Help With Half Reaction/Redox Equation

[erok81]

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.
Thanks.:biggrin:

[force]

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

[erok81]

Looks like I almost had it. By almost had it I mean not even close.:rofl:

Thanks for the help.:biggrin: