Question

Question #aec6a

Answer 1

Well, oxalato ion is OXIDIZED to give carbon dioxide gas.....

Explanation:

`"Oxidation:"` `C(+III) rarrC(+IV)`

`C_2O_4^(2-) rarr 2CO_2(g)+2e^(-)` `(i)`

Mass and charge are balanced so this is a valid oxidation reaction. And for every oxidation there is a corresponding reduction.

`"Reduction:"` `Fe(+III) rarrFe(+II)`

`Fe^(3+)+ e^(-)rarrFe^(2+)` `(ii)`

We add `(i)` and `(ii)` in such a way that the electrons are REMOVED from the equation, and in this way we formalize the electron transfer conceived to occur in a redox reaction, i.e. `(i) +2xx(ii)` gives......

`underbrace(2Fe^(3+)+C_2O_4^(2-) + cancel(2e^(-))rarr2Fe^(2+)+2CO_2(g)+cancel(2e^(-)))_"to give finally"`

`2Fe^(3+)+C_2O_4^(2-) rarr2Fe^(2+)+2CO_2(g)`

Ferric ion is reduced to ferrous ion, and oxalate ion is oxidized to carbon dioxide.

Do you follow?

Answer 2

Here's what I get.

Explanation:

You can find the general technique for balancing redox equations in acid solution here.

Step 1: Write the two half-reactions

`"C"_2"O"_4^"2-" → "CO"_2`
`"Fe"^"3+" → "Fe"^"2+"`

Step 2: Balance all atoms other than `"H"` and `"O"`

`"C"_2"O"_4^"2-" → "2CO"_2`
`"Fe"^"3+" → "Fe"^"2+"`

Step 3: Balance `"O"`

Done.

Step 4: Balance `"H"`

Done.

Step 5: Balance charge

`"C"_2"O"_4^"2-" → "2CO"_2 +2"e"^"-"`
`"Fe"^"3+" +"e"^"-" → "Fe"^"2+"`

We see that `bbcolor(red)("C"_2"O"_4^"2-"color(white)(l) "is oxidized")` because it loses electrons, and `bbcolor(red)("Fe"^"3+"color(white)(l)"is reduced")` because it gains an electron.

There are `bbcolor(red)("no disproportionations")`.

Step 6: Equalize electrons transferred

`1 × ["C"_2"O"_4^"2-" → "2CO"_2 +2"e"^"-"]`
`2 × ["Fe"^"3+" +"e"^"-" → "Fe"^"2+"]`

Step 7: Add the two half-reactions

`"C"_2"O"_4^"2-" → "2CO"_2 +color(red)(cancel(color(black)(2"e"^"-")))`
`ul("2Fe"^"3+" + color(red)(cancel(color(black)("2e"^"-"))) → "2Fe"^"2+"color(white)(mmmmm))`
`"C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+"`

Step 8: Check mass balance

`ul(mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right"))`
`color(white)(ml)"C"color(white)(mmmmm)2color(white)(mmmmmmm)2`
`color(white)(ml)"O"color(white)(mmmmm)4color(white)(mmmmmmm)4`
`color(white)(ml)"Fe"color(white)(mmmmll)2color(white)(mmmmmmm)2`

Step 9: Check charge balance

`ulbb("On the left"color(white)(m)"On the right")`
`color(white)(mmll)"4+"color(white)(mmmmmm)"4+"`

The balanced equation is

`color(red)("C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+")`