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Question

Balance the following chemical equation:Cr2O27(aq)+SO23(aq)Cr+3(aq)+SO24(aq) (Acidic medium)

A
Cr2O27(aq)+3SO23(aq)+H+(aq)2Cr+3(aq)+3SO24(aq)+4H2O(aq)
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B
Cr2O27(aq)+SO23(aq)+8H+(aq)2Cr+3(aq)+3SO24(aq)+4H2O(aq)
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C
Cr2O27(aq)+3SO23(aq)+8H+(aq)2Cr+3(aq)+3SO24(aq)+4H2O(aq)
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D
Cr2O27(aq)+3SO23(aq)+8H+(aq)Cr+3(aq)+3SO24(aq)+4H2O(aq)
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Solution

The correct option is C Cr2O27(aq)+3SO23(aq)+8H+(aq)2Cr+3(aq)+3SO24(aq)+4H2O(aq)
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 simplified n-factor values.
  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

Following the above mentioned steps:

Balance atom undergoing oxidation and reduction.
Cr2O27+SO232Cr+3+SO24
Cross multiply the oxidising or reducing agent with simplified n-factor number,
Cr2O27+3SO232Cr+3+3SO24
Balance oxygen atoms.
Cr2O27+3SO232Cr+3+3SO24++4H2O
Balance Hydrogen atoms.
Cr2O27+3SO23+8H+2Cr+3+3SO24++4H2O
balance charge
charge in reactant side = 0
charge in product side = 0
so the balanced equation is
Cr2O27+3SO23+8H+2Cr+3+3SO24+4H2O


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