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

1 Answer
Apr 19, 2017

Answer:

#2MnO_(4)^(-)(aq) + 5H_2C_2O_4(aq) + 6H^+(aq) rarr 2Mn^(2+)(aq) + 10CO_2(g) + 8H_2O(l)#

Explanation:

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

Then split the equation into its separate half-reactions.

#MnO_(4)^(-)(aq)rarrMn^(2+)(aq)#
#H_2C_2O_4(aq)rarrCO_2(g)#

Balance everything except the oxygen and hydrogen.

#MnO_(4)^(-)(aq)rarrMn^(2+)(aq)#
#H_2C_2O_4(aq)rarr2CO_2(g)#

Now balance the oxygen by adding #H_2O(l)#.

#MnO_(4)^(-)(aq)rarrMn^(2+)(aq)+4H_2O(l)#
#H_2C_2O_4(aq)rarr2CO_2(g)#

Since this is in an acidic solution, the hydrogen can be balanced by adding #H^(+)(aq)#.

#MnO_(4)^(-)(aq)+8H^+(aq)rarrMn^(2+)(aq)+4H_2O(l)#
#H_2C_2O_4(aq)rarr2CO_2(g)+2H^+(aq)#

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

#MnO_(4)^(-)(aq)+8H^+(aq)+5e^(-)rarrMn^(2+)(aq)+4H_2O(l)#
#H_2C_2O_4(aq)rarr2CO_2(g)+2H^+(aq)+2e^-#

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.

#2MnO_(4)^(-)(aq)+16H^+(aq)+10e^(-)rarr2Mn^(2+)(aq)+8H_2O(l)#
#5H_2C_2O_4(aq)rarr10CO_2(g)+10H^+(aq)+10e^-#

Finally, add the two half-reactions and simplify.

#2MnO_(4)^(-)(aq)+16H^+(aq)+10e^(-)+5H_2C_2O_4(aq)rarr2Mn^(2+)(aq)+8H_2O+10CO_2(g)+10H^+(aq)+10e^-#

#2MnO_(4)^(-)(aq) + 6cancel(16)H^+(aq) + cancel(10e^(-)) + 5H_2C_2O_4(aq) rarr 2Mn^(2+)(aq) + 8H_2O(l) + 10CO_2(g) + cancel(10H^+(aq)) + cancel(10e^-)#

#2MnO_(4)^(-)(aq) + 5H_2C_2O_4(aq) + 6H^+(aq) rarr 2Mn^(2+)(aq) + 10CO_2(g) + 8H_2O(l)#