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Question

For the redox reaction:
IO3(aq)+Rd(s)+H2ORdO4(aq)+I(aq)+H+
The correct coefficient of the IO3, H2O and H+ is :
  1. 5, 1, 10
  2. 7, 3, 6
  3. 1, 5, 10
  4. 7, 6, 4


Solution

The correct option is B 7, 3, 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 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
+5IO3(aq)+0Rd(s)+7RdO4(aq)+1I(aq)

0Rd(s)+7RdO4(aq)+  oxidation
nf=(|70|×1=7
+5IO3(aq)1I(aq)  reduction
nf=(|15|×1=6
Rd is reducing agent
IO3 is oxidising agent.

Balance atom undergoing oxidation and reduction.
Already balanced
IO3(aq)+Rd(s)RdO4(aq)+I(aq)


Cross mutiply the oxidising or reducing agent with n-factor.
7IO3(aq)+6Rd(s)6RdO4(aq)+7I(aq)

Balance atoms oxygen.
7IO3(aq)+6Rd(s)+3H2O6RdO4(aq)+7I(aq)

Balance hydrogen.
7IO3(aq)+6Rd(s)+3H2O6RdO4(aq)+7I(aq)+6H+

balance charge 
charge in reactant side = -7
charge in product side = -7
so the balanced equation is 7IO3(aq)+6Rd(s)+3H2O6RdO4(aq)+7I(aq)+6H+
 

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