Question #289b9

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

#"H"_ (2(g)) -> 2"H"_ ((aq))^(+) + 2"e"^(-)#

Notice that hydrogen gas, #"H"_2#, is being converted to hydrogen ions, #"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.

#1/2"O"_ (2(g)) + 2"H"_ ((aq))^(+) + 2"e"^(-) -> "H"_ 2"O"_ ((l))#

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 #color(blue)(+1)#.

On the products' side, hydrogen is once again in a #color(blue)(+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 #color(blue)(0)# on the reactants' side to #color(blue)(-2)# on the products' side.

#1/2 stackrel(color(blue)(0))("O") _ (2(g)) + 2"H"_ ((aq))^(+) + 2"e"^(-) -> "H"_ 2 stackrel(color(blue)(-2))("O") _ ((l))#

Finally, look at the third half-reaction.

#"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 #-># you are once again dealing with an oxidation half-reaction.

More specifically, cadmium, #"Cd"#, is being oxidized from cadmium metal to cadmium cations, #"Cd"^(2+)#.