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Question

What would be the coefficient of H+ ion in the balanced net ionic equations for the following reactions in acidic solution: S2O23(aq)+Cr2O27(aq)S4O26(aq)+Cr3+(aq)

A
10
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B
8
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C
14
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D
12
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Solution

The correct option is C 14
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
+2S2O23(aq)++6Cr2O27(aq)+2.5S4O26(aq)++3Cr3+(aq)

+2S2O23(s)+2.5S4O26(aq) oxidation
nf=(|+2.52|×2=1
+6Cr2O27(aq)+3Cr3+(aq) reduction
nf=(|36|×2=6
Cr2O27 is oxidising agent
S2O23 is reducing agent.

Balance atom undergoing oxidation and reduction.
2S2O23(aq)+Cr2O27(aq)S4O26(aq)+2Cr3+(aq)

Cross mutiply the oxidising or reducing agent with n-factor.
12S2O23(aq)+Cr2O27(aq)6S4O26(aq)+2Cr3+(aq)

Balance atoms oxygen.
12S2O23(aq)+Cr2O27(aq)6S4O26(aq)+2Cr3+(aq)+7H2O

Balance hydrogen.
12S2O23(aq)+Cr2O27(aq)+14H+6S4O26(aq)+2Cr3+(aq)+7H2O
Balance charge
charge in reactant side = -12
charge in product side = -6
so the balanced equation is 12S2O23(aq)+Cr2O27(aq)+14H+6S4O26(aq)+2Cr3+(aq)+7H2O+6e
the coefficient of H+ is 14.




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