Electrolytes and Net Ionic Equations

This video tutorial explains how to write the net ionic equation for the reaction between aluminum and copper(II) chloride. It begins with balancing the molecular equation, assigning states to the reactants and products, and checking the feasibility of the reaction using the activity series. The tutorial then covers splitting strong electrolytes into ions, determining their charges, and writing the complete ionic equation. Finally, it identifies and removes spectator ions to derive the net ionic equation.

What is the first step in writing a net ionic equation?

• Check the activity series

• Determine the states of matter

• Balance the molecular equation

• Identify spectator ions

Which of the following is a strong electrolyte?

• Copper(II) chloride

• Aluminum chloride

• Copper

• Aluminum

Why is it important to know the activity series of metals?

• To identify spectator ions

• To balance chemical equations

• To predict if a reaction will occur

• To determine the solubility of compounds

What charge does aluminum have in its elemental form?

• +2

• +1

• +3

• 0

How many chloride ions are present in the complete ionic equation for the reaction?

• 8

• 6

• 2

• 4

What is the charge of the copper ion in copper(II) chloride?

• +3

• +2

• +4

• +1

Which ions are considered spectator ions in this reaction?

• Hydrogen ions

• Chloride ions

• Copper ions

• Aluminum ions

What is the net ionic equation for the reaction between aluminum and copper(II) chloride?

• 2Al + 3Cu^2+ → 2Al^3+ + 3Cu

• Al^3+ + 3Cl^- → AlCl3

• Al + 3Cu^2+ → Al^3+ + 3Cu

• Al + CuCl2 → AlCl3 + Cu

What happens to the chloride ions in the net ionic equation?

• They react with aluminum

• They react with copper

• They remain unchanged as spectator ions

• They form a precipitate

What is the purpose of writing a net ionic equation?

• To show all reactants and products

• To balance the chemical equation

• To highlight the formation of precipitates

• To focus on the ions that participate in the reaction