# Question #289b9

Dec 25, 2016

Here's what I got.

#### Explanation:

Let's take a look at the first half-reaction

${\text{H"_ (2(g)) -> 2"H"_ ((aq))^(+) + 2"e}}^{-}$

Notice that hydrogen gas, ${\text{H}}_{2}$, is being converted to hydrogen ions, ${\text{H}}^{+}$, which, of course, implies that it is losing electrons. This is why electrons are being added to the products' side of the half-reaction.

In other words, you can say that the above half-reaction describes the oxidation of molecular hydrogen to hydrogen ions.

Move on to the second half-reaction.

$\frac{1}{2} {\text{O"_ (2(g)) + 2"H"_ ((aq))^(+) + 2"e"^(-) -> "H"_ 2"O}}_{\left(l\right)}$

This time, you have electrons added to the products' side, so right from the start, you can say that this represents a reduction half-reaction.

Since you're dealing with a half-reaction, only one element will change its oxidation state. In this case, hydrogen is present as ions on the products' side, which implies an oxidation state of $\textcolor{b l u e}{+ 1}$.

On the products' side, hydrogen is once again in a $\textcolor{b l u e}{+ 1}$ oxidation state as part of the water molecule.

You can thus say that this half-reaction describes the reduction of molecular oxygen to water; as a result, oxygen's oxidation state decreases from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{- 2}$ on the products' side.

$\frac{1}{2} {\stackrel{\textcolor{b l u e}{0}}{\text{O") _ (2(g)) + 2"H"_ ((aq))^(+) + 2"e"^(-) -> "H"_ 2 stackrel(color(blue)(-2))("O}}}_{\left(l\right)}$

Finally, look at the third half-reaction.

${\text{Cd"_ ((s)) + 2"OH"_ ((aq))^(-) -> "Cd"("OH") _(2(s)) + 2"e}}^{-}$

Once again, the electrons are added to the products' side, so you know that an element is losing electrons $\to$ you are once again dealing with an oxidation half-reaction.

More specifically, cadmium, $\text{Cd}$, is being oxidized from cadmium metal to cadmium cations, ${\text{Cd}}^{2 +}$.