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    2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g)

    Could anyone explain the oxidation numbers in depth please?
    Thanks, in advance, for taking your time
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    Oxidation states are based on electronegativity, which is the ability of a nuclei to withdraw a region of electrondensity to itself this is very important when dealing with the oxidation state for molecules. The oxidation state is the number of ELECTRONS an ion or atom needs to obtain a NEUTRAL CHARGE. So when you look at the example you have given it deals mainly with ions. Sr(NO3)2 is an ionic compound the NO3^-1 has an oxidation state of -1 since it needs to loose 1 electron to attain a neutral charge. Sr has a 2 charge hence it needs to gain 2 electrons to attain a neutral charge. A calculation can be used to find the oxidation state since 0-(-2)= 2 Therefore the oxidation state of Sr for the reactant is +2. The oxidation state of the N in the reactant can be calculated as follows: NO3^-1 has a -1 charge. Oxygen has a greater electronegativity therefore the oxidation state of oxygen is -2 since the shared region of electron density is withdrawn towards the oxygen nucleis. The oxidation state of Nitrogen is therefore -1-(-2(3))= +5 For SrO O=-2 Sr= 2. For NO2: O=-2 N= 4. N= 4 because 0-(-2(2))= 4. For O2 it is 0 because both atoms are oxygen atoms. I hope this helps
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    (Original post by TheTechnocrat)
    Oxidation states are based on electronegativity, which is the ability of a nuclei to withdraw a region of electrondensity to itself this is very important when dealing with the oxidation state for molecules. The oxidation state is the number of ELECTRONS an ion or atom needs to obtain a NEUTRAL CHARGE. So when you look at the example you have given it deals mainly with ions. Sr(NO3)2 is an ionic compound the NO3^-1 has an oxidation state of -1 since it needs to loose 1 electron to attain a neutral charge. Sr has a 2 charge hence it needs to gain 2 electrons to attain a neutral charge. A calculation can be used to find the oxidation state since 0-(-2)= 2 Therefore the oxidation state of Sr for the reactant is +2. The oxidation state of the N in the reactant can be calculated as follows: NO3^-1 has a -1 charge. Oxygen has a greater electronegativity therefore the oxidation state of oxygen is -2 since the shared region of electron density is withdrawn towards the oxygen nucleis. The oxidation state of Nitrogen is therefore -1-(-2(3))= +5 For SrO O=-2 Sr= 2. For NO2: O=-2 N= 4. N= 4 because 0-(-2(2))= 4. For O2 it is 0 because both atoms are oxygen atoms. I hope this helps
    Thank you for making my understanding better about oxidation numbers and how they are based from, Could you (or anyone else) also explain how oxygen goes from -2 to 0, hence being oxidised? When I do it, I get -2 to -4, hence being reduced....but that's not the case

    Incase, you need the pastpaper and markscheme:
    past paper - http://pmt.physicsandmathstutor.com/...et-B/Redox.pdf
    answers
    answer - http://pmt.physicsandmathstutor.com/...Redox%20MS.pdf


    I am looking at question 1.
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    (Original post by neluxsan)
    2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g)

    Could anyone explain the oxidation numbers in depth please?
    Thanks, in advance, for taking your time
    Simpler definition: Oxidation number tells you whether something has been reduced or oxidised.

    Oxidation: Loss of electrons
    Reduction: Gain of electrons

    You need to know these rules:
    The oxidation number of oxygen is usually -2.
    The oxidation number of hydrogen is usually +1.
    The oxidation number of a simple ion (i.e. Na+) is the charge of the ion.
    The oxidation of all elements is zero.
    The sum of all the oxidation states of compounds = 0

    Use these rules to form simple algebraic equations to find the oxidation number of the element you're unsure of. So, to find the oxidation state of sulphur in SO2:
    -You are trying to find the oxidation state of S in SO2, so the oxidation state is x.
    -You know the oxidation state of oxygen is -2

    So,
    x +(-2 *2)=0,
    so,
    x-4=0,
    so,
    sulphur has an oxidation state of 4.
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    (Original post by neluxsan)
    Thank you for making my understanding better about oxidation numbers and how they are based from, Could you (or anyone else) also explain how oxygen goes from -2 to 0, hence being oxidised? When I do it, I get -2 to -4, hence being reduced....but that's not the case

    Incase, you need the pastpaper and markscheme:
    past paper - http://pmt.physicsandmathstutor.com/...et-B/Redox.pdf
    answers
    answer - http://pmt.physicsandmathstutor.com/...Redox%20MS.pdf


    I am looking at question 1.
    The oxidation state of oxygen in O2 is 0.

    So it is oxidized to 0.

    The oxidation state of N in N2 is 0
    The oxidation state of H in H2 is 0.

    The answers are based on my first explanation.

    The electrons are equally shared between the two atoms. So no oxygen atom has a partial charge so no need to loose or gain an electron. At least that's my understanding of it.
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    (Original post by TheTechnocrat)
    The oxidation state of oxygen in O2 is 0.

    So it is oxidized to 0.

    The oxidation state of N in N2 is 0
    The oxidation state of H in H2 is 0.

    The answers are based on my first explanation.

    The electrons are equally shared between the two atoms. So no oxygen atom has a partial charge so no need to loose or gain an electron. At least that's my understanding of it.
    In the equation, 2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g) , I understand 02 is zero, but oxygen from nitrogen and strontium in the products side of the reaction are each -2 respectively, so do you not add them together to find total oxidation number of oxygen (which is how i got -4) or do you go for the lowest one?
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    (Original post by Kushala Daora)
    Simpler definition: Oxidation number tells you whether something has been reduced or oxidised.

    Oxidation: Loss of electrons
    Reduction: Gain of electrons

    You need to know these rules:
    The oxidation number of oxygen is usually -2.
    The oxidation number of hydrogen is usually +1.
    The oxidation number of a simple ion (i.e. Na+) is the charge of the ion.
    The oxidation of all elements is zero.
    The sum of all the oxidation states of compounds = 0

    Use these rules to form simple algebraic equations to find the oxidation number of the element you're unsure of. So, to find the oxidation state of sulphur in SO2:
    -You are trying to find the oxidation state of S in SO2, so the oxidation state is x.
    -You know the oxidation state of oxygen is -2

    So,
    x +(-2 *2)=0,
    so,
    x-4=0,
    so,
    sulphur has an oxidation state of 4.
    Thank you for the nice summary Improved my knowledge further
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    (Original post by neluxsan)
    In the equation, 2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g) , I understand 02 is zero, but oxygen from nitrogen and strontium in the products side of the reaction are each -2 respectively, so do you not add them together to find total oxidation number of oxygen (which is how i got -4) or do you go for the lowest one?
    You don't add or choose the lowest ones. You always state the oxidation state for atoms in each molecular formula seperatly from those of the others.
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    (Original post by TheTechnocrat)
    You don't add or choose the lowest ones. You always state the oxidation state for atoms in each molecular formula seperatly from those of the others.
    Ohhh........now I get it thanks
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    The oxidation state for all the other oxygen atoms in the products is the same as the reactant hence the oxygen atoms in those compounds are neither oxidized or reduced.
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    (Original post by TheTechnocrat)
    The oxidation state for all the other oxygen atoms in the products is the same as the reactant hence the oxygen atoms in those compounds are neither oxidized or reduced.
    Apparently that's not true according to the mark scheme, but it could be wrong

    Incase, you need the pastpaper and markscheme:
    past paper - http://pmt.physicsandmathstutor.com/...et-B/Redox.pdf
    answers
    answer - http://pmt.physicsandmathstutor.com/...Redox%20MS.pdf

    Question 1 pls
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    The markscheme is right to the best of my knowledge

    2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g)
    The oxidation state of oxygen for all of these is -2
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    But for O2 it is 0
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    (Original post by TheTechnocrat)
    But for O2 it is 0
    oh okay
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    (Original post by neluxsan)
    2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g)

    Could anyone explain the oxidation numbers in depth please?
    Thanks, in advance, for taking your time
    Basic rules for oxidation numbers: if it is a simple charged particle like Br- then the oxidation number is the charge hence Br- would have an oxidation number of -1; if it is an element in its naturally occurring form then the oxidation number will be 0 e.g. O2. Now, for more complex ones you need to look at things in terms of electronegativity (the ability of one mole of atom to lose one mole of electrons in its gaseous state). Fluorine is the most electronegative element therefore its oxidation state will always be -1. Next, oxygen will usually be -2 unless it is in a peroxide (oxidation state of -1) or bonded with fluorine. Chlorine will always be -1 unless bonded with oxygen or fluorine. Hydrogen will usually be +1 except in metal hydrides where it will be -1. Group 1 and 2 elements are easier as their oxidation numbers are the groups they are in (group 1 would have an oxidation number of +1 and group 2 +2).

    For the equation you have, first I would write out each element and then write down the oxidation numbers corresponding to the reaction. So, Sr would be +2 in the reactants and +2 in the products. This is because the overall charge of a nitrate is -1 and there are two nitrates so it would be -2. To balance this charge Sr would have to be +2. Then, the oxygen in SrO has an oxidation number of -2 so to balance this charge Sr must again be +2. Try this same method for all of the elements and you should be good with oxidation numbers.

    Hope this helped!

    EDIT:I realise I am a little late in answering your question whoops
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    (Original post by MainlyMathsHelp)
    Basic rules for oxidation numbers: if it is a simple charged particle like Br- then the oxidation number is the charge hence Br- would have an oxidation number of -1; if it is an element in its naturally occurring form then the oxidation number will be 0 e.g. O2. Now, for more complex ones you need to look at things in terms of electronegativity (the ability of one mole of atom to lose one mole of electrons in its gaseous state). Fluorine is the most electronegative element therefore its oxidation state will always be -1. Next, oxygen will usually be -2 unless it is in a peroxide (oxidation state of -1) or bonded with fluorine. Chlorine will always be -1 unless bonded with oxygen or fluorine. Hydrogen will usually be +1 except in metal hydrides where it will be -1. Group 1 and 2 elements are easier as their oxidation numbers are the groups they are in (group 1 would have an oxidation number of +1 and group 2 +2).

    For the equation you have, first I would write out each element and then write down the oxidation numbers corresponding to the reaction. So, Sr would be +2 in the reactants and +2 in the products. This is because the overall charge of a nitrate is -1 and there are two nitrates so it would be -2. To balance this charge Sr would have to be +2. Then, the oxygen in SrO has an oxidation number of -2 so to balance this charge Sr must again be +2. Try this same method for all of the elements and you should be good with oxidation numbers.

    Hope this helped!

    EDIT:I realise I am a little late in answering your question whoops
    Ohhh...Thank you so much for taking your time....It is very helpful...I will take your advice
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    (Original post by neluxsan)
    Ohhh...Thank you so much for taking your time....It is very helpful...I will take your advice
    Glad I can help
 
 
 
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