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Question

Consider the following redox reaction :

S4O26(aq)+Al(s)H2S(aq)+Al3+(aq)

What will be the net balanced equation in an acidic medium (by oxidation number method).
 


Your Answer
A
2S4O26(aq)+6Al(s)+15H+2H2S(aq)+6Al3+(aq)+6H2O(l)
Correct Answer
B
S4O26(aq)+6Al(s)+20H+4H2S(aq)+6Al3+(aq)+6H2O(l)
Your Answer
C
3S4O26(aq)+6Al(s)+2H+3H2S(aq)+6Al3+(aq)+6H2O(l)
Your Answer
D
S4O26(aq)+5Al(s)+20H+4H2S(aq)+5Al3+(aq)+4H2O(l)

Solution

The correct option is B S4O26(aq)+6Al(s)+20H+4H2S(aq)+6Al3+(aq)+6H2O(l)
+2.5S4O26(aq)+0Al(s)H22S(aq)++3Al3+(aq)

S4O26 is oxidising agent
Al is reducing agent.

nf=(|O.S.ProductO.S.Reactant|×number of atom

+2.5S4O26(aq)H22S(aq)  oxidation
nf=(|22.5|×4=18

0Al(aq)+3Al3+(aq)  reduction

nf=(|30|×1=3
 Ratio of nf  for oxidation to reduction is 6:1

Cross mutiply the oxidising and reducing agents with ratio of n-factors.

 S4O26(aq)+6 Al(s)H2S(aq)+Al3+(aq)
Balancing elements except O and H

S4O26(aq)+6Al(s)4H2S(aq)+6Al3+(aq)

Balance atoms oxygen by adding H2O.
S4O26(aq)+6Al(s)4H2S(aq)+6Al3+(aq)+6H2O(l)

Balance hydrogen by adding H+
S4O26(aq)+6Al(s)+20H+4H2S(aq)+6Al3+(aq)+6H2O(l)

Balancing charge:
 
charge in reactant side = -2+20=+18
charge in product side = +18
so the balanced equation is 

S4O26(aq)+6Al(s)+20H+4H2S(aq)+6Al3+(aq)+6H2O(l)

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