# Balancing other atoms in a redox reaction using half reaction method

I have this balanced redox reaction: $$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$ I think its unbalanced form should be (I guess water added to balanced it) $$\ce{Al + NaNO3 + NaOH -> NH3 + NaAlO2}$$ I wrote the half reactions as \begin{align} \ce {6H2O + 8e- + NO3- &-> NH3 + 9OH-}\label{rxn:1}\tag{1}\\ \ce {Al &-> Al^3+ + 3e-}\label{rxn:2}\tag{2} \end{align} Then by $3\cdot\eqref{rxn:1} + 8\cdot\eqref{rxn:2}$, final half reaction is: $$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$

Now I am really confused how to get above balanced redox reaction by the final half reaction I have obtained. What is the missing idea?

$$\ce{UPDATE:- Anybody please show me how to proceed from here,}$$ $$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$ $$\ce{to here}$$ $$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$

• What you call "final half reaction" is not a half; in fact it is a complete reaction, balanced and consistent. It is just that you don't want ions on either side, so you add counterions to them. Also, note that $\ce{Al^3+}$ reacts with $\ce{OH-}$. – Ivan Neretin Jan 25 '18 at 8:43
• @IvanNeretin yes. Thanks for correction. How to get NaAlO2 as aproduct by doing so? Could you please show the procedure? – stackerRook Jan 25 '18 at 9:00
• What procedure? Add as much $\ce{OH-}$ as it takes. Remove the water. – Ivan Neretin Jan 25 '18 at 9:05
• @IvanNeretin I think you didn't get what I said. So from the final reaction that I have, I don't see any obvious way to get the firstly I mentioned FULLY BALANCED reaction with all compounds (Not as IONS) . NO3- become NaNO3 and NH3 remain as same. no problem there. But how Al3+ become NaAlO2? What procedure to get that final complete reaction after having the reaction (with ions) – stackerRook Jan 25 '18 at 9:07
• You are supposed to "add" the reaction between Al3+ and OH-, which is also known in its own right (say, as AlCl3 + NaOH), not necessarily related to any redox. Can you balance it? – Ivan Neretin Jan 25 '18 at 9:14

$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH⟶3NH3 + 8NaAlO2}$$

But your choice of oxidation half reaction is:

$$\ce{Al -> Al^{3+} + 3e-}$$

Notice that the right side of the half reaction doesn't represent the right side of the total reaction (once you've accounted for the sodium spectator ion).

You must find the corresponding oxidation half reaction, and if it's not available, then you're out of luck.

Your half reaction will have $\ce{Al}$ on the left, and $\ce{AlO2-}$ and 3 electrons on the right; with water, hydrogen ions, and hydroxides to balance.

In comments, you also asked for why this is the oxidation product. That is the wrong question to ask. All you're doing is breaking down the total reaction into component half reactions. The simple half reaction ($\ce{Al|Al^{3+}}$) is totally valid. It's just gives a different total reaction from the one you started with. There's nothing wrong with that reaction. It's just different.

• so you mean I cannot "guess" the reaction products by considering the medium and reactants etc? Does that mean i have to remember the products (by heart)? This is a question for me. – stackerRook Jan 25 '18 at 15:35
• You can. You just might be wrong. The reaction will do what it wants to do, not what we would like it to do. With appropriate experience and insight, you might have a pretty good success rate, but again, you're just never guaranteed to be right. – Zhe Jan 25 '18 at 15:39
• The idea of matching the left & right of half reactions with that of the full reaction to be balanced was something I should've focused. Thanks for pointing that out. Great help and thank you very much sir! – stackerRook Jan 25 '18 at 15:42