Question

For the redox reaction: IO−3(aq)+Rd(s)+H2O→RdO−4(aq)+I−(aq)+H+ The correct coefficient of the IO−3, H2O and H+ is :5, 1, 107, 3, 61, 5, 107, 6, 4

Solution

The correct option is B 7, 3, 6Steps 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 +5IO−3(aq)+0Rd(s)→+7RdO−4(aq)+−1I−(aq) 0Rd(s)→+7RdO−4(aq)+  oxidation nf=(|7−0|×1=7 +5IO−3(aq)→−1I−(aq)  reduction nf=(|−1−5|×1=6 Rd is reducing agent IO−3 is oxidising agent. Balance atom undergoing oxidation and reduction. Already balanced IO−3(aq)+Rd(s)→RdO−4(aq)+I−(aq) Cross mutiply the oxidising or reducing agent with n-factor. 7IO−3(aq)+6Rd(s)→6RdO−4(aq)+7I−(aq) Balance atoms oxygen. 7IO−3(aq)+6Rd(s)+3H2O→6RdO−4(aq)+7I−(aq) Balance hydrogen. 7IO−3(aq)+6Rd(s)+3H2O→6RdO−4(aq)+7I−(aq)+6H+ balance charge  charge in reactant side = -7 charge in product side = -7 so the balanced equation is 7IO−3(aq)+6Rd(s)+3H2O→6RdO−4(aq)+7I−(aq)+6H+

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