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Solution
The correct option is CCr2O−27(aq)+3SO−23(aq)+8H+(aq)→2Cr+3(aq)+3SO−24(aq)+4H2O(aq) Steps for Balancing redox reactions:
Identify the oxidation and reduction half .
Find the oxidising and reducing agent.
Find the n-factor of oxidising and reducing agent.
Balance atom undergoing oxidation and reduction.
Cross multiply the oxidising or reducing agent with simplified n-factor values.
Balance atoms other than oxygen and hydrogen.
Balancing oxygen atoms
Balancing hydrogen atoms
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.Product−O.S.Reactant|×number of atom
Following the above mentioned steps:
Balance atom undergoing oxidation and reduction. Cr2O−27+SO−23→2Cr+3+SO−24
Cross multiply the oxidising or reducing agent with simplified n-factor number, Cr2O−27+3SO−23→2Cr+3+3SO−24
Balance oxygen atoms. Cr2O−27+3SO−23→2Cr+3+3SO−24++4H2O
Balance Hydrogen atoms. Cr2O−27+3SO−23+8H+→2Cr+3+3SO−24++4H2O
balance charge
charge in reactant side = 0
charge in product side = 0
so the balanced equation is Cr2O−27+3SO−23+8H+→2Cr+3+3SO−24+4H2O