[Hello_15]

Basically, I'm in AP Chem without having had any chemistry class before, so I am a little lost.

I understand how to determine spectactor ions in double replacement reactions, but I'm a little more confused when it comes to single replacement reactions. For example: Cl₂ + 2KI  I₂ + 2Cl-. Why is the pottassium a spectactor ion while Cl is not?

Also, I'm having some trouble understanding double replacement reactions where a precipitate is a reactant. My teacher tried to explain it, but its just not clicking.

Can anyone help?

Thanks,

Rachel.

[Astrokel]

spectator ion doesn't 'participate' in the reaction, so they basically don't undergo redox therefore their oxidation state remains unchanged. from your case K is the spectator since its oxidation remain as +1 before and after while Cl changes from 0 to -1.

[macman104]

It's not necessarily oxidation state.  To determine spectator ions:

1)  Write out the full ionic equation, include the state materials are in if given (so, gas, solid, liquid, aqueous, etc).
2)  Cancel any "pieces" that are the same on both sides.  This includes oxidation states AND physical states.

So, for example, in the reaction:

2KCl_(aq) + Pb(NO₃)_2(aq)  2KNO_3(aq) + PbCl_2(s)

The full ionic equation is:

2K⁺_(aq) + 2Cl^-_(aq) + Pb²⁺_(aq) + 2NO₃^-_(aq)  2K⁺_(aq) + 2NO₃^-_(aq) + 2PbCl_2(s)

So we see here that K⁺_(aq) is on both sides, as is NO₃^-_(aq).  However, on the left, we have Cl^-_(aq), but we don't have a Cl^-_(aq) on the right, even though in both cases, Cl is in a -1 oxidation state.  However, on the right, Cl is part of a solid (as PbCl₂ is not soluble in water, which is assumed since we say aqueous).

So!  If we were to write out the full ionic equation of your reaction, we have (*NOTE* I'm going to assume that I₂ forms a solid, but it might actually not...)

Cl_2(g) + 2K^-_(aq) + 2I^-_(aq)  I_2(s) + 2K^-_(aq) + 2Cl^-_(aq)

We see that while Cl and I change oxidation states which make them candidates for being included, you can also look and verify since there is no Cl_2(g) on the right, and there is no Cl^-_(aq) on the left, etc.  This is the complete thorough way to check for spectator ions.

[Hello_15]

Thank you. 

I figured out my main problem. I was trying to cancel spectator ions in compounds that weren't aqueous. When I realized that it was kind of a slap on the head moment.