# Write a balanced redox equation for the following in acidic solution?

## ${C}_{2} {O}_{4}^{2 -} + C {r}_{2} {O}_{7}^{2 -} \setminus \to C {O}_{2} + C {r}^{3 +}$ I understand that ${C}_{2} {O}_{4}^{2 -} \setminus \to 2 C {O}_{2} + 2 {e}^{-}$ but I don't know how the $C {r}_{2} {O}_{7}^{2 -}$ part works out. My attempt (where do the electrons go?) $C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} \ldots \setminus \to 2 C {r}^{3 +} + 7 {H}_{2} O$

Jul 17, 2018

Well you gots a redox reaction here, which we address by the methods of half-equations that utilize electrons as stoichiometric particles...

#### Explanation:

Oxalate ion is oxidized to carbon dioxide...

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

We gots here the transition, $C \left(+ I I I\right) \rightarrow C \left(+ I V\right)$

And meanwhile dichromate ion is REDUCED to $C {r}^{3 +}$..the difference in oxidation numbers is as always the number of electrons we invoke....

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \left(l\right)$ $\left(i i\right)$

${\underbrace{C r \left(V I +\right)}}_{\text{orange-red}}$ is REDUCED to

${\underbrace{C r \left(I I I +\right)}}_{\text{green}}$

And for the balanced redox equation we simply add the half-equations in such a way as to eliminate the electrons....i.e. $3 \times \left(i\right) + \left(i i\right)$

$C {r}_{2} {O}_{7}^{2 -} + {C}_{2} {O}_{4}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 6 C {O}_{2} + 6 {e}^{-} + 7 {H}_{2} O \left(l\right)$

....but we may cancel the electrons....

$C {r}_{2} {O}_{7}^{2 -} + {C}_{2} {O}_{4}^{2 -} + 14 {H}^{+} + \cancel{6 {e}^{-}} \rightarrow 2 C {r}^{3 +} + 6 C {O}_{2} + \cancel{6 {e}^{-}} + 7 {H}_{2} O \left(l\right)$

...to give after cancellation...

$C {r}_{2} {O}_{7}^{2 -} + 3 {C}_{2} {O}_{4}^{2 -} + 14 {H}^{+} \rightarrow 2 C {r}^{3 +} + 6 C {O}_{2} + 7 {H}_{2} O \left(l\right)$

And so is this balanced with respect to MASS and charge...?

We could even further simplify this...by treating the $C r \left(V I\right)$ as chromic acid, ${H}_{2} C {r}_{2} {O}_{7}$, and the oxalate ion as oxalic acid...$H \left(O =\right) C - C \left(= O\right) O H$...and deploying ${H}^{+}$ as $H C l$.

${H}_{2} C {r}_{2} {O}_{7} + 3 {C}_{2} {O}_{4} {H}_{2} + 6 H C l \rightarrow 2 C r C {l}_{3} + 6 C {O}_{2} + 7 {H}_{2} O \left(l\right)$