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Help With Half Reaction/Redox Equation

  1. Oct 28, 2005 #1
    Can anyone help me with this? I have my entire homework section done except for this. For some reason, even on the previous test, I cannot figure out how to do these.
    Here is the question. I hope I get this right, I haven't used LaTeX on this site much.:confused:

    Balance the following oxidation-reduction reactions by the half-reaction method.

    Ok, the LaTeX keeps saying reload page, I have been trying for the last 15 minutes and I can't get it to show up. So, I'll try it without.:tongue2:

    FeI3(aq) + Mg(s) ---> Fe(s) + MgI2(aq)

    Be sure to indicate the spectator ions, the oxidation reaction, the reduction reaction, the reducing agent, the oxidizing agent.

    Here is what I have so far:

    Half reactions:
    Fe^3+(aq) ---> Fe(s) + 3e- = Oxidizing Reaction/Reducing Agent.
    2e- + Mg(s) ---> Mg(aq) = Reducing Reaction/Oxidizing Agent.

    Balanced equation:
    2FeI3(aq) + 3Mg(s) ---> 2Fe(s) + 3MgI2(aq)

    I have the spectator ion as Iodine.

    I am pretty sure most of this is wrong, as I have no clue how to do these. If someone could help me out/point me in the right direction, I would appreciate it.
    If you have any questions about it, or need more info, let me know.
    Last edited: Oct 29, 2005
  2. jcsd
  3. Oct 29, 2005 #2
    scince FeI3 and MgI2 are in an aqueous solution they split like this Fe^+3 + I^-1 and Mg^+2 + I^-1 the I^-1 are the spectator ions so they cancle out
    your left with Fe^+3(aq) + Mg(s) ===> Fe(s) + Mg^+2(aq), the half reactions are as follows

    Fe^+3(aq) + 3e- ===> Fe(s)/reducing half reaction therefore Fe^+3(aq) is the oxidizing agent

    Mg(s) ===> Mg^+2(aq) + 2e-/oxidizing half reaction therefore Mg(s) is the reducing agent

    the electrons gained must = electrons lost, so we multiply the half reactions by the lcd of the electrons/lost/gained before combining them

    2*(Fe^+3(aq) + 3e- ===> Fe(s))
    3*(Mg(s) ===> Mg^+2(aq) + 2e-)
    2Fe^+3(aq) + 6e- ===> 2Fe(s)
    3Mg(s) ===> 3Mg^+2(aq) + 6e-
    2Fe^+3(aq) + 3Mg(s) ===> 2Fe(s) + 3Mg^+2(aq)
    to check we have the same net-charge(+6) and elements on both sides of the chemical equation
    Last edited by a moderator: Oct 29, 2005
  4. Oct 29, 2005 #3
    Looks like I almost had it. By almost had it I mean not even close.:rofl:

    Thanks for the help.:biggrin:
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