In the an initial half-reaction, the oxidation number of hydrogen goes from #color(blue)(+1)# ~ above the reactants" next to #color(blue)(0)# ~ above the products" side, therefore you have the right to say the water is being reduced to hydrogen gas.

You are watching: Complete and balance the following half-reaction: o2(g)→h2o(l) (basic solution)

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

Now, every atom of hydrogen is gaining #1# electron, therefore you can say the #2# atoms of hydrogen will 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 atom of oxygen, you need to include water molecule to the side that demands atoms the oxygen and also protons, #"H"^(+)#, to the next that needs atoms that 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 proton by including hydroxide anions, #"OH"^(-)#, to both sides of the equation. The hydroxide anions and also the protons will incorporate to kind 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 the the half-reaction is well balanced in terms of charge since 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 should assume the it"s taking place in basic medium because that"s really not the case. Methanol can be oxidized come formaldehyde, but the reaction must take ar in acidic medium.

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

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

In the second half-reaction, the oxidation number of carbon goes indigenous #color(blue)(-2)# ~ above the reactants" side to #color(blue)(0)# top top the products" side, so you have the right to say that methanol is gift oxidized come formaldehyde.

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

In this case, every atom the carbon loser #2# electrons, and since you have an atom of carbon top top both political parties 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 atom of oxygen are currently balanced, however the atom 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 location in basic medium, so add hydroxide anions to both political parties 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))#