When finding the half equation and overall redox equation for the following:

a) $\ce{Fe^2+(aq) + Cl2(g) <=>}$
b) $\ce{N2(g) + H2(g) <=>}$

The solution from the textbook is given below. I do not understand why the products (in the overall reaction) for a) are 2 individual ions (and not $\ce{FeCl3}$ whilst the product for b) is a compound (and not $\ce{2N3- + 6H+}$).

One difference I noted was, when the electrons are balanced for a) it gives $\ce{2Fe3+}$ and $\ce{2Cl-}$ , which cannot go together unless its $\ce{2Fe3+}$ and $\ce{6Cl-}$, whilst for b), when the electrons are balanced we have $\ce{6H+}$ and $\ce{N2}$ which combine as is to form $\ce{2NH3}$. Is this the reason?

The solution given is:

Oxidation: $\ce{2Fe^2+(s) -> 2Fe^3+ + 2e- }$
Reduction: $\ce{Cl2(g) + 2e- -> 2Cl- }$
Overall: $\ce{2Fe^2+ (aq) + Cl2(g) <=> 2Fe^3+ + 2Cl- }$

Oxidation: $\ce{3H2 -> 6H+ + 6e-}$
Reduction: $\ce{N2 + 6e- -> 2N^3-}$
Overall: $\ce{N2(g) + 3H2(g) <=> 2NH3}$


1 Answer 1


While FeCl3 is a possible product, the reaction in A is aqueous. Wikipedia shows that is readily dissolves in water (92 g/100 mL @ 20C). The products exist as ions, and are written as such.

I would suspect you're not supposed to have solubility constants memorized though, so in general if it is in water they will exist as ions.

B is a gaseous reaction and the product is a compound, not an ion set like A.


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