Write balanced molecular and net ionic equations for the reactions of (d) acetic acid, CH3COOH, with zinc.
Hey guys, here we're going to have zinc reacting with acetic acid. So we have zinc solid plus acetic acid which they write as CH3COOH. Remember another way we could've written acetic acid is HC2H3O2. Both are technically correct. So here this is aqueous, we have to check on the activity series chart and see is zinc above hydrogen? So we take a look, let's see, so zinc is right here. Yes, it is above hydrogen so we can displace the hydrogen from the acid to produce H2 gas. Zinc here is +2 and it's the hydrogen that's connected to the oxygen that's getting displaced so your acetate ion looks like this. Remember another way it could've looked would be this, it's the same thing. So the numbers are different so they criss-cross when they combine, this is aqueous plus the hydrogen gas that we displaced. We need to balance this so we throw a 2 here.
Now we're moving to the total or complete ionic equation. Remember we only break up aqueous compounds. So this is zinc solid plus 2 acetate ions plus 2 H+ ions produces zinc 2+ ions plus 2 acetate ions again plus H2 gas. So we're going to get rid of spectator ions, spectator ions get cancelled out so we have left at the end is zinc solid plus H+ aqueous produces zinc 2+ aqueous plus H2 gas. So that represents our net ionic equation, once we've gotten rid of the spectator ions. So remember to be able to do this type of question, you need to be able to look at the activity series within your book. Using that allows us to see if we displace hydrogen or not. Remember, we say that this reaction because we're producing H2 as a gas we can say that this is called a hydrogen displacement reaction or we can say that it represents a single displacement reaction or a replacement reaction because we have one element displacing or kicking out another element so that it forms a new compound.