Answer to Question #108250 in Inorganic Chemistry for Komila Loona

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 1

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:

I_2(aq) + H₂O_{2 (aq)} → IO₃⁻_(aq) + H₂O_(l)

Сonsider this reaction under acidic conditions:

half-reactions:

I_2(aq)+6H₂O_(l)-10e^-→2IO₃^-_(aq)+12H⁺_{(aq )} -oxidation

H₂O_{2 (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:

I_2(aq)+6H₂O_(l)+5H₂O_{2 (aq)}+10H⁺_(aq)→2IO₃^-_(aq)+12H⁺_{(aq )}+10H₂O_(l)

or after the restriction

I_2(aq)+5H₂O_{2 (aq)}→2IO₃^-_(aq)+2H⁺_{(aq )}+4H₂O_(l)

And finally, we have the balanced redox equation:

I_2(aq) + 5H₂O_2(aq) →2HIO_3(aq) + 4H₂O_(l)

Сonsider this reaction under basic conditions:

half-reactions:

I_2(aq)+12OH^-_(aq)-10e^-→2IO₃^-_(aq)+6H₂O_(l) -oxidation

H₂O_{2 (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:

I_2(aq)+12OH^-_(aq)+5H₂O_{2 (aq)} →2IO₃^-_(aq)+6H₂O_(l) +10OH^-_(aq)

or after the restriction

I_2(aq)+2OH^-_(aq)+5H₂O_2(aq) →2IO₃^-_(aq)+6H₂O_(l)

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

I_2(aq)+2KOH_(aq)+5H₂O_2(aq) →2KIO_3(aq)+6H₂O_(l)