# Question 412b9

Nov 22, 2017

Here's what I got.

#### Explanation:

The first thing to notice here is that copper is being reduced from copper(II) cations to copper(I) cations, which implies that nitrogen is being oxidized in this redox reaction.

More specifically, the oxidation number of nitrogen goes from $\textcolor{b l u e}{+ 4}$ in nitrogen dioxide to $\textcolor{b l u e}{+ 5}$ in the nitrate anions.

Also, notice that you have some protons present on the products' side; this tells you that the reaction is taking place in acidic medium.

stackrel(color(blue)(+2))("Cu") ""_ ((aq))^(2+) + "e"^(-) -> stackrel(color(blue)(+1))("Cu") ""_ ((aq))^(+)

Each copper(II) cations gains $1$ electron to form copper(I) cations. The half-reaction is balanced in terms of charge because you have

$\left(2 +\right) + \left(1 -\right) \to \left(1 +\right)$

The oxidation half-reaction looks like this

stackrel(color(blue)(+4))("N")"O"_ (2(aq)) -> stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-)

Here each nitrogen atom loses $1$ electron to go from an oxidation state of $\textcolor{b l u e}{+ 4}$ on the reactants' side to an oxidation state of $\textcolor{b l u e}{+ 5}$ on the products' side.

In order to balance the atoms of oxygen, add water molecules to the side that needs oxygen and protons, ${\text{H}}^{+}$, to the other side.

In your case, you will have

${\text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(+4))("N")"O"_ (2(aq)) -> stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-)+ 2"H}}_{\left(a q\right)}^{+}$

The half-reaction is balanced in terms of charge because you have

$0 \to \left(1 -\right) + \left(1 -\right) + 2 \times \left(1 +\right)$

You can thus say that the two half-reactions are

{(color(white)(aaaaa)stackrel(color(blue)(+2))("Cu") ""_ ((aq))^(2+) + "e"^(-) -> stackrel(color(blue)(+1))("Cu") ""_ ((aq))^(+)), ("H"_ 2"O"_ ((l)) + stackrel(color(blue)(+4))("N")"O"_ (2(aq)) -> stackrel(color(blue)(+5))("N")"O"_ (3(aq))^(-) + "e"^(-)+ 2"H"_ ((aq))^(+)) :}#