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Question

Consider the redox reaction giben below :
IO3(aq)+Rd(s)+H2ORdO4(aq)+I(aq)+H+
What will be the correct coefficient of the IO3, H2O and H+ respectively.

A
1,5,10
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B
5,1,10
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C
7,3,6
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D
7,6,4
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Solution

The correct option is C 7,3,6
The redox reaction is :
IO3(aq)+Rd(s)+H2ORdO4(aq)+I(aq)+H+
Rd is reducing agent.
IO3 is oxidising agent.
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
Equalising the decrease/increase in oxidation number:
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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