MnO4 - + C2O4 2- + H----->  Mn2+ + CO2 + H2O

Balance the above redox reaction and find the n-factor and equivalent weight of -

  1. MnO4
  2. C2O4 2- 
  3.  Mn2+ 
  4.  CO2 

Dear Student,

We will balance this equation by ion-electron method
Oxidation half reaction is C2O42- CO2 
Oxidation state of C changes from +3 (C2O42-) to +4 (CO2)
Reduction half reaction is MnO4-  Mn+2 
Oxidation state of C changes from +7 (MnO4-) to +2 (Mn+2)

In this process, we balance both the half reaction separately

Oxidation:  C2O42- 2CO2 + 2e-  .............(1)
(Balance C atom by multiplying 2. O atoms get balanced automatically. To balance the charge we add 2e to right hand side)

Reduction: MnO4- + 8H+ + 5e Mn+2 + 4H2O  ...........(2)
(Mn atoms are balanced. Add 4 H2O to balance O-atoms. H atoms was added as H+ on deficient side. Charge was balanced by adding 5e on left hand side )

Since, electrons are not balanced in both the equation, we multiply equation (1) by 5 and equation (2) by 2

      5 C2O42-                   10 CO2 + 10e-
2MnO4- + 16H+ + 10e 2Mn+2 + 8H2O
_________________________________
​​​​​​5 C2O42-  + 2MnO4- + 16H+ 2Mn+2 + 10 CO2 + 8H2O (Overall balanced equation)

n-factor = No. of electrons involved in oxidation/reductionStoichiometric co-effecient
n-factor of  C2O42- = 10/5 = 2
n-factor of MnO4- = 10/2 = 5
n-factor of Mn+2 = 10/2 = 5
n-factor of CO2 = 10/10 = 1

Equivalent weight =molar massn-factor
EMnO4- = 119/5 =23.8
EC2O42- = 88/2 = 44
EMn+2 = 55/5 = 11
ECO2 = 44/1 = 44

Regards

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