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    Hi,

    I'm trying to revise redox equations but I have become a bit stuck on question 2b on this site:

    http://www.chemguide.co.uk/inorganic...xequations.pdf

    I get up to the bit where you add the 12 H+ ions on the left side (because there's 12 hydrogens on the left so you need to balance them), but I don't know why and how you add the electrons? The mark scheme is here:

    http://www.chemguide.co.uk/inorganic...xequations.pdf

    Thank you
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    (Original post by Sameer599)
    Hi,

    I'm trying to revise redox equations but I have become a bit stuck on question 2b on this site:

    http://www.chemguide.co.uk/inorganic...xequations.pdf

    I get up to the bit where you add the 12 H+ ions on the left side (because there's 12 hydrogens on the left so you need to balance them), but I don't know why and how you add the electrons? The mark scheme is here:

    http://www.chemguide.co.uk/inorganic...xequations.pdf

    Thank you
    Hey!

    You have to add the electrons on left hand side of the equation because its reduction. Remember that the electrons are used to balance the change, so:

    2ClO3- + 12H -----> Cl2 + 6H2O

    The charge on the left hand side of the equation is +10 and on the right hand side its 0. So to make the charges balance, you have to add 10 electrons to the right hand side of the equation which makes it neutral.

    So the final answer is : 2ClO3- + 12H+ + 10e ------> Cl2 + 6H2O

    Hope I helped!
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    (Original post by Nav_Mallhi)
    Hey!

    You have to add the electrons on left hand side of the equation because its reduction. Remember that the electrons are used to balance the change, so:

    2ClO3- + 12H -----> Cl2 + 6H2O

    The charge on the left hand side of the equation is +10 and on the right hand side its 0. So to make the charges balance, you have to add 10 electrons to the right hand side of the equation which makes it neutral.

    So the final answer is : 2ClO3- + 12H+ + 10e ------> Cl2 + 6H2O

    Hope I helped!
    Thanks for your reply!

    But I don't understand how the charge on the left is +10? Isn't it +12 because of the 12 hydrogen ions?
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    (Original post by Sameer599)
    Thanks for your reply!

    But I don't understand how the charge on the left is +10? Isn't it +12 because of the 12 hydrogen ions?
    You also have to take into account the ClO3 which is - and there are two of them, so +12 add -2 is 10
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    (Original post by Nav_Mallhi)
    Hey!

    You have to add the electrons on left hand side of the equation because its reduction. Remember that the electrons are used to balance the change, so:

    2ClO3- + 12H -----> Cl2 + 6H2O

    The charge on the left hand side of the equation is +10 and on the right hand side its 0. So to make the charges balance, you have to add 10 electrons to the right hand side of the equation which makes it neutral.

    So the final answer is : 2ClO3- + 12H+ + 10e ------> Cl2 + 6H2O

    Hope I helped!
    I was struggling as well and that explained it well, thanks

    its 10+ not 12+ because of the 2- from the 2ClO3-?
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    (Original post by Sameer599)
    Thanks for your reply!

    But I don't understand how the charge on the left is +10? Isn't it +12 because of the 12 hydrogen ions?
    You are right that's its +12 but your are forgetting you take into account the charge on 2ClO3- which is -2 because 2x-1= -2

    So look,

    2ClO3- + 12H+ + 10e- ------> Cl2 + 6H2O

    Charge on 12H+ is +12 (1x12)
    The charge on the 2ClO3- is -2 (2x-1)
    So the charge so far is: 12-2 = +10
    And then we add the 10 electrons to make the charge negative overall.

    Are you following me through? And don't hesitate to ask me if you are not sure about what I've done!
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    (Original post by Nav_Mallhi)
    You are right that's its +12 but your are forgetting you take into account the charge on 2ClO3- which is -2 because 2x-1= -2

    So look,

    2ClO3- + 12H+ + 10e- ------> Cl2 + 6H2O

    Charge on 12H+ is +12 (1x12)
    The charge on the 2ClO3- is -2 (2x-1)
    So the charge so far is: 12-2 = +10
    And then we add the 10 electrons to make the charge negative overall.

    Are you following me through? And don't hesitate to ask me if you are not sure about what I've done!
    Oh okay I think I get it, so if it's reduction the electrons go on the left but if it's oxidation it goes on the right? Also, on question 2c, is it 2e- on the left because it's reduction and because on the right there is 2+ on Mn and 4H+ on the left so you need 2e- to balance it out?
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    (Original post by Sameer599)
    Oh okay I think I get it, so if it's reduction the electrons go on the left but if it's oxidation it goes on the right? Also, on question 2c, is it 2e- on the left because it's reduction and because on the right there is 2+ on Mn and 4H+ on the left so you need 2e- to balance it out?
    Yep, that's correct. In reduction the electron are in the left hand side and in oxidation electrons are on the right hand side. And remember you are not always told whether is an oxidation or reduction reaction to work out if the reaction is a oxidation or reduction reaction you have to look at the oxidation states.

    And yes, you need 2 electrons on the left hand side to make the charge +2 overall.
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    (Original post by Nav_Mallhi)
    Yep, that's correct. In reduction the electron are in the left hand side and in oxidation electrons are on the right hand side. And remember you are not always told whether is an oxidation or reduction reaction to work out if the reaction is a oxidation or reduction reaction you have to look at the oxidation states.

    And yes, you need 2 electrons on the left hand side to make the charge +2 overall.
    Thank you so much for your help, I understand it now Do you know ehre I can get questions like these but without the equations given?
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    (Original post by Sameer599)
    Thank you so much for your help, I understand it now Do you know ehre I can get questions like these but without the equations given?
    It's okay
    Exam papers are always a really good resource to use.
 
 
 
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