Question

Question #e27c7

Answer 1

`4"Fe"_ ((aq))^(2+) + "O"_ (2(g)) + 4"H"_ ((aq))^(+) -> 4"Fe"_ ((aq))^(3+) + 2"H"_ 2"O"_((l))`

Explanation:

You're dealing with a redox reaction, so right from the start you know that you should look for the species that is being oxidized and for the species that is being reduced.

In this case, the problem tells you that the iron(II) cations, `"Fe"^(2+)`, are being oxidized to iron(III) cations, `"Fe"^(3+)`. Iron's oxidation number will go from `+2` on the reactants' side to `+3` on the products' side.

The oxidation half-reaction will look like this

`"Fe"_ ((aq))^(2+) -> "Fe"_ ((aq))^(3+) + "e"^(-)`

Now focus on finding the species that is being reduced.

Notice that the reaction is said to take place in the presence of dissolved oxygen, `"O"_2`, and in acidic solution, since you also have aqueous hydrogen ions, `"H"^(+)`, present.

In acidic solution, oxygen will be reduced to water. Oxygen's oxidation number will go from `0` on the reactants' side to `-2` on the products' side.

A total of `4` electrons will be gained here, `2` for every atom of oxygen.

The reduction half-reaction will look like this

`"O"_ (2(g)) + 4"H"_ ((aq))^(+) + 4"e"^(-) -> 2"H"_ 2"O"_((l))`

Now, redox reactions must have equal numbers of electrons transferred, i.e. gained in the reduction half-reaction and lost in the oxidation half-reaction.

To get this to happen, multiply the oxidation half-reaction by `4`

`{ (color(white)(aaaaaaaaaaaa)["Fe"_ ((aq))^(2+) -> "Fe"_ ((aq))^(3+) + "e"^(-) ] xx 4), ("O"_ (2(g)) + 4"H"_ ((aq))^(+) + 4"e"^(-) -> 2"H"_ 2"O"_((l))) :}`

Add the two half-reactions to get

`4"Fe"_ ((aq))^(2+) + "O"_ (2(g)) + 4"H"_ ((aq))^(+) + color(red)(cancel(color(black)(4"e"^(-)))) -> 4"Fe"_ ((aq))^(3+) + color(red)(cancel(color(black)(4"e"^(-)))) + 2"H"_ 2"O"_((l))`

The balanced net ionic equation will thus be

`color(green)(|bar(ul(color(white)(a/a)color(black)(4"Fe"_ ((aq))^(2+) + "O"_ (2(g)) + 4"H"_ ((aq))^(+) -> 4"Fe"_ ((aq))^(3+) + 2"H"_ 2"O"_((l)))color(white)(a/a)|)))`