Find Oxidation State of S in S2O32- Reducing Cl2

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The oxidation state of sulfur in S2O32- is determined to be +2. In the reaction, one mole of S2O32- reduces four moles of Cl2, releasing a total of 8 electrons. The oxidation state of chlorine in the resulting compound is -1, as each chlorine atom is reduced from an oxidation state of 0. To maintain charge balance, the number of electrons released by S2O32- must equal the number of electrons gained by chlorine. Understanding the electron transfer is key to solving the problem.
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Homework Statement



One mole of aqueous S2O32- ions reduces four moles of Cl2 molecules. What is the oxidation state of the sulphur containing product of this reaction?

Homework Equations





The Attempt at a Solution



oxidation state in S2O32- = +2

since 4moles of Cl2 are reduced, 8 electrons are released by S2O32- ions.

these are all information i can get, but i have no ideas how to proceed, please help me..
 
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Write down the electronic balance and use the fact that the total nr of chlorine atoms should be conserved.

S_{2}O_{3}^{2-} + 4 Cl_2-----> ...

What's the oxydation state of chlorine in the resulting compound ? How many electrons are changes for all 8 ions ?
 
dextercioby said:
Write down the electronic balance and use the fact that the total nr of chlorine atoms should be conserved.

S_{2}O_{3}^{2-} + 4 Cl_2-----> ...

What's the oxydation state of chlorine in the resulting compound ? How many electrons are changes for all 8 ions ?

Thanks.. but i still don't understand..

I know that number of e- received by chlorine = number of e- released by S2O32-

but the problem is how i find the number of e- released by S2O32- from the information given?

Please guide me..:confused:
 
You should start with the chlorine. You know that 8 <pieces> of chlorine go from oxydation state 0 to oxydation state -1. How many electrons are captured ?
 
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