Find Oxidation State of S in S2O32- Reducing Cl2

[Michael_Light]

Homework Statement

One mole of aqueous S₂O₃^2- 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 S₂O₃^2- = +2

since 4moles of Cl₂ are reduced, 8 electrons are released by S₂O₃^2- ions.

these are all information i can get, but i have no ideas how to proceed, please help me..

 

[dextercioby]

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 ?

 

[Michael_Light]

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 S₂O₃^2-

but the problem is how i find the number of e^- released by S₂O₃^2- from the information given?

Please guide me..

 

[dextercioby]

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 ?

 

[DeeNos]

To find the oxidation state of the sulfur-containing product in this reaction, we first need to identify what the product is. The product of this reaction is likely to be sulfur dioxide (SO2) since it is a common product of the reduction of sulfur-containing compounds.

To determine the oxidation state of sulfur in SO2, we can use the following equation:

Oxidation state of sulfur + 2 x oxidation state of oxygen = 0

Since oxygen has an oxidation state of -2 in SO2, we can solve for the oxidation state of sulfur:

Oxidation state of sulfur + 2 x (-2) = 0
Oxidation state of sulfur = +4

Therefore, the oxidation state of sulfur in the product of this reaction is +4. This indicates that sulfur has been oxidized in the reaction.