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Redox question

  1. May 29, 2009 #1
    Hi there,

    I have some trouble understanding redox reactions. Here is the question:

    Sulphuric acid is one of the strongest inorganic acides which in some reaction shows oxidation properties. In which of reactions below, sulphuric acid acts as oxidating tool?

    A.[tex]
    \text{H}_2{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2{O}
    [/tex]
    B.[tex]
    2 \text{H}_2{SO}_4 + \text{Cu} \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2 \text{H}_2{O}
    [/tex]
    C.[tex]
    \text{H}_2{SO}_4 + \text{BaCO}_3 \rightarrow \text{BaSO}_4 + \text{CO}_2 + \text{H}_2{O}
    [/tex]
    D.[tex]
    \text{H}_2{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2 \text{H}_2{O}
    [/tex]

    Thank you.
     
  2. jcsd
  3. May 29, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

  4. May 30, 2009 #3
    Hi Borek,

    I tried first this way:

    [tex]

    \text{H}_2{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2{O}

    [/tex]

    ON THE LEFT SIDE:

    For [tex] \text{H}_2{SO}_4[/tex] (oxidation numbers are: H is +1, S is +6 and O4 -8)
    For [tex] \text{CaCO}_3[/tex] (oxidation numbers are: Ca is +2, C is +4, O3 is -6)

    ON THE RIGHT SIDE:

    For [tex] \text{Ca}_2{SO}_4[/tex] (oxidation numbers are: Ca is +2, S is +6 and O4 -8)
    For [tex] \text{CO}_2[/tex] (oxidation numbers are: C is +4, O2 is -4)
    For [tex] \text{H}_2{O}[/tex] (oxidation numbers are: H2 is +2, O is -2)

    So what changed, the sum of oxidation numbers on left is equal with ox. numbers on right side so we don't have an redox reaction? Am I right?

    Thank you.
     
  5. May 30, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    H&S are OK, but you should not assign -8 to O4 - but -2 to O. Oxidation number is a property of an inividual atom (note: this is not a real property, this is just a useful way of dealing with th eproblem, ON doesn't exist and can't be measured in the real world).

    No, comparing sum of oxidation numbers is not enough to conclude there were no redox reaction taking place. For that you have to compare oxidation numbers of individual atoms. Take S for example - +6 on the left, +6 on the right, so it was neither oxidized not reduced. Do the same comparison for all other atoms.

    For properly balanced reaction equation sums of ON on both sides will be always equal.
     
  6. May 30, 2009 #5
    So it doesn't matter how atoms has O, oxidation number is always -2. This rule appears for all other elements right?

    So for O4, O3, and O2, we have -2, what am I gonna do for H2SO4, because H2 is +2 and O4 -2, so S is not going to have any oxidation number?

    I would be very happy if you do first for me and give a conclusion and then i would study it and maybe understand it.

    Thanks again.

    Btw, sulphuric acid acts as oxidating tool if number of electrons on right side is less?
     
  7. May 30, 2009 #6

    Borek

    User Avatar

    Staff: Mentor

    Ignoring elemental oxygen (O2 with oxidation number 0) and peroxides (ON -1) - yes.

    Honestly, I don't understand what you wrote, so I will just explain details not adressing your question directly.

    Lat's take a look at H2SO4. Sum of ON must be zero, because the molecule is neutral.

    Hydrogen in most compounds is +1, oxygen is -2.

    What is ON of S? Molecule is neutral so sum of all oxidation numbers is

    2*(+1) + ONS + 4*(-2) = 0

    Note, that 4*(-2) - sum of oxidation numbers of oxygen - is -8. That's the same number you wrote earlier, you just should not assign it to O4, as it suggests you are making a mistake (and not a rare one). It is not O4 being -8, but 4 individual O's being -8 in total.

    In the first case S is +6 on boths isdes, H is +1, O is -2, Ca is +2 and C is +4 - so neither oxidation number changes. No redox reaction.

    IMHO you are on the right track from the very beginning, just your thinking needs some fine tuning :wink:
     
  8. May 30, 2009 #7
    Borek, (it means "pie" in my language), thanks for the answer. Im just a type of autodidact, (i left school) so I'm just trying to get some things right. (Including tuning of my thinking).

    Thank you again. :)
     
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