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When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation

$$\ce{NO3- → NH4+}$$

is balanced into:

$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$

Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. the balanced equation becomes

$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$

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  • $\begingroup$ Most of these redox reactions are done in aqueous medium, thus, there are plenty of $\ce{H2O}$ molecules around to use. $\endgroup$ – Mathew Mahindaratne Jan 20 at 9:12
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It's not that we really have a choice in this case between water or hydroxide or pick one or another arbitrarily to fit the stoichiometry of the whole redox reaction. The redox reaction is obviously taking place in quite a strong acidic medium (makes sense, e.g. if $\ce{Zn}$ is used as reducing agent) as you supply protons $\ce{H+}$ to the nitrate according to your reaction.

In fact, 10 protons for each reduced nitrate group is needed; and even if you propose hydroxide ion formation, you'll still need 7 protons (second reaction). Either way, there is no chance for $\ce{OH-}$ to coexist with $\ce{H+}$ in acidic medium and equilibrium will be shifted towards water formation instead.

Note that it doesn't mean that reduction of nitrate is forbidden if hydroxide ions are formed. For example, you can use another reducing agent such as $\ce{Al}$ and do perform the reduction in basic media instead. In this case you are going to end up with hydroxide ions all right:

$$ \begin{align} \ce{Al + 6 OH- &→ [Al(OH)6]^3- + 3 e-} &|\cdot 8 \tag{ox}\\ \ce{NO3- + 8 e- + 6 H2O &→ NH3 + 9 OH-}&|\cdot 3 \tag{red}\\ \hline \ce{8 Al + 3 NO3- + 21 OH- + 18 H2O &→ 8 [Al(OH)6]^3- + 3 NH3} \tag{redox} \end{align} $$

Just as with the formation of human personality, environment also plays gigantic role in chemistry.

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