Question

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

`1)` `"NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + "H"_2"O"(g)`
`2)` `"Cu"(s) + "AgNO"_3(aq) -> "Ag"(s) + "Cu"("NO"_3)_2(aq)`

Answer 1

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 `"N"_2"O"` is just `1`.

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

`color(blue)("NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + \mathbf(2)"H"_2"O"(g))`

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 `"NO"_3^(-)` polyatomic ions on the products side, so there must be `2` equivalents of `"AgNO"_3` to balance it (you can't change the number of `"NO"_3` on the reactants side without touching `"Ag"`).

Then, we have unbalanced the `"Ag"`, requiring `2` equivalents of `"Ag"(s)` on the products side as well to rebalance `"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:

`color(green)("Cu"(s) + \mathbf(2)"Ag"^(+)(aq) -> \mathbf(2)"Ag"(s) + "Cu"^(2+)(aq))`

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 (`"Ag"^(+)(aq) + e^(-) -> "Ag"(s)` 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 (`"Cu"(s) -> "Cu"^(2+) + 2e^(-)` is the half reaction), sending electrons through the conductive wire rightwards, thus forming a cation in solution, at the anode (left beaker).