Question regarding redox process

[Mathman23]

Hi

I got the following redox-process:

$IO_3^{-} (aq) + HSO_3^{-}(aq) \rightarrow I_2(aq) + SO_4^{2-}(aq)$

How do I balance it?

Any hits will be apriciated :)

Sincerely

Fred

 

[chem_tr]

It is not hard, provided that you correctly find the oxidation states of all atoms involved in the process. You can easily assume that hydrogen is almost always 1+, while oxygen is almost 2-. The remaining atoms are the ones involved in the redox reaction.

In iodate, as oxygen is 2- and the total charge is 1-, you can calculate iodine's oxidation state. Do the same for iodine, bisulfite, and sulfate. Find how many electrons are exchanged between atoms.

Show your work, and we'll go on from there.

 

[Mathman23]

It is not hard, provided that you correctly find the oxidation states of all atoms involved in the process. You can easily assume that hydrogen is almost always 1+, while oxygen is almost 2-. The remaining atoms are the ones involved in the redox reaction.

In iodate, as oxygen is 2- and the total charge is 1-, you can calculate iodine's oxidation state. Do the same for iodine, bisulfite, and sulfate. Find how many electrons are exchanged between atoms.

Show your work, and we'll go on from there.

Hi

Here is what I have been able to come up as far:


$IO_3^{-} (aq) + HSO_{3}^{-}(aq) \rightarrow I_{2}(aq) + SO_4^{2-}(aq)$

Step1.

$IO_3^{-} \rightarrow I_2$

$HSO_3^{-} \rightarrow SO_4^{2-}$

Step2.
$IO_3^{-} + 4H^{+} \rightarrow I_{2} + 2H_2O$
$HSO_3^{-} + H_2 O \rightarrow SO_4^{2-} + 3H^+$

Step3.

$IO_3^{-} + 4H^{+} + 4e^{-} \rightarrow I_2 + 2H_2O$
$HSO_3^{-} + H_2O \rightarrow SO_4^{2-} + 3H^{+} + 3e^{-}$

From here I'm stuck. How do I proceed?

/Fred

 

[so-crates]

Now just balance the e- and H+ ions and add the two reactions together.

 

[Mathman23]

Then what I have done until now is correct?

/Fred

Now just balance the e- and H+ ions and add the two reactions together.

 

[chem_tr]

So-crates is right, maybe you can try one additional step before writing the actual ions: Write them in their plain ionic type (like I⁵⁺ and I⁰) to see how many electrons are leaving exactly. Then convert them to the actual ions.

$$I^{5+} +5e^-\longrightarrow I^0$$

As you know that I⁵⁺ is IO₃^-, you can write it now. But iodine is diatomic, so ten electrons must be processed:

$$2IO_3^- + 10e^- \longrightarrow I_2$$

Now that we balanced the electrons, only atomic balance is left to complete this half redox reaction. Reaction proceeds in acidic medium, so we must include H⁺ on the left side; as oxygen is no longer present in the right side, and we put H⁺, you can easily conclude that water will be present on the right side:

$$2IO_3^-+10e^-+12H^+\longrightarrow I_2+6H_2O$$

At last, you must be sure that both electrons and atoms are balanced. Do the same for bisulfate-sulfate redox and show your work.

 

[Mathman23]

So-crates is right, maybe you can try one additional step before writing the actual ions: Write them in their plain ionic type (like I⁵⁺ and I⁰) to see how many electrons are leaving exactly. Then convert them to the actual ions.

$$I^{5+} +5e^-\longrightarrow I^0$$

As you know that I⁵⁺ is IO₃^-, you can write it now. But iodine is diatomic, so ten electrons must be processed:

$$2IO_3^- + 10e^- \longrightarrow I_2$$

Now that we balanced the electrons, only atomic balance is left to complete this half redox reaction. Reaction proceeds in acidic medium, so we must include H⁺ on the left side; as oxygen is no longer present in the right side, and we put H⁺, you can easily conclude that water will be present on the right side:

$$2IO_3^-+10e^-+12H^+\longrightarrow I_2+6H_2O$$

At last, you must be sure that both electrons and atoms are balanced. Do the same for bisulfate-sulfate redox and show your work.

Hi

I guess the second reaction must then be:

$HSO_3^- \rightarrow SO_4^{2-}$

$S^{5+} + 5e^- \rightarrow S^{4+} + 4e^-$


$HSO_3^{-} + 5e^- \rightarrow SO_4^{2-} + 4e^-$

Correct?

Thanks in advance.

Sincerely
Fred

 

[chem_tr]

I'm afraid these are incorrect; in redox reactions, you place the electrons in one side, not two. If an ion is to be reduced, electrons go into this ion, so electrons are included in the left side of half reaction. The same can be said for oxidations, you write the electron(s) on the right side.

It seems that you have not determined the oxidation states correctly. Take oxygen 2- and hydrogen 1+ and recalculate.

If I understand it correctly, you first convert the ion to be oxidized into its neutral state by reduction, then oxidize. This is not logical, subtracting just two electrons are enough.

 

[pack_rat2]

Recheck the oxidation number of S in the bisulfite!

 

[Mathman23]

Recheck the oxidation number of S in the bisulfite!

Thanks for Your answer.

Then the correct balance must be since its an oxidation:

$H_{2}O + HSO_{3}^{-} \rightarrow SO_{4}^{2-} + 3H+$

But why isn't the oxdidation number for $HSO_{3}^{-}$ 5 since Sulphur is group number 6 and the as You write the oxidation number for O is -2 and for Hydrogen 1 ?

/Fred

 

[chem_tr]

Your redox is incorrect again

If this is an oxidation, where is your electron? The correct one should be like this:

$$H_2O + HSO_3^- \longrightarrow H_2SO_4 + 2e^- + H^+$$

In bisulfite, put 1+ for hydrogen, and 2- for oxygen. You will find 4+ for sulfite sulfur. Do the same thing for sulfate to find 6+. Two electrons must leave bisulfite ion.

 

[pack_rat2]

Suggestion: Determine the oxidation number for each element that is oxidized or reduced, ie, S and I. Remember that O is (almost) always -2 and H is (almost) always +1. Then write the half-reactions for those elements...don't bother including H and O in them.