Question #5e7c6
1 Answer
Here's what I got.
Explanation:
Start by writing the unbalanced chemical equation and assigning oxidation numbers to all the atoms that take part in the reaction
#stackrel(color(blue)(+7))("Cl")stackrel(color(blue)(-2))("O")""_ (4(aq))^(-) + stackrel(color(blue)(+4))("N")stackrel(color(blue)(-2))("O")_ (2(aq)) -> stackrel(color(blue)(-1))("Cl")""_ ((aq))^(-) + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O")"" _(3(aq))^(-)#
Notice that the oxidation number of chlorine is going from
On the other hand, the oxidation number of nitrogen is going from
The reduction half-reaction looks like this--I'll skip the states from this point on to keep the equations simple
#stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)#
Since you're in acidic solution, you can balance the oxygen atoms by adding water molecules to the side that needs oxygen and hydrogen ions, or protons,
You should end up with
#8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"#
Notice that the charge is balanced because you have
#8 * (1+) + (1-) + 8 * (1-) = (1-)#
The oxidation half-reaction looks like this
#stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-)#
Once again, balance the oxygen atoms by adding water and the hydrogen atoms by adding protons.
You should end up with
#"H"_ 2"O" + stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-) + 2"H"^(+)#
Once again, the charge is balanced because you have
#0 = (1-) + (1-) + 2 * (1+)#
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 the electrons out, multiply the oxidation half-reaction by
#{(8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"), (color(white)(aaaaaaa)"H"_ 2"O" + stackrel(color(blue)(+4))("N") "O"_ 2 -> stackrel(color(blue)(+5))("N") "O"_ 3^(-) + "e"^(-) + 2"H"^(+)color(white)(aaaa)| xx 8) :}#
and add the two half-reactions to get
#{(8"H"^(+) + stackrel(color(blue)(+7))("Cl")"O"_ 4^(-) + 8"e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-) + 4"H"_ 2"O"), (color(white)(aaaaa)8"H"_ 2"O" + 8stackrel(color(blue)(+4))("N") "O"_ 2 -> 8stackrel(color(blue)(+5))("N") "O"_ 3^(-) + 8"e"^(-) + 16"H"^(+)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#8"H"^(+) + "ClO"_ 4^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 8"H"_ 2"O" + 8"NO"_ 2 -> "Cl"^(-) + 4"H"_ 2"O" + 8"NO"_ 3^(-) + color(red)(cancel(color(black)(8"e"^(-)))) + 16"H"^(+)#
This is equivalent to
#"ClO"_ (4(aq))^(-) + 4"H"_ 2"O"_ ((l)) + 8"NO"_ (2(aq)) -> "Cl"_ ((aq))^(-) + 8"NO"_ (3(aq))^(-) + 8"H"_ ((aq))^(+)#