1
$\begingroup$

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:

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

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.