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Balancing redox equation

  1. Nov 19, 2005 #1
    Hello. I have a question on balancing redox equations. I’ve tried to follow the steps in my book but I don’t understand how to do it. Can someone please walk me through one.

    H2S(g) + NO3{-1}(aq) > NO(g) + S(s) (acidic solution)

    Now the book says I have to add electrons but how?

    Thanks
     
  2. jcsd
  3. Nov 19, 2005 #2
    1. Get the half-reactions:

    H2S --> S
    NO3- --> NO

    2. Balance any atoms that are not O or H:

    H2S --> S
    NO3- --> NO

    3. Balance O atoms using H2O:

    H2S --> S
    NO3- --> NO + 2H2O

    4. Balance H atoms using H+:

    H2S --> S + 2H+
    4H+ + NO3- --> NO + 2H2O

    5. Balance charge with electrons:

    H2S --> S + 2H+ + 2e-
    4H+ + 3e- + NO3- --> NO + 2H2O

    6. Multiply equations so that you have the same amount of electrons being transfered in each reaction:

    3[H2S --> S + 2H+ + 2e-] = 3H2S --> 3S + 6H+ + 6e-
    2[4H+ + 3e- + NO3- --> NO + 2H2O] = 8H+ + 6e- + 2NO3 --> 2NO + 4H2O

    7. Add up equations:

    3H2S --> 3S + 6H+ + 6e-
    8H+ + 6e- + 2NO3 --> 2NO + 4H2O
    ------------------------------------
    3H2S + 8H+ + 6e- + 2NO3- --> 3S + 6H+ + 6e- + 2NO + 4H2O

    8. Cancel out species that occur on opposite sides:

    3H2S + 2H+ + 2NO3- --> 3S + 2NO + 4H2O

    9. Check for errors. Make sure atoms are balanced and the overall charge on each side is balanced.
    10. This is your answer because it is in acidic solution. If it is was in basic solution, you would have to neutralize the hydrogen atoms with hydroxide ions.
     
  4. Nov 19, 2005 #3
    Thank You so much:smile: !! I wish my book had explained it in steps like that.
     
  5. Nov 19, 2005 #4
    What would I do if it was in a basic solution. For example:

    Cr{+3}(aq) + MnO2(s) > Mn{+2}(aq) + CrO4{-2}(aq) in a basic solution
     
  6. Nov 19, 2005 #5
    Ok well let's just pretend the first problem you posted was done in basic solution. For acidic solution, we had:

    3H2S + 2H+ + 2NO3- --> 3S + 2NO + 4H2O

    So we have 2H+ that we need to neutralize with 2OH-, giving 2H2O on the left side and 2OH- on the right side:

    3H2S + 2H2O + 2NO3- --> 2S + 2NO + 4H2O + 2OH-

    Now we can cancel out the 2H2O on the left with the 4H2O on the right:

    3H2S + 2NO3- --> 2S + 2NO + 2H2O + 2OH-

    This reaction could never occur in basic solution, but it's just an example.
     
  7. Nov 20, 2005 #6
    One last question. I was doing this one and I got stuck:

    IO3{-1}(aq) + I{-1}(aq) > I2(s)

    IO3{-1} + I{-1} > I2
    6H{+1} + IO3{-1} + I{-1} > I2 + 3H2O
    4e- + 6H{+1} + IO3{-1} + I{-1} > I2 + 3H2O

    But now what do I do because there is not another e on the other side.

    Thanks
     
  8. Nov 20, 2005 #7
    This one is a bit tricky. Like always, first split it up into two distinct half-reactions:

    IO3- --> I2
    I- --> I2

    and then balance the I atoms:

    2IO3- --> I2
    2I- --> I2

    Continue from here and I think you should be able to get it. Tell me if you have problems.
     
    Last edited: Nov 20, 2005
  9. Nov 20, 2005 #8
    Yeah I can finish that one. Would I do the same thing for one like this:

    P4 > PH3 + H2PO2{-1}

    And how do I know what is the reducing agent and oxidizing agent in the final balanced equation (you can use the 1st one for an example since it is already balanced).
     
    Last edited: Nov 20, 2005
  10. Nov 20, 2005 #9
    For this one

    P4 > PH3 + H2PO2{-1}

    Would I do this:

    P4 > PH3
    P4 > H2PO2{-1}
     
  11. Nov 20, 2005 #10
    Yes.

    IO3- --> I2
    IO3-: I has an oxidation state of +5.
    I2: I has an oxidation state of 0.

    I is moving towards a more negative state, so it is reduced. Therefore, IO3- is the oxidizing agent.

    I- --> I2
    I-: I has an oxidation state of -1
    I2: I has an oxidation state of 0

    I is moving towards a more positive state, so it is oxidized. Therefore, I- is the reducing agent.

    Same procedure applies to your P4 one.
     
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