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Redox equations

  1. Jan 6, 2004 #1

    Balancing Redox equations (pg. 607-8)

    Step 4: Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.
    Practice problem 10b.

    KClO (s)==>KCl(aq) + O2(g)

    When I did step 4 of this equation, I was unable to understand why and how the fraction 3/2 is put in front of the oxygen gas in the product side. And also why oxygen is chosen from all the other compounds to place the coefficient infront. I would like this point clarified because its been hindering me of comprehending and solving similar equations.
  2. jcsd
  3. Jan 6, 2004 #2


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    Would you please SHOW us the various steps so we will know what "step 4" is and perhaps be able to see where "3/2" comes in?

    It should be clear that "KClO (s)==>KCl(aq) + O2(g)" is incorrect because the left hand side has only one oxygen atom while the right hand side has two.

    It has been a long time since I did chemistry and I am not absolutely sure I remember what "oxidation numbers" are but it seems clear to me that the formula has to be
    2KClO (s)==>2KCl(aq) + O2(g)
    in order to balance the number of atoms on each side.
    Last edited: Jan 6, 2004
  4. Jan 7, 2004 #3
    Thank you for responding and sorry for being unclear, I thought you might have the textbook I named within reach so that you would be able to review the section and the practice problems better. Anyways I typed out the Steps to the best of my ability, and if there is anything unclear tell me so.

    Step 1: Assign oxidation numbers to all the atoms in the equation.

    -1 +5 -2
    +1 -1 0
    increase(-2==>0) (OXIDATION)

    Step 2: Identify which atoms are oxidized and which are reduced..

    Look at the above example.

    Step 3: Use a line to connect the atoms that undergo oxidation. Use a separate line to connect the atoms that undergo reduction.

    I was unable to do it here, but if you revise the textbook I named you will find it done.

    Step 4: Use appropriate coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number.

    Here is the difficulty
    Their answer is

    Step 5: Do a final check to insure the equation is balanced for both atoms and charge.

    The final answer:

    Thanks again
  5. Jan 7, 2004 #4


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    Yes, now I see MY difficulty at least!

    Your original post said:
    KClO (s)==>KCl(aq) + O2(g)

    NOT KClO3!!! Probably if it had not been so long since I took chemistry, I would have recognized that there is no such compound as "KClO" but that potassium chlorate is KClO3.

    Okay, here's how I would "balance" that equation:

    Since O2 has two oxygen atoms, if I multiply O2 by n, the right side will have 2n oxygen atoms. If I multiply the left side of the equation by m, there will be 3m oxygen atoms on the left: we must have 2n= 3m. Obviously, the simplest thing to do is take n=3, m= 2.

    Given that we multiply the left side by 2, there are 2 potassium and 2 Chlorine atoms on the left and, to have that number on the right, I have to multiply KCl by 2:

    2KClO3= 2KCl+ 3O2.

    Although I am talking about atoms rather than "oxidation numbers", it works out to the same thing.
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