tassium 1odate (KIO,) is a strong oxidizing agent and will oxidize Fe2+ ions to

Question

tassium 1odate (KIO,) is a strong oxidizing agent and will oxidize Fe2+ ions to Fe3+. In doing so, iodate ion (IO; ) is reduced to elemental iodine (13). a. Use the oxidation state rules (see the Background section) to assign oxidation states to the iodine atoms in iodate ion (IO; ) and iodine (1,). b. The half-reaction for the reduction of iodate is shown below. Use the difference in oxidation states for the iodine atoms in IO;- and I, to determine the number of electrons gained in this half-reaction. Hint: Hydrogen ions (H+) and water molecules (H,0) are required to balance mass and charge. IO; + 6H+ + e- AI, + 3H,0 3. Combine the oxidation half-reaction for Fe2+ (see the Background section) with the reduction half-reaction for iodate (Question 2b) and write the balanced equation for the overall redox reaction of Fe2+ with IO,-. Hint: The number of electrons on each side must cancel out.

Explanation / Answer

Answer – 2) a) We are given iodate ion IO3- is reduced to elemental iodine and need to assign oxidation states to the iodine atoms in iodate ion and in iodine

We know oxidation state rules like for free state element oxidation state is zero, so oxidation state for iodine in elemental iodine has zero oxidation state.

The oxidation state is nothing but the charge on the element in the ion or compound. The oxidation state of oxygen is -2 except in OF2 it is +2, H2O2 it is -1. We know for ion there is sum of the oxidation state is equal to it charge.

So the oxidation state of iodine in iodate ion is as follow

I + (3*-2) = -1

I -6 = -1

I = +5

So, oxidation state of iodine in iodate ion is +5

b) We are given half reaction for the reduction of iodate and need to determine the number of electrons gained in this given half reaction.

IO3- + 6 H+ + ___e- ------> ½ I2 + 3 H2O

We need to balance the charge by adding the number of electrons, so in the reactant side there are

Reactant side = -1+6 = +5 charge

product side = 0 charge

we need same charge on both side, so we need to add 5 electrons on reactant side for making zero charge on right side also

Reactant side = -1+6 = +5 -5 = 0 charge

IO3- + 6 H+ + 5e- ------> ½ I2 + 3 H2O

So, there are 5 electrons gained in this given half reaction.

3) In this we are given half reaction for oxidation from Fe2+ to Fe3+ and from the part 2b we got the half reduction reaction

Oxidation half reaction -   Fe2+ ----> Fe3+ + e-

Reduction half reaction- IO3- + 6 H+ + 5e- ------> ½ I2 + 3 H2O

Now we need to same electrons from both half reaction, so we need multiply oxidation half reaction by 5 and then add both electrons

5 Fe2+ ----> 5 Fe3+ + 5e-

IO3- + 6 H+ + 5e- ------> ½ I2 + 3 H2O

______________________________

5 Fe2+ + IO3- + 6 H+ -----> 5 Fe3+ + ½ I2 + 3 H2O

Overall balanced reaction is as follow –

5 Fe2+ + IO3- + 6 H+ -----> 5 Fe3+ + ½ I2 + 3 H2O

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