Question aec6a

Sep 11, 2017

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

Explanation:

$\text{Oxidation:}$ $C \left(+ I I I\right) \rightarrow C \left(+ I V\right)$

${C}_{2} {O}_{4}^{2 -} \rightarrow 2 C {O}_{2} \left(g\right) + 2 {e}^{-}$ $\left(i\right)$

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

$\text{Reduction:}$ $F e \left(+ I I I\right) \rightarrow F e \left(+ I I\right)$

$F {e}^{3 +} + {e}^{-} \rightarrow F {e}^{2 +}$ $\left(i i\right)$

We add $\left(i\right)$ and $\left(i i\right)$ 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. $\left(i\right) + 2 \times \left(i i\right)$ gives......

${\underbrace{2 F {e}^{3 +} + {C}_{2} {O}_{4}^{2 -} + \cancel{2 {e}^{-}} \rightarrow 2 F {e}^{2 +} + 2 C {O}_{2} \left(g\right) + \cancel{2 {e}^{-}}}}_{\text{to give finally}}$

$2 F {e}^{3 +} + {C}_{2} {O}_{4}^{2 -} \rightarrow 2 F {e}^{2 +} + 2 C {O}_{2} \left(g\right)$

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

Do you follow?

Sep 12, 2017

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

${\text{C"_2"O"_4^"2-" → "CO}}_{2}$
$\text{Fe"^"3+" → "Fe"^"2+}$

Step 2: Balance all atoms other than $\text{H}$ and $\text{O}$

${\text{C"_2"O"_4^"2-" → "2CO}}_{2}$
$\text{Fe"^"3+" → "Fe"^"2+}$

Step 3: Balance $\text{O}$

Done.

Step 4: Balance $\text{H}$

Done.

Step 5: Balance charge

$\text{C"_2"O"_4^"2-" → "2CO"_2 +2"e"^"-}$
$\text{Fe"^"3+" +"e"^"-" → "Fe"^"2+}$

We see that $\boldsymbol{\textcolor{red}{\text{C"_2"O"_4^"2-"color(white)(l) "is oxidized}}}$ because it loses electrons, and $\boldsymbol{\textcolor{red}{\text{Fe"^"3+"color(white)(l)"is reduced}}}$ because it gains an electron.

There are $\boldsymbol{\textcolor{red}{\text{no disproportionations}}}$.

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"^"-")))#
$\underline{\text{2Fe"^"3+" + color(red)(cancel(color(black)("2e"^"-"))) → "2Fe"^"2+} \textcolor{w h i t e}{m m m m m}}$
$\text{C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+}$

Step 8: Check mass balance

$\underline{m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)}$
$\textcolor{w h i t e}{m l} \text{C} \textcolor{w h i t e}{m m m m m} 2 \textcolor{w h i t e}{m m m m m m m} 2$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m m} 4 \textcolor{w h i t e}{m m m m m m m} 4$
$\textcolor{w h i t e}{m l} \text{Fe} \textcolor{w h i t e}{m m m m l l} 2 \textcolor{w h i t e}{m m m m m m m} 2$

Step 9: Check charge balance

$\underline{\boldsymbol{\text{On the left"color(white)(m)"On the right}}}$
$\textcolor{w h i t e}{m m l l} \text{4+"color(white)(mmmmmm)"4+}$

The balanced equation is

$\textcolor{red}{\text{C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+}}$