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Homework Help: Redox equation: confirmation

  1. Jul 28, 2008 #1
    Hello guys. I have been a lurker at these forums for quite a while now, they are very helpful if you have run into a snag in your homework. :)

    What I have here is a redox reaction, and I will be very grateful if someone could confirm my answers.

    Equation: Ag + (NO3-) -> (Ag+) + NO

    Substance reduced : Nitrate
    Substance oxidized : Silver
    Half reaction for oxidation : Ag - > (Ag+) + (e-)
    Half reaction for reduction: NO3- + (4H+) + (e-) -> 2H2O + NO
    Net balanced equation : (NO3-) + (4H+) + Ag -> 2H2O + NO + (Ag+)
    Reduced equation : (NO3-) + (4H+) + Ag -> 2H2O + NO + (Ag+)

    I am particularly unsure of the net balanced equation. I know I am supposed to cancel out spectator ions, but I don't think any exist within this particular reaction. Please excuse any of my blunders, I am terrible at Chemistry. :(

    Edit: Is there a way to remove forum formatting ? I had everything lined up neatly...
  2. jcsd
  3. Jul 28, 2008 #2


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    Your nitrate half-reaction seems wrong. NO3- should have an "oxidation" state of +5, since the three oxygens contribute a -6; so unknown + (-6) = -1, the unknown for Nitrogen is +5. Your product of NO (nitrogen monoxide?) would yield a +2 for the Nitrogen. Nitrogen then changes from +5 to +2, a gain of 3 electrons.

    You seem to have enough skill in balancing for the Hydrogens, the Oxygens, and the Waters, so you can restart from here.
  4. Jul 28, 2008 #3
    Hmm. I must be mistaken, because I was under the impression that when balancing charges, only the sum of the net charge of all the molecules has to be taken.

    For example, given this reaction:
    8H+ + MnO4¯ ---> Mn (2+) + 4H2O

    The left side of the equation would have a net charge of 7, as 8 hydrogen ions yield a charge of 8, plus -1 from permanganate. The right side would yield a charge of 2, with permanganate yielding 2 and the water molecules being neutral. So the charge decreases from 7 to 2, in other words, a gain of 5 electrons, making the balanced half reaction:

    (5e-) + 8H+ + MnO4¯ ---> Mn (2+) + 4H2O

    When applied to this:

    (NO3-) -> NO
    Balancing the oxygen:
    (NO3-) -> NO + 2H2O
    Balancing the hydrogen:
    (4H+) + (NO3-) -> NO + 2H2O

    The left side of the reaction has a net charge of 3, as 4 hydrogen ions yield a charge of +4, plus the -1 from nitrate. The right side is neutral, as the water molecules yield a charge of 0 and as oxygen in NO has a charge of -2, and 1x-2+unknown = 0 = 2, which cancel out to be 0. So the charge goes from 3 to 0, which means 3 electrons are gained:

    (4H+) + (NO3-) + (3e-) -> NO + 2H2O

    Is that correct?
  5. Jul 28, 2008 #4


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    That's OK. In your previous attempt charges were not balanced - you had +2 on the left, 0 on the right. Now it is zero on both sides.

    Note that you don't need zero on both sides, you need identical charge on both sides. Could be 0 = 0, -2 = -2, +3 = +3 and so on. If it is not zero, you are just missing some spectators.
  6. Jul 28, 2008 #5
    Ah yes, I realized that while I was re-doing the half reaction for nitrate. So factoring in the correct half reaction, I have this for the entire reaction. Can someone kindly confirm this?

    3Ag + 4H + (NO3-) -> (3Ag+) + NO + 2H2O
  7. Jul 28, 2008 #6


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