Ionic Equations and Half-Reactions

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Ionic Equations and Half-Reactions Electrochemistry SCH4C

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

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)

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) +2 0 0 +2 Oxidation: Mg(s)  Mg2+(aq) Reduction: Cu2+(aq)  Cu(s)

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.

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)

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)

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) 0 +2 +4 0 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!