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Topic: Redox Reactions, some conceptual explaining.  (Read 4603 times)

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Offline Yusuf

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Redox Reactions, some conceptual explaining.
« on: September 09, 2012, 07:38:15 PM »
MnO4‾(aq) + S2-(aq) → MnS(s) + S(s)



MnO4- ---> Mn2+
S2- ---> S

Why/How can someone remove the S from MnS?

Offline UG

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Re: Redox Reactions, some conceptual explaining.
« Reply #1 on: September 09, 2012, 07:47:28 PM »
I am not sure what you mean, please elaborate.

Offline Yusuf

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Re: Redox Reactions, some conceptual explaining.
« Reply #2 on: September 09, 2012, 07:51:38 PM »
I am not sure what you mean, please elaborate.

In the solution for the redox problem, the guy split up the equations in to two, but he also removed the S from the MN when he splited them up.

Offline UG

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Re: Redox Reactions, some conceptual explaining.
« Reply #3 on: September 09, 2012, 08:03:28 PM »
In the solution for the redox problem, the guy split up the equations in to two, but he also removed the S from the MN when he splited them up.
This redox reaction is not balanced, so in order to balance it, you need to split the reaction into two half reactions, an oxidation half and a reduction half, that is what this guy has done. The reduction half equation is MnO4- ---> Mn2+ and the oxidation half equation is S2- :rarrow: S. Then you balance these two half reactions individually.
Some of the (non oxidised) sulphur remains with Mn in MnS (where S is in the -2 oxidation state) and some of the sulphur exists as elemental sulphur, where is it in the 0 oxidation state.

Offline Yusuf

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Re: Redox Reactions, some conceptual explaining.
« Reply #4 on: September 09, 2012, 08:37:11 PM »
In the solution for the redox problem, the guy split up the equations in to two, but he also removed the S from the MN when he splited them up.
This redox reaction is not balanced, so in order to balance it, you need to split the reaction into two half reactions, an oxidation half and a reduction half, that is what this guy has done. The reduction half equation is MnO4- ---> Mn2+ and the oxidation half equation is S2- :rarrow: S. Then you balance these two half reactions individually.
Some of the (non oxidised) sulphur remains with Mn in MnS (where S is in the -2 oxidation state) and some of the sulphur exists as elemental sulphur, where is it in the 0 oxidation state.


MnO4- ---> Mn2+

So, what you are saying is that if something is not a oxidizing agent or a reducing agent then their is no need to include it in the 2 split up equations?

Offline Vidya

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Re: Redox Reactions, some conceptual explaining.
« Reply #5 on: September 09, 2012, 10:47:45 PM »
MNO4- ---> Mn 2+
it is the reduction reaction and MNO4- is the oxidizing agent.Check how the oxidation number of Mn is changing from left to right.

Offline UG

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Re: Redox Reactions, some conceptual explaining.
« Reply #6 on: September 10, 2012, 12:27:38 AM »
MnO4- ---> Mn2+

So, what you are saying is that if something is not a oxidizing agent or a reducing agent then their is no need to include it in the 2 split up equations?
Yes, most of the time it is not necessary to put in the counter ions which do not take part in the reactions.

Offline Yusuf

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Re: Redox Reactions, some conceptual explaining.
« Reply #7 on: September 10, 2012, 02:06:26 PM »
MnO4- ---> Mn2+

So, what you are saying is that if something is not a oxidizing agent or a reducing agent then their is no need to include it in the 2 split up equations?
Yes, most of the time it is not necessary to put in the counter ions which do not take part in the reactions.

Thank you! Do you know of any Chemistry book that teaches Chemistry like this? Straight forward and understandable.

Offline Borek

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Re: Redox Reactions, some conceptual explaining.
« Reply #8 on: September 10, 2012, 02:11:24 PM »
Every general chemistry book I have seen contained a chapter on net ionic equations.
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