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    question: complete the ionic equation between iodine and thiosulfate ions.

    I2(aq) + 2S2O32-(aq) ---> ???

    this is in an as level exam and we are not taught this equation, therefore there must be a method of working out what the products are?

    Any ideas? thanks

    I would like the method of how to work it out.. I am so bad at these..
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    (Original post by upthegunners)
    question: complete the ionic equation between iodine and thiosulfate ions.

    I2(aq) + 2S2O32-(aq) ---> ???

    this is in an as level exam and we are not taught this equation, therefore there must be a method of working out what the products are?

    Any ideas? thanks

    I would like the method of how to work it out.. I am so bad at these..
    2S2O32¯(aq) + I2(aq) → S4O62¯(aq) + 2I¯(aq)

    The thiosulfate is a reducing agent, that's just something to learn. Hence it's reducing the iodine to iodide ions.
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    (Original post by LeonVII)
    2S2O32¯(aq) + I2(aq) → S4O62¯(aq) + 2I¯(aq)

    The thiosulfate is a reducing agent, that's just something to learn. Hence it's reducing the iodine to iodide ions.
    I thought iodine got oxidised easily so therefore the Iodine would be the reducing agent? how on earth do you decide what number to put after S and O on the RHS of the equation? :lol: This is why I hate Chemistry..
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    (Original post by upthegunners)
    I thought iodine got oxidised easily so therefore the Iodine would be the reducing agent? how on earth do you decide what number to put after S and O on the RHS of the equation? :lol: This is why I hate Chemistry..
    As thisoulfate is the reduing agent, it itself gets oxidised: in the simplest sense it gains oxygen.

    Now if you look closely, the S406 is double the S2O3. This is essentially intuitive, you wouldnt want to start mixing up weird ratios like making the product S3O7 or something! Granted, a lot of chemistry is just gut feeling and being familiar with equations
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    (Original post by LeonVII)
    As thisoulfate is the reduing agent, it itself gets oxidised: in the simplest sense it gains oxygen.

    Now if you look closely, the S406 is double the S2O3. This is essentially intuitive, you wouldnt want to start mixing up weird ratios like making the product S3O7 or something! Granted, a lot of chemistry is just gut feeling and being familiar with equations
    but no more S or O atoms have been added to the equation?
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    (Original post by upthegunners)
    but no more S or O atoms have been added to the equation?
    They have to the product and then you put a 2 infront of the initial reactant to balance it out.
 
 
 
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