1. Ionic Equations and Half-Reactions
Electrochemistry
SCH4C

2. Total Ionic Equations
When investigating redox reactions, it is useful to show all dissociated ions to better predict oxidation and reduction.
Example: CuSO4(aq) + Mg(s)  Cu(s) + MgSO4(aq)
Cu2+(aq) + SO42-(aq) + Mg(s)  Cu(s) + Mg2+(aq) + SO42-(aq)
Any ions that appear on both sides are spectator ions and aren’t actually involved in the reaction

3. Net Ionic Equations
Remove all spectator ions to form the net ionic equation, showing only the ions that are involved in the reaction
Example:
TIE: Cu2+(aq) + SO42-(aq) + Mg(s)  Cu(s) + Mg2+(aq) + SO42-(aq)
NIE: Cu2+(aq) + Mg(s)  Cu(s) + Mg2+(aq)

4. Predicting Redox
Once we have established the net ionic equation, assigning oxidation numbers is an easier task:
Cu2+(aq) + Mg(s)  Cu(s) + Mg2+(aq)
Oxidation: Mg(s)  Mg2+(aq)
Reduction: Cu2+(aq)  Cu(s)

5. Half-Reactions
Half-reactions are separate representations of the oxidation and reduction reactions.
They are used to follow the transfer of electrons in redox reactions, and so we must balance the charges of half-reactions using electrons
Example:
Oxidation: Mg(s)  Mg2+(aq) + 2e-
Reduction: Cu2+(aq) + 2e-  Cu(s)
Magnesium atoms lose electrons, while copper ions gain electrons.

6. Example
Determine if the following reaction is a redox reaction by showing the net ionic equation and assigning oxidation numbers. If it is a redox reaction, write the half-reactions.
Pb(s) + Cu(NO3)2(aq)  Pb(NO3)4(aq) + Cu(s)

7. Example: Pb(s) + Cu(NO3)2(aq)  Pb(NO3)4(aq) + Cu(s)
Step 1: Balance the equation
Pb(s) + 2Cu(NO3)2(aq)  Pb(NO3)4(aq) + 2Cu(s)
Step 2: Write the Total Ionic Equation
Pb(s) + 2Cu2+(aq) + 4NO3-(aq)  Pb4+(aq) + 4NO3-(aq) + 2Cu(s)
Step 3: Write the Net Ionic Equation
Pb(s) + 2Cu2+(aq)  Pb4+(aq) + 2Cu(s)

8. Example: Pb(s) + Cu(NO3)2(aq)  Pb(NO3)4(aq) + Cu(s)
Step 4: Use Oxidation Numbers to Predict Redox
Pb(s) + 2Cu2+(aq)  Pb4+(aq) + 2Cu(s)
Oxidation: Pb(s)  Pb4+(aq)
Reduction: 2Cu2+(aq)  2Cu(s)
Step 5: Balance the charges using electrons
Oxidation: Pb(s)  Pb4+(aq) + 4e-
Reduction: 2Cu2+(aq) + 4e-  2Cu(s)
Please note that each ½ reaction should have the same number of electrons!