[lucasloredo]

The problem is as follows: "Calculate E_cell values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions that are not already balanced."

a.) MnO₄^– (aq) + I^–(aq) ——> I₂(aq) + Mn²⁺(aq)

So I set up my two half-reactions:    5 [ 2I^– ——> I₂ + 2e^–      E_cell = –0.54v

                                                  2 [ 5e^– + 8H⁺ + MnO₄^– ——> Mn²⁺ + 4H₂O ]      E_cell=1.51v


and when you combine them you get:  16H⁺ + 10I^– + 2MnO₄^– ——> 5I₂ + 2Mn²⁺ + 8H₂O   E_cell = 0.97v

So I'm fairly positive that part of the question is right. However, our teacher hasn't yet explained how to find out if the reaction is spontaneous. I would guess that it might have something to do with the reactivity series, but I wasn't sure and didn't want to chance it.

Thanks in advance for your *delete me*
Lucas                 

[mike]

If the reduction potential is positive it is spontaneous in the direction written. If the reduction potential was negative the reverse reaction would be spontaneous.

[lucasloredo]

Thanks for the quick reply, but I'm still a bit confused. So, for instance, the reduction potential for the first half reaction, as I wrote it, is –0.54v, so does that make the whole equation spontaneous? I am very confused    Could you please elaborate a little further? We just learned this concept a couple days ago and I'm still not comfortable with the material.

Thanks again,
Lucas

[mike]

Remember that reduction is a gain in electron (OILRIG - oxidation is loss/reduction is gain).

So reduction potentials are for the reaction in the reduction direction, if you reverse the reaction (like you did with iodine) it becomes an oxidation reaction.

The reaction with the higher more positive reduction potential will proceed spontaneously in the reduction direction when reacted with another compound with a smaller reduction potential (in which case it will proceed in the oxidation direction).