What is the net ionic equation for rection of liquid bromide with sodium iodide crystals?

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Technically you are correct. How can there be any ions when the reactants are not in aqueous medium ? But here it is a matter of convention and maybe an illustration of how to write the ionic equations from molecular ones. Likewise, if you have done balancing of redox reactions , you might notice half of the reactions given to balance are completely impossible according to the laws of chemical combination. You'll always be at a loss when you overthink in chemistry, even if your argument is right. Taking it as just a convention would be a better idea .

Hope this helped !

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kullalok
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kullalok

high school student

Updated on August 01, 2022

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  • kullalok
    kullalok over 1 year

    I have this problem,

    Liquid bromine is added to a container of sodium iodide crystals

    I interpreted it with this molecular formula $\ce{Br_{2(l)} + 2NaI_{(s)} \rightarrow 2NaBr_{(s)} + I_{2(g)}}$.

    From my understanding, since none of the products nor any of the reactants is aqueous, there shouldn't be a net ionic equation. However, my friend has proceeded with the problem as follows:

    $\ce{Br_2 + 2Na^+ + 2I^- \rightarrow 2Na^+ + 2Br^- + I_2}$

    $\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$.

    Could somebody explain which one of us is correct and why?

    • Greg E.
      Greg E. about 10 years
      First, iodine probably shouldn't be a gas, because it exists as a solid under standard conditions. As for how the net ionic equation should be written, I think it's a matter of convention. When I took general chemistry I, we were instructed to separate soluble ionic compounds and cancel spectator ions ($\ce{Na+}$) irrespective of whether the reaction was explicitly taking place in aqueous solution. This is a redox reaction, and sodium is not undergoing change in oxidation state, so that seems right to me. Hopefully somebody weighs in with a conclusive answer as to what the convention is.