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

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

Now, each atom of hydrogen is getting #1# electron, so you can say that #2# atoms of hydrogen will certainly gain a complete 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 must add water molecules to the side that needs atoms of oxygen and proloads, #"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 fundamental medium, you need to neutralize the prolots by including hydroxide anions, #"OH"^(-)#, to both sides of the equation. The hydroxide anions and the proloads will incorporate to create 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 certainly 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 have the right to be streamlined 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-reactivity 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 standard medium because that"s really not the case. Methanol can be oxidized to formaldehyde, however the reactivity should take area in acidic medium.

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In this conmessage, balancing the second half-reaction in a simple tool is not extremely valuable, to put it mildly.

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

In the second half-reactivity, the oxidation variety of carbon goes from #color(blue)(-2)# on the reactants" side to #color(blue)(0)# on the products" side, so you have the right to 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 situation, each atom of carbon loses #2# electrons, and also given that you have actually an atom of carbon on both sides of the half-reaction, you have the right to 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 currently 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 reactivity takes location in standard medium, so include hydroxide anions to both sides to neutralize the proloads.

#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 certainly obtain 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))#