[PIEHelp]

Question is to end up with the net ionic equation for, "Aqueous Cesium Iodide plus an aqueous solution of bromine."

I'm getting a molecular equation of:
2CsI (aq) + Br₂(aq)  2CsBr (aq) + I₂(g)

Which I then (skipping a step) find a net ionic equation of:
2I^-(aq) + Br₂(aq)  2Br^- + I₂(g).

But I'm not quite sure. At first I was thinking that bromine would disassociate entirely and the net ionic would be just the iodine, but I've changed from that. Just not entirely convinced I've got it correct there.

[Borek]

The only thing that is tricky here is the final state of iodine, and IMHO there is no way to give an unambiguous answer without knowing more about initial concentrations and amounts of all substances involved. I would probably go for I₂(aq). As long as there is iodide present I₂ reacts with it and becomes dissolved as I₃^-, once all I^- gets oxidized some I₂ stays in water and some moves out of the solution.

[PIEHelp]

So, its just a general exercise without any amounts involved. Just given the reactants, find the products and then balance the equation and complete the rest of the question. Interesting that the iodine would stay aqueous in its final form, but it does make sense.

Thanks for the answer.