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    I have two questions

    10 (c) (i) Write the two half-equations for the following redox reaction.
    2H+ + 2Br– + H2SO4 -----> Br2 + SO2 + 2H2O

    10 (c) (ii) Write an equation for the reaction of solid potassium chloride with concentrated
    sulfuric acid.

    These are from q10c on http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF



    Now for the first part, I don't understand how I am meant to find out how to write these half equations.

    I wrote down:
    2Br- + H2+ ----> Br2 + H2

    and

    2H+ + SO4^2- ----> SO2 + 2O2

    The actual answers are

    2Br- -----> Br2 + 2e―
    4H+ + SO4^2- + 2e- -----> SO2 + 2H2O


    How am I meant to figure that out :eek: :confused:



    Also for the second part How should I know what the reactants are? The equation isn't even taught at A level, so we have to figure it out somehow

    Actual answer to that is
    KCl + H2SO4 ---> KHSO4 + HCl
    OR
    2KCl + H2SO4 -------> K2SO4 + 2HCl




    Any help will be MUCH appreciated

    (at this rate I am going to fail chem2 REALLY badly :cry2:)
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    (Original post by user1-4)
    I have two questions

    10 (c) (i) Write the two half-equations for the following redox reaction.
    2H+ + 2Br– + H2SO4 Br2 + SO2 + 2H2O

    10 (c) (ii) Write an equation for the reaction of solid potassium chloride with concentrated
    sulfuric acid.

    These are from q10c on http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF



    Now for the first part, I don't understand how I am meant to find out how to write these half equations.

    I wrote down:
    2Br- + H2+ ----> Br2 + H2

    and

    2H+ + SO4^2- ----> SO2 + 2O2

    The actual answers are

    2Br― Br2 + 2e―
    4H+ + SO42― + 2e― SO2 + 2H2O

    How am I meant to figure that out :eek: :confused:



    Also for the second part How should I know what the reactants are? The equation isn't even taught at A level, so we have to figure it out somehow

    Actual answer to that is
    KCl + H2SO4 KHSO4 + HCl
    OR
    2KCl + H2SO4 K2SO4 + 2HCl



    Any help will be MUCH appreciated

    (at this rate I am going to fail chem2 REALLY badly :cry2:)
    Right, what I did to get the correct answers was realise which elements are being oxidised and which are being reduced. In the first one, bromine is oxidised from bromide ions which are negative, to bromine atoms. So, as bromine is diatomic, you need 2Br- to balance out the Br2 on the right hand side. Also, on both sides of the equation, you need to balance the charge. 2Br- will give you an overall charge of -2 on the left side of the equation. To balance this on the right side, you add 2 electrons to the right side. So both sides have overall charges of -2.
    So we’ve established that Bromide ions are oxidised, and there’s another element that gets reduced (so its oxidation state decreases). This is sulphur, from +6 in H2SO4 to +4 in SO2. Basically, you need to learn oxidation number rules, which will help a lot. I’ll attach a copy for you here.

    For the second one, I actually wrote 2H+ + 2e- + H2SO4 ----> SO2 + 2H2O so I don’t get why it’s 4H+.
    For the second question, I think it’s just one of those equations you have to learn. I’ve seen it in textbooks; it’s part of Unit 2 Group 7 reactions with Group 1 metals and acid. It comes up in Multiple choice a lot. It’s because halogens are more oxidising as you go up the group, so it reduces the Sulphuric acid to HSO4 and evolves white misty fumes of HCl gas. Have a look in the CGP revision guide if you have it on page 82, it’s a reaction of the chloride ion.

    Unit 2 inorganic is quite dull, but don’t worry. I failed it the first time and retook, so it can be done.

    Here are some general guidelines to write a redox half equation.
    • If it’s oxidation, electrons go on the right side, as electrons have been removed from the atom/species.
    • If it’s reduction, electrons are on the left hand side, as electrons have been added to the species.
    You must always balance charge on both sides of the equation.
    You can do that in the following ways:
    • By adding H+ to give a more positive charge
    • By adding electrons to lower the charge
    • You can balance oxygen species by adding H2O to the opposite side of the equation
    • You can balance hydrogens by adding H+ and if the H+ makes the charge too positive on one side, you can add electrons to lower that excess charge.
    • Cancel the equation down the stoichiometry where necessary (As in, if the balancing numbers are 2, 4, 8 and 6, you can cancel these to 1, 2, 4 and 3)
    • Start by balancing the element that changes oxidation state.

    For example, let’s say we need to write the half equation for
    VO2 + -----> V 3+

    Firstly, we’d need to balance the vanadiums because that's the element changing oxidation state. But in this case, we don’t have to because there’s one on either side. What else is out of proportion? Oxygen.
    We balance oxygen by adding a water on the right side, so we add 2H2Os to balance 2 oxygens on the right side with the 2 oxygens on the left side.

    Now, hydrogens are out of proportion as there aren’t any on the left side. We correct this by adding 4H+ because there’s a total of 4 hydrogens on the right side now. Now, there’s an inequality of charge. The overall charge on the left is +5, and the overall charge on the right is +3. If we are to balance the charges out, we add 2 electrons onto the left to bring the overall +5 charge down to +3. We’ve now finished balancing the equation because we’ve balanced the charges.

    The final equation is
    2e- + 4H+ +VO2+ -----> V 3+ + 2H2O

    You can see this is right because all charges are balanced, and all elements are balanced as both sides of the equation need to add up to the same numbers of elements and the same charge.
    Attached Files
  1. File Type: docx Oxidation Number Rules[1].docx (13.3 KB, 85 views)
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    (Original post by Mollymod)

    For the second one, I actually wrote 2H+ + 2e- + H2SO4 ----> SO2 + 2H2O so I don’t get why it’s 4H+.
    Because sulphuric acid dissociates to give two hydrogen ions, so you can use these as acid on the LHS and only reduce the sulphate ion.

    H2SO4 --> 2H+ + SO42-



    .. and please be called Molly in real life, 'cause your UserName is that way TPM!!
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    (Original post by charco)
    Because sulphuric acid dissociates to give two hydrogen ions, so you can use these as acid on the LHS and only reduce the sulphate ion.

    H2SO4 --> 2H+ + SO42-



    .. and please be called Molly in real life, 'cause your UserName is that way TPM!!
    Oh right, I see, that makes sense. Thanks
    And no, I'm not called Molly, sorry

    What does TPM mean :P
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    wow, thanks A LOT!! :eek:

    I am still reading through it and trying to understand it. THANK YOU

    EDIT: read through a few times. That helps SOOOOO MUCH! I think every person stuggling on redox needs to see this post.
 
 
 
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