S2O32-
ethanedioate C2O42-
Manganate (VII) MnO4-
Iodate (V) IO3….
Sulfate (IV) SO32-
Sulfate (VI) SO42-
dichromate (VI) Cr2O72- Nitrate (V) NO3-
What is the method for deducing half equations?
-Write what you know
-Balance the main element
-Balance Oxygens with H2O’s
-Balance H’s with H+’s
-Balance charge with electrons (add to more positive side)
The redox reaction of copper (II) ions with iodide. Note – in this reaction, you form CuI(s) – you will need some additional iodide to react with the Cu (I).
2Cu2+ + 2I- → 2Cu+ + I2
or
2Cu2+ + 4I- → 2CuI + I2
How do you work out half equations from overall equations?
-Write out the starting and end charge of the species.
- Work out whether electrons have been lost or gained
- Add in the correct number of electrons to balance the charge
- Remember oxidation: electrons on RHS; reduction electrons on LHS
How do you construct redox equations using oxidation numbers?
-Write a ‘draft’ equation based on the information given in the question.
-Identify the changing oxidation numbers
-Balance the equation so that changes in oxidation number are balanced,
i.e. total increase in oxidation number = total decrease in oxidation number.
Since copper increases by 2 and nitrogen decreases by 3, we must multiply Cu/Cu2+ by 3 and HNO3/NO by 2
(check notes)
What can you use titration data and half equations to do?
to calculate the number of moles of electrons being added or taken from a species, this will allow you to calculate changes in oxidation state
Example: 0.180 moles of dichromate ions will react with 0.540 moles of nickel forming chromium (III) ions. What is the new oxidation number of nickel?
Write down or construct a half equation for the known change:
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Step 2: Calculate the number of moles of electrons being added to the dichromate ions.
Moles of electrons = 0.180 x 6 = 1.08 moles
Step 3: Calculate how many moles of electrons are being lost from each mole of nickel,
i.e. calculate the ratio of electrons transferred to nickel
1.08/0.540 = 2.00
Step 4: Each mole of nickel is losing 2 moles of electrons, i.e. each atom of nickel is losing 2
electrons, and therefore the oxidation number will have increased by 2.
The new oxidation number of nickel will be +2
If 3 moles of manganate(VII) react with 5 moles of X+ to give manganese(II) ions,
what will be the new oxidation state of X?
MnO4
- + 5e- + 8H+ → Mn2+ + 4H2O
moles of electrons transferred to MnO4
- = 3 x 5 = 15 moles
moles of electrons lost from X+ per mole of X+ = 15/5 = 3 moles
each X+ loses 3 electrons, so new oxidation state = +4
(check notes)
