# How to know which order to put the equations in

• leroyjenkens
In summary, the problem asks for the reduction potential of two metals, but does not provide a way to determine which will be reduced and which will be oxidized. The answer puts the Zn half reaction as Zn → Zn2+ + 2e- and the Copper half reaction as Cu → Cu2+ + 2e- but they never tell the reader how they know which half reaction will be oxidized.
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.

If you put Zn metal in a blue Cu-solution, the Zn metal dissolves and Cu metal is deposited.
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...

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.

## 1. How do I know which order to put the equations in?

The order in which equations should be arranged depends on the specific problem at hand. In general, it is recommended to start with the most fundamental equations and then work towards more complex equations. This will help to build a strong foundation for solving the problem.

## 2. Is there a specific rule for arranging equations?

There is no specific rule for arranging equations, as it depends on the problem and the equations involved. However, it is important to ensure that the equations are logically organized and follow a logical flow.

## 3. Should I arrange the equations in a particular order for every problem?

No, the order in which equations should be arranged may vary depending on the problem. It is important to carefully analyze the problem and determine the most appropriate order for the equations.

## 4. Are there any strategies for determining the order of equations?

One strategy is to identify the key variables and relationships in the problem and then arrange the equations accordingly. Another strategy is to start with the most basic equations and then move on to more complex ones.

## 5. Can I change the order of equations while solving a problem?

Yes, it is possible to change the order of equations while solving a problem. This may be necessary if new information is discovered or if the initial arrangement of equations is not yielding a solution. However, it is important to maintain a logical flow and ensure that all equations are relevant to the problem.

• Electrical Engineering
Replies
3
Views
578
• Biology and Chemistry Homework Help
Replies
6
Views
2K
• Chemistry
Replies
5
Views
1K
• Biology and Chemistry Homework Help
Replies
1
Views
3K
• Chemistry
Replies
15
Views
3K
• Biology and Chemistry Homework Help
Replies
12
Views
6K
• Biology and Chemistry Homework Help
Replies
1
Views
7K
• Biology and Chemistry Homework Help
Replies
1
Views
3K
• Chemistry
Replies
1
Views
2K
• Biology and Chemistry Homework Help
Replies
6
Views
2K