CHEMISTRY URGENT PLZ HELP

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Write a balanced equation for the following reduction-oxidation reaction.

SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)

can someone help me step by step??

  • CHEMISTRY URGENT PLZ HELP -

    I answered this earlier by saying S goes from oxidation state of +4 on the left to +6 on the right (loss of 2 electrons). Mn goes from an oxidation state of +7 on the left to +2 on the right (gain of 5 electrons). You want to make the electron loss = electron gain. What about this don't you understand or can you finish?

  • CHEMISTRY URGENT PLZ HELP -

    First of all I am not the same person as before...his name was Max, my name is Kevin so i do not know that you answer this question before. Secondly, I do not know what you mean about about loss of 2 electrons, how would i use hydrogen to balance the charges when I have:
    SO3(2-)-- SO4(2–)+ 2e(-)+?H(+)

  • CHEMISTRY URGENT PLZ HELP -

    I did it brute force as follows (not a chemist)
    O3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
    first deal with S, Mn atoms
    there must be the same number on each side

    nSO3(2-) + mMnO4(-) --- nSO4(2–) + mMn(2+)
    now try 2jH(+) on the left and jH2O on the right
    nSO3(2-) + mMnO4(-) +2jH(+) --- nSO4(2–) + mMn(2+) + jH2O
    O atoms
    3 n + 4 m = 4 n + j
    or
    n+j = 4 m
    + charges
    -2 n - 1 m + 2 j = -2 n + 2 m
    so from the second one
    3 m = 2 j
    j = (3/2) m
    from the first one
    n + (3/2) m = 4 m
    or
    n = (5/2) m
    Now go back using m = 2 to make the fraction go away
    then j = 3
    and n = 5
    5 SO3(-2) + 2 MnO4(-1) + 6 H(+1) --> 3 H2O + 5 SO4(-2) + 2 Mn(+2)

  • CHEMISTRY URGENT PLZ HELP -

    uh i don't think that's how u do it because i think you need to add and remove electrons from two separate reactions first before multiplying it out...but thank you anyway

  • CHEMISTRY URGENT PLZ HELP -

    Now to do it right (for my own entertainment )
    S(+4) ---> S(+6) + 2 e(-)
    M(+7) + 5 e(-) ---> M(+2)
    to balance the electrons multiply the first equation by 5 and the second by two.
    5 S(+4) ---> 5 S(+6) + 10 e(-)
    2 M(+7) + 10 e(-) ---> 2 M(+2)
    There, that gives you the 5 and the 2 which is the hard part. (First time I have done one of these since about 1957)

  • CHEMISTRY URGENT PLZ HELP -

    Kevin--The numbers Damon provided are correct. Let me know what you don't understand about it and I can help you through it (as a chemist).

  • CHEMISTRY URGENT PLZ HELP -

    Kevin, you may not have written the question the first time and perhaps it was some other person named Max, BUT both of you were/are using the same computer because the IP address is the same. But I still encourage you to tell me what you don't understand if you don't get how Damon balanced the equation for you.

  • CHEMISTRY URGENT PLZ HELP -

    thanks damon

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