Write balanced molecular and net ionic equations for the reactions of (b) chromium with hydrobromic acid
In this case, we have chromium reacting with a hydrobromic acid. So we have chromium, solid, plus HBr, aqueous. So we're trying to displace hydrogen from this acid, so what we do is we look at the list, here goes chromium here, it is above hydrogen, so we will be able to displace the hydrogen from the acid to produce H2. Now the chromium that we're looking at here is chromium 3+, so chromium kicks out the hydrogen and remember, chromium is 3+ here and then bromine is in group VII so minus one so, their charges criss-cross, so I have CrBr3, here this will be aqueous, plus H2 gas being produced.
We have to balance this out, so we put, let's see, we put a three here, so that we have, actually we can't put a three here, let's see, we put a three here to give us six, we put a two here, a six here, and a two here. Now our equation is balanced. Now we're going to say we have to break it up into it's ions, but remember for the total or complete ionic equation only aqueous compound break up. If it's not aqueous, it stays together. So we'll have two Cr, solid, plus six H+, aqueous, plus six Br-, aqueous, giving us two Cr 3+ ,aqueous, plus six Br- ,aqueous, plus three H2, gas. Remember, your spectator ions are the ions that look the same on both sides, so here it would be the bromide ion. Those are your spectator ions, they don't come down.
So our net ionic equation in the end would be two Cr, solid, plus six H+, aqueous, gives us two Cr 3+, aqueous, plus 3 H2, gas. So remember, we look at our activity series chart to see if the metal is above hydrogen, if it is it can displace the H within a compound to produce H2 gas, if it's below it it cannot displace the hydrogen from the given compound