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### Topic: Quick Redox question  (Read 11734 times)

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#### kct

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##### Quick Redox question
« on: May 21, 2005, 01:02:30 PM »
So.. If I have a polyatomic ion, when do I break it up and when do I keep it as the polyatomic ion?

Ex. IF I had sulfite ion, SO3 2- + MnO4 - --> SO4 2- + Mn 2+ , do I immedietely break it up so that I have
SO3 2- -- SO4 2-        and MnO4- --> Mn 2+
Or do I have to break this up somemore before doing anything? I'm getting a little confused . Thanks

#### Garneck

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##### Re:Quick Redox question
« Reply #1 on: May 21, 2005, 01:05:10 PM »
You shouldn't break up any polyatomic anions.

And yes. You have to balance the two equations first, before you write the whole equation.
« Last Edit: May 21, 2005, 01:07:21 PM by Garneck »

#### woelen

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##### Re:Quick Redox question
« Reply #2 on: May 23, 2005, 03:41:11 AM »
So.. If I have a polyatomic ion, when do I break it up and when do I keep it as the polyatomic ion?

Ex. IF I had sulfite ion, SO3 2- + MnO4 - --> SO4 2- + Mn 2+ , do I immedietely break it up so that I have
SO3 2- -- SO4 2-        and MnO4- --> Mn 2+
Or do I have to break this up somemore before doing anything? I'm getting a little confused . Thanks

If we are talking about aqueous redox reactions, just look at the input and output-products. No breaking apart of ions.

For making half reactions correct, you have to do the following:
1) Account for all changes of oxidation number. The net change is translated into a number of electrons, either given or taken.
2) Oxo-groups released take up H(+) from solution or release OH(-) in solution. If reaction is in alkaline media, then use H2O/OH(-) for accounting the release of oxogroups. If reaction is in acidic media, then use H(+)/H2O for accounting the release of oxo-groups.
3) Oxo-groups taken produce H(+) or take OH(-).

The reaction of sulfite and permanganate to sulfate and manganous ions is in acidic environment, so bookkeeping must be done with H(+) ions and H2O.

For sulfite:

S goes from ox.state +4 to +6. O does not change oxidation state. Net result: release of 2 electrons.
SO3(2-) changes to SO4(2-). Uptake of one oxo-group. This releases two H(+).
Formal half-reaction:

SO3(2-) + H2O --> SO4(2-) + 2H(+) + 2e(-)

For permanganate:

Mn goes from ox.state +7 to ox.state +2. O does not change oxidation state. Net result: uptake of 5 electrons.
MnO4(-) changes to Mn(2+). Release of four oxo-groups. This requires eight H(+).
Formal half reaction:

MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O

Now, for setting up the whole reaction, multiply both half reactions with the smallest possible numbers, such that there are integral coefficients and the number of electrons on both sides of the equations is the same. We need to multiply the equation for sulfite with 5 and the equation for permanganate with 2. In that case we have 10 electrons on both sides of the arrow. This cancels the charge. So, we get:

5SO3(2-) + 5H2O     +        2MnO4(-) + 16H(+)   -->  5SO4(2-) + 10H(+)       +       2Mn(2+) + 8H2O

This equation can be simplyfied by leaving out H2O and H(+) which appears at both sides of the arrow.

5SO3(2-)  +  2MnO4(-) + 6H(+)   -->  5SO4(2-)  +  2Mn(2+) + 3H2O

I hope, this helps you somewhat.

Want to wonder? See http://www.oelen.net/science