Grade 11Physical Chemistry
Mno42-+So3- gives Mn2+ + So42- balance eqnPlease balance this redox reaction by half reaction method
1 Answer
keshav agarwal
Oxidation half
SO3- ==>> SO42-
1st step : balance all atoms of rxn except O and H
2nd step of balancing: add H2O molecules on side of Rxn where it is deficiency of oxygen atoms
(H2O molecules = diffrence of oxygen atoms)
so our Rxn would be
SO3- + H2O ==>> SO42-
3rd step : balance hydrogen atoms in Rx half by adding H+ ions on required side
so Rxn will be
SO3- + H2O ==>> SO42- + 2H+
4th step : balance the charge on both sides of rxn half by adding required no. of e-
SO3- + H2O ==>> SO42- + 3e- + 2H+
So our oxidation half is balanced
Reduction half
MnO42-==>> Mn2+
as done in oxidation half
balance reduction half
MnO42- + 8H+ + 4e- ==>> Mn2+ + 4H2O
FINAL Rxn
multiply both halfs by such factors to get equal no. of e- in both halves
like in this question
mutiply ox. half by 4 and red. half by 3
you will get 12e- in both halves
add multiplied rxn halves
you will get
3MnO42- + 4SO3- +16H+ ==>> 3Mn2+ + 4SO42- + 8H2O
Please approve my answer..........................
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