Answer to Question #176425 in General Chemistry for anonymous

Question

Write the balanced equation for the following reaction, using the half reaction method.

Please note that you must write down all the steps that you took to balance the reaction in detail and explain your working.

 

HCOOH + MnO₄^- → CO₂ + Mn²⁺ in acidic solution

Answer 1

Separating the whole reaction into two half-reactions

HCOOH → CO2 MnO4- → Mn2+

Balancing the non-hydrogen and non-oxygen elements first

HCOOH → CO2 MnO4- → Mn2+

(C is balanced) (Mn is balanced)

Balancing oxygen through adding H2O(l) to the side that needs oxygen (1 O : 1 H2O)

HCOOH → CO2 MnO4- → Mn2+ + 4H2O

Balancing hydrogen though adding H+ to the side which needs hydrogen (1 H: 1 H+)

HCOOH → CO2 + 2H+ 8H+ + MnO4- → Mn2+ + 4H2O

Balancing the charges: adding electrons to the more positive side (or less negative side)

HCOOH → CO2 + 2H+ 8H+ + MnO4- → Mn2+ + 4H2O

Reactants Products Reactants Products

HCOOH = 0 2H+ = +2 8H+ =+8 Mn2+ =+2

__MnO4- = -1________________________

Overall = 0 +2 +7 overall = +2

+2e- +5e-

0 +2

Balancing electrons on the two half-reactions

5[HCOOH → CO2 + 2H+ +2e-]

2[5e- + 8H+ + MnO4- → Mn2+ + 4H2O]

5HCOOH → 5CO2 + 10H+ + 10e-

10e- + 16H+ + 2MnO4- → 2Mn2+ + 8H2O

Getting the overall reaction through adding the 2 reaction.

5HCOOH → 5CO2 + 10H+ + 10e-

10e- + 16H+ + 2MnO4- → 2Mn2+ + 8H2O

5HCOOH + 6H+ + 2MnO4- → 2Mn2+ + 8H2O + 5CO2

Balanced Redox Reaction;

5HCOOH + 6H+ + 2MnO4- → 2Mn2+ + 8H2O + 5CO2