Summary 3.1 and 3.2 Electrodes and Redox Reactions
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Course
3.1 and 3.2
Institution
WJEC
A summary booklet with all the content needed for 3.1 and 3.2 WJEC A level Chemistry. Useful for learning the content and creating flashcards.
Includes:
- Oxidation states
- Forming and combining half equations
- Electrode potentials and cells
- DANIEL and SHE cells
- The electrochemical ser...
3.1 and 3.2 Electrodes and Redox Reactions
Oxidation – Loss of electrons / becomes more positive.
Reduction – Gain electrons / becomes more negative.
Oxidising agent – Gains electrons.
Reducing agent – Loses electrons.
Oxidation States
Rules:
Elements 0
Hydrogen +1
-1 if a metal hydride (e.g NaH)
Group 1 +1 always
Group 2 +2 always
Aluminium +3 always
Oxygen -2
-1 if Peroxides (e.g H2O2)
+2 if Fluorine
Fluorine -1 always
Chlorine -1
+ve if Fluorine and Oxygen
- Sum of oxidation states in a compound must equal 0.
- Sum of oxidation states in an ion must equal the charge.
Half Equations
Half equations are used to show what happens to each species in a redox reaction separately in
terms of electrons.
One equation will show the oxidation process and the other will show the reduction process.
Forming half equations:
- Balance the main atom
- Balance Oxygen with water
- Balance Hydrogens using H+
- Balance the charges with electrons (the same side as H+)
- Check the right number of electrons have been used using oxidation states
, Combining Half Equations
Example 1 – in acid solution, dichromate (VI) ions (Cr2O72-) will oxidise sulphate (IV) ions (so32-) to
sulphate (VI) ions (SO42-) and are themselves reduced.
Know from the question that SO32-
will be oxidised and therefore lose
electrons and Cr2O72- is being
reduced so will gain electrons.
Need to multiply SO32-
equation by 3 so that both
equations have the same
number of electrons.
Combine both equations
and cancel like terms.
Electrode Potentials and Cells
Electricity – The movement of electrons
- Each metal has the potential to lose electrons.
- If two metals are put together in a solution connected by a wire, then the metal that has a
greater reducing potential will give up its electrons more easily to the other one.
Zn2+(aq) + 2e- ⇌ Zn
Cu2+ + 2e- ⇌ Cu
- When Zinc is placed in a solution of its own ions equilibrium will lie more heavily to the left
as Zn is a reactive metal so gives up electrons more easily than Copper where equilibrium is
more heavily to the right as it is less reactive so less likely to give up electrons.
- Electrode potentials give a value to the different positions of equilibrium.
Potential Difference
- The difference between the negativity of a metal and the positivity of the solution
- Recorded as voltage.
- The greater the difference the bigger the voltage
- The move negative the potential the more likely the substance is to lose electrons and be
oxidised.
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