Write the total ionic and the net ionic equations for the following:
3 H2 (g) + 2 AuCl3 (aq) → 6 HCl (aq) + 2 Au (s)
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To write the total ionic and net ionic equations, you need to break down the reactants and products into their respective ions.
The given balanced equation is:
3 H2 (g) + 2 AuCl3 (aq) → 6 HCl (aq) + 2 Au (s)
To write the total ionic equation:
- Write all the compounds that are soluble (aqueous) as their individual ions.
- Write any insoluble compounds as their molecular formula.
Breaking down the reactants and products:
3 H2 (g) → 6 H+ (aq) + 6 e- (oxidation half-reaction)
2 AuCl3 (aq) → 6 Cl- (aq) + 2 Au3+ (aq) (disproportionation of Au atoms)
Note: The Au atoms in AuCl3 undergo a redox reaction, where some atoms are oxidized from a 0 oxidation state to +3 (Au3+), and the rest are reduced to elemental gold (Au).
6 HCl (aq) (already in ionic form)
2 Au (s) (solid elemental gold)
Writing the total ionic equation:
3 H2 (g) + 2 AuCl3 (aq) → 6 H+ (aq) + 6 Cl- (aq) + 2 Au (s)
To write the net ionic equation:
- Remove the spectator ions (the ions that appear on both sides of the equation and do not participate in the reaction).
- Keep only the ions and molecules that are directly involved in the reaction.
For the net ionic equation:
The spectator ions in this reaction are H+ and Cl- since they are both present on both sides of the equation.
Removing the spectator ions, we get:
Net ionic equation:
2 AuCl3 (aq) → 2 Au (s) + 6 Cl- (aq)
Therefore, the total ionic equation is: 3 H2 (g) + 2 AuCl3 (aq) → 6 H+ (aq) + 6 Cl- (aq) + 2 Au (s)
And the net ionic equation is: 2 AuCl3 (aq) → 2 Au (s) + 6 Cl- (aq)