Balance each of the following half-reactions, assuming that they occur in basic solution?

In the first half-reaction, the oxidation number of hydrogen goes from #color(blue)(+1)# on the reactants' side to #color(blue)(0)# on the products' side, so you can say that water is being reduced to hydrogen gas.

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

Now, each atom of hydrogen is gaining #1# electron, so you can say that #2# atoms of hydrogen will gain a total of #2# electrons.

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

Now, in order to balance the atoms of oxygen, you need to add water molecules to the side that needs atoms of oxygen and protons, #"H"^(+)#, to the side that needs atoms of hydrogen.

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

In a basic medium, you must neutralize the protons by adding hydroxide anions, #"OH"^(-)#, to both sides of the equation. The hydroxide anions and the protons will combine to form water.

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

This will get you

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

which can be simplified to

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

Notice that the half-reaction is balanced in terms of charge because you have

#2 xx 0 + 2 xx (1-) -> 2xx 0 + 2 xx (1-)#

#color(white)(a)#
#color(white)(aaaaaaaaaaaa)/color(white)(a)#

SIDE NOTE: For the second half-reaction, you really need to assume that it's taking place in basic medium because that's really not the case. Methanol can be oxidized to formaldehyde, but the reaction must take place in acidic medium.

In this context, balancing the second half-reaction in a basic medium is not very practical, to put it mildly.

#color(white)(aaaaaaaaaaaa)/color(white)(a)#

In the second half-reaction, the oxidation number of carbon goes from #color(blue)(-2)# on the reactants' side to #color(blue)(0)# on the products' side, so you can say that methanol is being oxidized to formaldehyde.

#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _((aq))#

In this case, each atom of carbon loses #2# electrons, and since you have an atom of carbon on both sides of the half-reaction, you can say that you have

#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _((aq)) + 2"e"^(-)#

The atoms of oxygen are already balanced, but the atoms of hydrogen are not, so add #2# protons on the products' side.

#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + 2"H"_ ((aq))^(+)#

Once again, the reaction takes place in basic medium, so add hydroxide anions to both sides to neutralize the protons.

#2"OH"_ ((aq))^(-) + stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + overbrace(2"H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-))^(color(red)(=2"H"_ 2"O" _((l))))#

This will get you

#2"OH"_ ((aq))^(-) + stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + 2"H"_ 2"O"_ ((l))#

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

#2 xx (1-) + 1 xx 0 -> 1 xx 0 + 2 xx (1-) + 2 xx 0#