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Question

What would be the coefficient of H+ and H2O respectively in the balanced net ionic equations for the given redox reactions in acidic solution:
S4O26(aq)+Al(s)H2S(aq)+Al3+(aq)

A
20, 6
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B
6, 20
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C
10, 4
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D
4, 10
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Solution

The correct option is A 20, 6
Steps for Balancing redox reactions:
  1. Identify the oxidation and reduction half .
  2. Find the oxidising and reducing agent.
  3. Find the n-factor of oxidising and reducing agent.
  4. Balance atom undergoing oxidation and reduction.
  5. Cross multiply the oxidising or reducing agent with n-factor.
  6. Balance atoms other than oxygen and hydrogen.
  7. Balancing oxygen atoms
  8. Balancing hydrogen atoms
  9. Balance charge
For acidic medium:
As soon as we add x H2O units, we add 2x H+ ions on the opposite side.

nf=(|O.S.ProductO.S.Reactant|×number of atom
+2.5S4O26(aq)+0Al(s)H22S(aq)++3Al3+(aq)

+2.5S4O26(aq)H22S(aq) oxidation
nf=(|22.5|×4=18
0Al(aq)+3Al3+(aq) reduction
nf=(|30|×1=3
S4O26 is oxidising agent
Al is reducing agent.

Balance atom undergoing oxidation and reduction.
S4O26(aq)+Al(s)4H2S(aq)+Al3+(aq)

Ratio of n- factor's =18:3
dividing the n-factors by common value in them. which is 3 in this case.
Ratio of n- factor's =6:1

Cross mutiply the oxidising and reducing agents with ratio of n-factors. and balancing the main elements other than oxygen and hydrogen.

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

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

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

balance 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(aq)
the coefficient of H+ and H2O respectively are 20 and 6.


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