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    I will get straight to the question.

    "Equation below represents the unbalanced equation for this reaction

    Cu + H+ + NO3- ----> Cu2+ + NO + H2O

    By considering oxidation numbers, balance this equation"

    I know the oxidation numbers. For N it is has gone from +5 to +2..... for Cu it's gone from 0 to +2 but i'm not really sure how to approach this, i know you have to balance out the charges but i have forgotten about this.

    Thanks in advance.
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    (Original post by boromir9111)
    I will get straight to the question.

    "Equation below represents the unbalanced equation for this reaction

    Cu + H+ + NO3- ----> Cu2+ + NO + H2O

    By considering oxidation numbers, balance this equation"

    I know the oxidation numbers. For N it is has gone from +5 to +2..... for Cu it's gone from 0 to +2 but i'm not really sure how to approach this, i know you have to balance out the charges but i have forgotten about this.

    Thanks in advance.
    1. Write out the half equations to get the number of electrons involved in each.
    2. Then multiply the half equations respectively to make the electrons equal in both and then..
    3. Add them together.
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    (Original post by charco)
    1. Write out the half equations to get the number of electrons involved in each.
    2. Then multiply the half equations respectively to make the electrons equal in both and then..
    3. Add them together.
    Yeah, sorry not sure what you mean by electrons involved in each?...... we have H+, NO3- and Cu2+..... i need to do half equations that get me to those?
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    (Original post by boromir9111)
    I will get straight to the question.

    "Equation below represents the unbalanced equation for this reaction

    Cu + H+ + NO3- ----> Cu2+ + NO + H2O

    By considering oxidation numbers, balance this equation"

    I know the oxidation numbers. For N it is has gone from +5 to +2..... for Cu it's gone from 0 to +2 but i'm not really sure how to approach this, i know you have to balance out the charges but i have forgotten about this.

    Thanks in advance.
    4H+ + NO3- + e- ----> NO + 2H2O

    You should be able to figure out the other half eqn and combine them, that completes your redox reaction, then just do what charco stated.
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    (Original post by boromir9111)
    Yeah, sorry not sure what you mean by electrons involved in each?...... we have H+, NO3- and Cu2+..... i need to do half equations that get me to those?
    Yes:

    Cu + H+ + NO3- ----> Cu2+ + NO + H2O

    For example on the LHS you have Cu and on the RHS you have Cu2+ so the half equation is:

    Cu --> Cu2+ + 2e

    the NO3- half equation is a little more complicated as you have to add in H+ ions to absorb the oxygens you don't want (making H2O)

    NO3- + 4H+ + 3e --> NO + 2H2O
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    (Original post by shengoc)
    4H+ + NO3- + e- ----> NO + 2H2O

    You should be able to figure out the other half eqn and combine them, that completes your redox reaction, then just do what charco stated.
    Shengoc, this half equation is wrong,. the sum of the charges on both sides are not equal...

    it should read

    4H+ + NO3- + 3e- ----> NO + 2H2O
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    (Original post by charco)
    Shengoc, this half equation is wrong,. the sum of the charges on both sides are not equal...

    it should read

    4H+ + NO3- + 3e- ----> NO + 2H2O
    hold on, i thought you said it was 4H+ + NO3- + e- ---> NO + 2H2O?
    edit - i see why it's 3e-, is it cause the charge on the right hand side is +2?
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    (Original post by boromir9111)
    hold on, i thought you said it was 4H+ + NO3- + e- ---> NO + 2H2O?
    Read MY posts and you'll see I didn't...

    The half equation needs 3 electrons to balance the charges.
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    (Original post by boromir9111)
    hold on, i thought you said it was 4H+ + NO3- + e- ---> NO + 2H2O?
    edit - i see why it's 3e-, is it cause the charge on the right hand side is +2?
    There is no charge on the RHS - both molecules are neutral (by definition)
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    (Original post by charco)
    Shengoc, this half equation is wrong,. the sum of the charges on both sides are not equal...

    it should read

    4H+ + NO3- + 3e- ----> NO + 2H2O
    i just realised that, i hate that i still don't know how to use the subs/super script on the site, hmm.:rolleyes:
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    Oh ok, so is the overall equation this:

    3Cu + 8H+ 2NO3- ------> 3Cu2+ 2NO + 4H2O?
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    (Original post by shengoc)
    i just realised that, i hate that i still don't know how to use the subs/super script on the site, hmm.:rolleyes:
    use square brackets with [su b] something here [/ sub]

    without the spaces obviously !

    For superscript use 'sup instead of 'sub'
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    (Original post by boromir9111)
    Oh ok, so is the overall equation this:

    3Cu + 8H+ 2NO3- ------> 3Cu2+ 2NO + 4H2O?
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    (Original post by boromir9111)
    Oh ok, so is the overall equation this:

    3Cu + 8H+ 2NO3- ------> 3Cu2+ 2NO + 4H2O?

    What board are you with?
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    (Original post by charco)
    Shengoc, this half equation is wrong,. the sum of the charges on both sides are not equal...

    it should read

    4H+ + NO3- + 3e- ----> NO + 2H2O
    I know i got the answer right but could you just please explain this part again just in case another question like this comes up but in a different way? thanks in advance.
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    (Original post by doginthesky)
    What board are you with?
    OCR
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    (Original post by boromir9111)
    I know i got the answer right but could you just please explain this part again just in case another question like this comes up but in a different way? thanks in advance.
    4H+ + NO3- + 3e- ----> NO + 2H2O

    You are given NO3- on the LHS and you know that it turns to NO on the RHS.

    By inspection it has lost two oxygen atoms, so you must use hydrogen ions to 'mop up' these oxygen atoms and leave water.


    This gives you:


    NO3- + 4H+ --> NO + 2H2O

    Now if you count up the charges on both sides you see that the LHS has 3+ charges and the RHS has none. To make these equal you must add 3 electrons to the LHS.

    This gives you the balanced half equation:

    NO3- + 4H+ + 3e --> NO + 2H2O

    The technique of using water or hydrogen ions to 'provide' or 'mop up' oxygen atoms is something to get used to.

    For example: in the use of SO2 as a reducing agent you get SO42- as the product.

    So the half equation requires you to PROVIDE oxygen atoms (leaving hydrogen ions in the process):

    SO2 + 2H2O --> SO42- + 4H+

    Now you have to balance the charges:

    LHS has no charge but the RHS has 2+ overall, so you have to add 2 electrons to the RHS

    SO2 + 2H2O --> SO42- + 4H+ + 2e

    All sorted..
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    (Original post by charco)
    4H+ + NO3- + 3e- ----> NO + 2H2O

    You are given NO3- on the LHS and you know that it turns to NO on the RHS.

    By inspection it has lost two oxygen atoms, so you must use hydrogen ions to 'mop up' these oxygen atoms and leave water.


    This gives you:


    NO3- + 4H+ --> NO + 2H2O

    Now if you count up the charges on both sides you see that the LHS has 3+ charges and the RHS has none. To make these equal you must add 3 electrons to the LHS.

    This gives you the balanced half equation:

    NO3- + 4H+ + 3e --> NO + 2H2O

    The technique of using water or hydrogen ions to 'provide' or 'mop up' oxygen atoms is something to get used to.

    For example: in the use of SO2 as a reducing agent you get SO42- as the product.

    So the half equation requires you to PROVIDE oxygen atoms (leaving hydrogen ions in the process):

    SO2 + 2H2O --> SO42- + 4H+

    Now you have to balance the charges:

    LHS has no charge but the RHS has 2+ overall, so you have to add 2 electrons to the RHS

    SO2 + 2H2O --> SO42- + 4H+ + 2e

    All sorted..
    Okay, i understand in terms of hydrogen being used to "mop up" oxygen atoms and how that relates to le chatliers principle in a way for me that is. Thanks once again for helping me, much appreciated!!!
 
 
 
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