Answer to Question #108250 in Inorganic Chemistry for Komila Loona

Question #108250
Write the oxidation and reduction half-reactions for this reaction and use them to
balance the redox equation under acidic and basic conditions.
I2(aq) + H2O2 (aq) → IO3

(aq) + H2O(l)
1
Expert's answer
2020-04-07T13:47:32-0400

Question: Write the oxidation and reduction half-reactions for this reaction and use them to

balance the redox equation under acidic and basic conditions.

I2(aq) + H2O2 (aq) → IO3

(aq) + H2O(l)+

Answer:

I2(aq) + H2O2 (aq) → IO3(aq) + H2O(l)

Сonsider this reaction under acidic conditions:

half-reactions:

I2(aq)+6H2O(l)-10e-→2IO3-(aq)+12H+(aq ) -oxidation

H2O2 (aq)+2H+(aq)+2e-→2H2O(l) -reduction

The total number of electrons is 10 - we apply a multiplier of 5 to the second half-reaction.

Full ionic equation:

I2(aq)+6H2O(l)+5H2O2 (aq)+10H+(aq)→2IO3-(aq)+12H+(aq )+10H2O(l)

or after the restriction

I2(aq)+5H2O2 (aq)→2IO3-(aq)+2H+(aq )+4H2O(l)

And finally, we have the balanced redox equation:

I2(aq) + 5H2O2(aq) →2HIO3(aq) + 4H2O(l)


Сonsider this reaction under basic conditions:

half-reactions:

I2(aq)+12OH-(aq)-10e-→2IO3-(aq)+6H2O(l) -oxidation

H2O2 (aq))+2e-→2OH-(aq) -reduction

The total number of electrons is 10 - we apply a multiplier of 5 to the second half-reaction.

Full ionic equation:

I2(aq)+12OH-(aq)+5H2O2 (aq) →2IO3-(aq)+6H2O(l) +10OH-(aq)

or after the restriction

I2(aq)+2OH-(aq)+5H2O2(aq) →2IO3-(aq)+6H2O(l)

And finally, we have the balanced redox equation (KOH as a basic agent):

I2(aq)+2KOH(aq)+5H2O2(aq) →2KIO3(aq)+6H2O(l)

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