# How de we balance the reaction of iodide and NO to give iodine and ammonium ion?

Jan 2, 2017

You balance $\text{(i) mass}$, and $\text{(ii) charge.}$

#### Explanation:

If charge and mass are not balanced, you know that your equation does not represent reality, because every chemical reaction ever observed conserves mass and charge.

For the redox reaction, (of which of course this is an example), we can consider the oxidation of iodide, and reduction of $N O$ separately:

${I}^{-} \rightarrow \frac{1}{2} {I}_{2} + {e}^{-}$ $\left(i\right)$

Iodide is oxidized to iodine.

$N O + 6 {H}^{+} + 5 {e}^{-} \rightarrow N {H}_{4}^{+} + {H}_{2} O$ $\left(i i\right)$

Nitrogen monoxide, ${N}^{+ I I}$, is reduced to ammonium ion, $N {H}_{4}^{+}$, $\left({N}^{- I I I}\right)$. As always, the difference in oxidation numbers is accounted for by (conceptual) electron transfer.

And the overall equation, $5 \times \left(i\right) + \left(i i\right)$

$N O + 6 {H}^{+} + 5 {I}^{-} \rightarrow N {H}_{4}^{+} + \frac{5}{2} {I}_{2} + {H}_{2} O$ $\left(i i\right)$

So are mass and charge balanced? If not, they should be. What did you see in this reaction?

There are many other examples here of redox reactivity (of course you have to find them!). If there are queries, ask, and someone will help you.