# How do I balance these reactions? Also, how can I write the net ionic reaction for the second one?

## 1) $\text{NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + "H"_2"O} \left(g\right)$ 2) "Cu"(s) + "AgNO"_3(aq) -> "Ag"(s) + "Cu"("NO"_3)_2(aq)

Mar 14, 2016

1)

color(red)(?)"NH"_4"NO"_3(s) stackrel(Delta" ")(->) color(red)(?)"N"_2"O"(g) + color(red)(?)"H"_2"O"(g)

This is a thermal decomposition reaction, with gases formed, so it likely involves heating the pure solid. That's why I put $\Delta$, which means add heat.

Note that the nitrogens are already balanced. There is also an odd number of oxygens. That tells you that the coefficient for $\text{N"_2"O}$ is just $1$.

Therefore, we only touch $\text{H"_2"O}$ and give it a coefficient of $2$. That indeed gives $4$ equivalents of $\text{H}$ on both sides, and $3$ equivalents of $\text{O}$ on both sides.

$\textcolor{b l u e}{\text{NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + \mathbf(2)"H"_2"O} \left(g\right)}$

2)

color(red)(?)"Cu"(s) + color(red)(?)"AgNO"_3(aq) -> color(red)(?)"Ag"(s) + color(red)(?)"Cu"("NO"_3)_2(aq)

We know that there are two ${\text{NO}}_{3}^{-}$ polyatomic ions on the products side, so there must be $2$ equivalents of ${\text{AgNO}}_{3}$ to balance it (you can't change the number of ${\text{NO}}_{3}$ on the reactants side without touching $\text{Ag}$).

Then, we have unbalanced the $\text{Ag}$, requiring $2$ equivalents of $\text{Ag} \left(s\right)$ on the products side as well to rebalance $\text{Ag}$.

color(blue)("Cu"(s) + \mathbf(2)"AgNO"_3(aq) -> \mathbf(2)"Ag"(s) + "Cu"("NO"_3)_2(aq))

This is the complete molecular equation of a redox reaction. You can see that if you wrote this as a net ionic equation, you would get:

$\textcolor{g r e e n}{{\text{Cu"(s) + \mathbf(2)"Ag"^(+)(aq) -> \mathbf(2)"Ag"(s) + "Cu}}^{2 +} \left(a q\right)}$

Suppose we have a copper anode and a silver cathode in an galvanic cell.

In this reaction, what we have happening is that aqueous silver cation is getting reduced in solution ($\text{Ag"^(+)(aq) + e^(-) -> "Ag} \left(s\right)$ is the half-reaction) due to the flow of electrons through the conductive wire rightwards, and thus forming more solid silver at the cathode (right beaker).

The solid copper is getting oxidized (${\text{Cu"(s) -> "Cu}}^{2 +} + 2 {e}^{-}$ is the half reaction), sending electrons through the conductive wire rightwards, thus forming a cation in solution, at the anode (left beaker).