Question #92bc1
1 Answer
Explanation:
Start by assigning oxidation numbers to the atoms that take part in the half-reaction
#stackrel(color(blue)(+7))("I"_2) stackrel(color(blue)(-2))("O"_7) -> stackrel(color(blue)(+3))("I") stackrel(color(blue)(-2))("O"_2^(-))#
You're dealing with a reduction half-reaction in which iodine's oxidation state changes from
The problem tells you to note the charges because you need to make sure that once you balance the half-reaction, the overall charge on the reactants' side must be equal to the overall charge on the products' side.
So, start by balancing the iodine atoms
#stackrel(color(blue)(+7))("I"_2)"O"_7 -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#
Now, each iodine atom will gain
#stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#
Since you're in acidic solution, you can balance the oxygen atoms by adding water molecules to the side that needs oxygen, and the hydrogen atoms by adding protons,
In this case, you have
#stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-) + 3"H"_2"O"#
To balance the hydrogen atoms, add
#6"H"^(+) + stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#
Now make sure that the charge is balanced. On the reactants' side, you have
On the products' side, you have
The balanced half-reaction will thus be
#6"H"^(+) + "I"_2"O"_7 + 8"e"^(-) -> 2"IO"_2^(-) + 3"H"_2"O"#