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How would you balance the following reaction using the oxidation number method? $$\ce{Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O}$$

My approach:
Half reactions:
\begin{align} \ce{Zn &-> Zn(NO3)2} & \text{Change in oxidation state of }\ce{Zn} &= 2\\ \ce{2HNO3 &-> NH4NO3}& \text{Change in oxidation state of }\ce{N} &= 8\\ \end{align}

Multiplying by the L.C.M of the change of the oxidation number. \begin{align} \ce{4Zn &-> 4Zn(NO3)2}\\ \ce{2HNO3 &-> NH4NO3} \end{align}

Adding them, $$\ce{4Zn + 2HNO3 -> 4Zn(NO3)2 + NH4NO3}$$

I don't know how to proceed further.

Am I going wrong in any of the above steps? (Please don't convert the reactants and products to ions, and solve it using the method that I am using, so that I can identify my mistake)

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  • $\begingroup$ Your half reactions are not balanced, that's the problem. $\endgroup$ – Ivan Neretin Nov 30 '17 at 10:24
  • $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン Nov 30 '17 at 11:10
  • $\begingroup$ You should convert the reactants and products to ions to identify what actually changes. And you must not not forget the electrons! $\ce{Zn -> Zn^2+ +2e-}$, etc. $\endgroup$ – Martin - マーチン Nov 30 '17 at 11:16
  • $\begingroup$ But my book says when writing half reactions in oxidation number method, only the atoms whose oxidation number changes should be balanced. Well, that's what I did. $\endgroup$ – Piano Land Nov 30 '17 at 11:23
  • $\begingroup$ Still, try to balance the half-reactions before adding them together. I believe that's the easier way. $\endgroup$ – Ivan Neretin Nov 30 '17 at 11:32

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