How to know which order to put the equations in

[leroyjenkens]

Homework Statement

The problem asks me for the reduction potential of two metals: Zn/Zn²⁺ and Cu/Cu²⁺

All I need to know is how do I determine which one will be reduced and which one will be oxidized.

The answer to the question puts the Zinc half reaction as Zn → Zn²⁺ + 2e^-

And of course the Copper half reaction is put the opposite way so that it's reduced.

But they never tell me how they know which way to arrange the equations. Like why didn't they make the copper half reaction go Cu → Cu²⁺ + 2e^- so that it is oxidized?

I know how to answer the rest of the question, I just can't find anywhere in the text that tells me this vital detail.

Thanks.

 

[janhaa]

If you put Zn metal in a blue Cu-solution, the Zn metal dissolves and Cu metal is deposited.
And the blue color fades.
Because Cu ions attracts electrons and form a metal. While solid Zn goes into solution and gives away electrons.
thus:
$$E^o(Cu/Cu^{2+}) >E^o(Zn/Zn^{2+})$$
that is
$$Cu^{2+}(aq)+Zn(s)=>Zn^{2+}(aq)+Cu(s)$$
====

the opposite reaction is not possible...

 

[Yanick]

There are two ways to figure out which metal will be the oxidant/reductant. One never made sense to me and I don't like memorizing so I never remembered it. Its something like E°_cell = E°_anode - E°_cathode but don't quote me on that because, as I've stated its annoying that you need to memorize stuff. Also it requires you to already know which is the anode and which is the cathode, or you have to just guess and check (only two possibilities so its not so bad, but still).

What I do is I remember that reactions with ΔG° < 0 will proceed forward spontaneously, as written. I also remember that ΔG° = -nFE°, where n is the moles of electrons being transferred in the balanced reaction and F is Faraday's constant. Without doing any explicit calculations you can see that ΔG° < 0 when E° > 0. So you need to just sum the standard half reaction potentials to get a E° > 0.

For example the reduction of elemental iodine (I₂) to iodide (I^-) has an E° = +0.53V. The reduction of ferrous iron (Fe²⁺) to iron metal (Fe⁰_(s)) has an E° = -0.44V. Obviously we have to have one of those being oxidized so we need to flip one of the chemical equations, which leads to us multiplying the E° of that half reaction by -1. Once we have flipped the chemical equation we simply sum up the E° of each half reaction. So here goes:

We flip the iodine half reaction so that it is now being oxidized, therefore we multiply the E° of the reduction of iodine by -1 and we get E°_Iodine = -0.53V. We then add the reduction potential of ferrous iron and the oxidation potential of iodide to get E° of the cell. So -0.53V + (-0.44V) = -0.97V. Going back to ΔG° = -nFE° we see that with a negative redox potential of the overall cell we will get ΔG° > 0 and the reaction will not proceed forward (as written) spontaneously. From here we can simply just realize that we need to multiply each E° by -1, which is simply just flipping the half reactions so that the iodine is being reduced and the iron metal is being oxidized.

So we have E° for iron metal being oxidized to ferrous iron as being +0.44V and we have an E° of elemental iodine being reduced to iodide ions as +0.53V. Adding together we have 0.44V + 0.53V = 0.97V. Going to ΔG° = -nFE° we get a ΔG° < 0 with an E°_cell > 0 so we conclude that the reaction will proceed forward, as written. Therefore we then say that elemental iodine (I₂) will oxidize iron metal (Fe⁰).

That is how I work this stuff out from scratch because it only requires you to know the fundamentals which is really what you remember from any class unless your work directly uses the nitty gritty of any given topic.

A quick way to figure out what is oxidized and what is reduced is to remember that the higher up a reduction half reaction is on a reduction potential table (in other words, the more positive a reduction half reaction is) the stronger the oxidizing "power" of that substance is. Another way of putting it is, the thing with the higher reduction potential will oxidize the thing with the lower reduction potential.

Try and see if you can work out the Cu and Zn problem on your own and make sure that it is consistent whichever way you decide to do it.