How do you balance the following reaction in acidic solution MnO_4^-(aq) + H_2C_2O_4(aq) -> Mn^(2+)(aq) + CO_2(g)?

Apr 19, 2017

$2 M n {O}_{4}^{-} \left(a q\right) + 5 {H}_{2} {C}_{2} {O}_{4} \left(a q\right) + 6 {H}^{+} \left(a q\right) \rightarrow 2 M {n}^{2 +} \left(a q\right) + 10 C {O}_{2} \left(g\right) + 8 {H}_{2} O \left(l\right)$

Explanation:

First off, you need to recognize this reaction as a REDOX reaction.

Then split the equation into its separate half-reactions.

$M n {O}_{4}^{-} \left(a q\right) \rightarrow M {n}^{2 +} \left(a q\right)$
${H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow C {O}_{2} \left(g\right)$

Balance everything except the oxygen and hydrogen.

$M n {O}_{4}^{-} \left(a q\right) \rightarrow M {n}^{2 +} \left(a q\right)$
${H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 C {O}_{2} \left(g\right)$

Now balance the oxygen by adding ${H}_{2} O \left(l\right)$.

$M n {O}_{4}^{-} \left(a q\right) \rightarrow M {n}^{2 +} \left(a q\right) + 4 {H}_{2} O \left(l\right)$
${H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 C {O}_{2} \left(g\right)$

Since this is in an acidic solution, the hydrogen can be balanced by adding ${H}^{+} \left(a q\right)$.

$M n {O}_{4}^{-} \left(a q\right) + 8 {H}^{+} \left(a q\right) \rightarrow M {n}^{2 +} \left(a q\right) + 4 {H}_{2} O \left(l\right)$
${H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 {H}^{+} \left(a q\right)$

Add electrons (${e}^{-}$) in order to balance the overall charge of the two sides of either half-reaction.

$M n {O}_{4}^{-} \left(a q\right) + 8 {H}^{+} \left(a q\right) + 5 {e}^{-} \rightarrow M {n}^{2 +} \left(a q\right) + 4 {H}_{2} O \left(l\right)$
${H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 {H}^{+} \left(a q\right) + 2 {e}^{-}$

Multiply each half-reaction by an integer that will allow the electrons in each half-reaction to cancel out when the half-reactions are added.

$2 M n {O}_{4}^{-} \left(a q\right) + 16 {H}^{+} \left(a q\right) + 10 {e}^{-} \rightarrow 2 M {n}^{2 +} \left(a q\right) + 8 {H}_{2} O \left(l\right)$
$5 {H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 10 C {O}_{2} \left(g\right) + 10 {H}^{+} \left(a q\right) + 10 {e}^{-}$

Finally, add the two half-reactions and simplify.

$2 M n {O}_{4}^{-} \left(a q\right) + 16 {H}^{+} \left(a q\right) + 10 {e}^{-} + 5 {H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 M {n}^{2 +} \left(a q\right) + 8 {H}_{2} O + 10 C {O}_{2} \left(g\right) + 10 {H}^{+} \left(a q\right) + 10 {e}^{-}$

$2 M n {O}_{4}^{-} \left(a q\right) + 6 \cancel{16} {H}^{+} \left(a q\right) + \cancel{10 {e}^{-}} + 5 {H}_{2} {C}_{2} {O}_{4} \left(a q\right) \rightarrow 2 M {n}^{2 +} \left(a q\right) + 8 {H}_{2} O \left(l\right) + 10 C {O}_{2} \left(g\right) + \cancel{10 {H}^{+} \left(a q\right)} + \cancel{10 {e}^{-}}$

$2 M n {O}_{4}^{-} \left(a q\right) + 5 {H}_{2} {C}_{2} {O}_{4} \left(a q\right) + 6 {H}^{+} \left(a q\right) \rightarrow 2 M {n}^{2 +} \left(a q\right) + 10 C {O}_{2} \left(g\right) + 8 {H}_{2} O \left(l\right)$