You’re dealing with a redox reaction in which nitric acid oxidizes elemental sulfur to sulfuric acid, #”H”_2″SO”_4#, while being reduced to nitrogen dioxide, #”NO”_2#, in the process.
Start by assigning oxidation numbers to the atoms that take place in the reaction
Bạn đang xem: How would you balance the following equation: #"S" + "HNO"_3 -> "H"_2"SO"_4 + "NO"_2 + "H"_2"O"# ?
#stackrel(color(blue)(0))(“S”)_ ((s)) + stackrel(color(blue)(+1))(“H”) stackrel(color(blue)(+5))(“N”) stackrel(color(blue)(-2))(“O”)_ (3(aq)) -> stackrel(color(blue)(+1))(“H”)_ 2 stackrel(color(blue)(+6))(“S”) stackrel(color(blue)(-2))(“O”)_ (4(aq)) + stackrel(color(blue)(+4))(“N”) stackrel(color(blue)(-2))(“O”)_ (2(g)) + stackrel(color(blue)(+1))(“H”)_ 2 stackrel(color(blue)(-2))(“O”)_ ((l))#
Notice that the oxidation state of nitrogen goes from #color(blue)(+5)# on the products’ side to #color(blue)(+4)# on the reactants’ side, which means that nitrogen is being reduced.
On the other hand, the oxidation state of sulfur goes from #color(blue)(0)# on the reactants’ side to #color(blue)(+6)# on the products’ side, which means that sulfur is being oxidized.
The oxidation half-reaction looks like this
#stackrel(color(blue)(0))(“S”)_ ((s)) -> “H”stackrel(color(blue)(+6))(“S”) “O”_ (4(aq))^(-) + 6″e”^(-)#
Balance the oxygen atoms by using water molecules.
#4″H”_ 2″O”_ ((l)) + stackrel(color(blue)(0))(“S”)_ ((s)) -> “H”stackrel(color(blue)(+6))(“S”) “O”_ (4(aq))^(-) + 6″e”^(-)#
To balance the hydrogen atoms, add protons, #”H”^(+)#, to the side that needs hydrogen atoms.
#4″H”_ 2″O”_ ((l)) + stackrel(color(blue)(0))(“S”)_ ((s)) -> “H”stackrel(color(blue)(+6))(“S”) “O”_ (4(aq))^(-) + 6″e”^(-) + 7″H”_ ((aq))^(+)#
The reduction half-reaction looks like this
#stackrel(color(blue)(+5))(“N”)”O”_ (3(aq))^(-) + “e”^(-) -> stackrel(color(blue)(+4))(“N”) “O”_ (2(g))#
Once again, balance the oxygen atoms by adding water molecules.
#stackrel(color(blue)(+5))(“N”)”O”_ (3(aq))^(-) + “e”^(-) -> stackrel(color(blue)(+4))(“N”) “O”_ (2(g)) + “H”_ 2″O”_ ((l))#
Balance the hydrogen atoms by adding protons.
#2″H”_ ((aq))^(+) + stackrel(color(blue)(+5))(“N”)”O”_ (3(aq))^(-) + “e”^(-) -> stackrel(color(blue)(+4))(“N”) “O”_ (2(g)) + “H”_ 2″O”_ ((l))#
Now, in any redox reaction ,the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.
To balance out the number of electrons transferred, multiply the reduction half-reaction by #6#. Add the two half-reactions to get
#{(color(white)(aaaaaaa.)4″H”_ 2″O”_ ((l)) + stackrel(color(blue)(0))(“S”)_ ((s)) -> “H”stackrel(color(blue)(+6))(“S”) “O”_ (4(aq))^(-) + 6″e”^(-) + 7″H”_ ((aq))^(+)), (2″H”_ ((aq))^(+) + stackrel(color(blue)(+5))(“N”)”O”_ (3(aq))^(-) + “e”^(-) -> stackrel(color(blue)(+4))(“N”) “O”_ (2(g)) + “H”_ 2″O”_ ((l))” “| xx 6) :}# #color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#4″H”_ 2″O”_ ((l)) + “S”_ ((s)) + 12″H”_ ((aq))^(+) + 6″NO”_ (3(aq))^(-) + color(red)(cancel(color(black)(6″e”^(-)))) -> “HSO”_ (4(aq))^(-) + 6″NO”_ (2(g)) + color(red)(cancel(color(black)(6″e”^(-)))) + 7″H”_ ((aq))^(+) + 6″H”_ 2″O”_ ((l))#
This will be equivalent to
#color(green)(bar(ul(|color(white)(a/a)color(black)(“S”_ ((s)) + 6″NO”_ (3(aq))^(-) + 5″H”_ ((aq))^(+) -> “HSO”_ (4(aq))^(-) + 6″NO”_ (2(g)) uarr + 2″H”_ 2″O”_ ((l)))color(white)(a/a)|)))#
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