Question

HCOOH + MnO, CO, + Mn2+ (in acidic solution) *

Answer 1

Step 1: Separate the whole reaction into two half-reactions

HCOOH → CO2 MnO4- → Mn2+

Step 2: Balance the non-hydrogen and non-oxygen elements first

HCOOH → CO2 MnO4- → Mn2+

(C is balanced) (Mn is balanced)

Step 3: Balance oxygen by adding H2O(l) to the side that needs oxygen (1 O : 1 H2O)

HCOOH → CO2 MnO4- → Mn2+ + 4H2O

Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)

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

Step 5: Balance the charges: add 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

Step 6: Balance 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

Step 7: Get the overall reaction by adding the two 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