TBR electrochemistry: voltage

[hmania]

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Well I was able to narrow down from 4 answers to 2.

Question: which of the following cells has the GREATEST voltage?
a) zn(s)/0.1M Zn 2+(aq)//1.00M Cu2+(aq)/ Cu(s)
b) Zn(s)/ 1.00 M Zn 2+ (aq)//0.1M Cu 2+ (aq)/ Cu(s)


I know that cells must have more reactants than products to result in a higher voltage. I thought Zn(s) --> Zn 2+ is oxidation (anode), therefore since the electrons flow from anode to cathode than Zinc would be the designated reactant. However, the solution states that Copper is the reactant because it is being reduced.

I understand that Copper is reduced and Zinc is Oxidized, so using RED CAT and AN OX. I thought copper would be the taken out of solution and metal plated on the electrode. 😕

As you can see, I am confused. Somebody help me!

 

[Pigglyjuff]

V=IR

Crap, the question doesn't provide I or R!
Shootshootshoot
*gets a 6 on PS*

 

[hmania]

Still in need of help!

 

[V5RED]

It is customary to post what you think the answer is along with what the source material says the answer is. Without that, you won't get many replies.

 

[hmania]

Well I was able to narrow down from 4 answers to 2.

Question: which of the following cells has the GREATEST voltage?
a) zn(s)/0.1M Zn 2+(aq)//1.00M Cu2+(aq)/ Cu(s)
b) Zn(s)/ 1.00 M Zn 2+ (aq)//0.1M Cu 2+ (aq)/ Cu(s)


I know that cells must have more reactants than products to result in a higher voltage. I thought Zn(s) --> Zn 2+ is oxidation (anode), therefore since the electrons flow from anode to cathode than Zinc would be the designated reactant. However, the solution states that Copper is the reactant because it is being reduced.

I understand that Copper is reduced and Zinc is Oxidized, so using RED CAT and AN OX. I thought copper would be the taken out of solution and metal plated on the electrode. 😕

As you can see, I am confused. Somebody help me!

I am thinking that the answer should be B because I believe that Zinc is the reactant since electrons flow from anode to cathode and Zinc, obviously, is oxidizing at the anode. However, the solution states that Copper is the reactant since it is reducing.

 

[cloak25]

Zn(s) --> Zn2+ + 2e-
Cu2+ + 2e- ---> Cu(s)

The answer choices show the concentrations for the cations. When you combine these two half reactions, Zn2+ ends up on the product side and Cu2+ on the reactant side. You know you need more reactants to get a higher voltage so you want more Cu2+. A should be the answer since it shows a higher concentration of Cu2+.

 

[V5RED]

I am thinking that the answer should be B because I believe that Zinc is the reactant since electrons flow from anode to cathode and Zinc, obviously, is oxidizing at the anode. However, the solution states that Copper is the reactant since it is reducing.

Ignoring any spectator ions, the reaction is Cu2+ + Zn(s) -> Cu(s) + Zn2+

Copper ions are reactants, and solid zinc is a reactant.

Solid copper is a product, and zinc ions are products.

The voltage is proportional to how far from equilibrium the starting state of the cell is. Since solids are not factored into equilibrium calculations, we only need to look at the ions. Since we know the reaction is very favorable, we know that at equilibrium, there should be a lot more product ions than reactant ions.

In "A", you are starting with a lot of reactant ions and very few product ions.

In "B", you are starting with a lot of product ions and very few reactant ions.

Therefore, "A" is much farther from equilibrium and has a higher voltage than "B".

If you want to get mathy about it as opposed to considering the equilibrium, you can use the nernst equation where E(real)=E(std)-RT/nF*lnQ.

In "A" Q is less than 1, so RT/nF*lnQ is a negative number, so E(real)>E(std) because you are subtracting a negative value from E(std).

In "B" Q is greater than 1, so RT/nF*lnQ is a positive number, so E(real)<E(std) because you are subtracting a positive value from E(std).

 

[sciencebooks]

I know that cells must have more reactants than products to result in a higher voltage.

Could somebody explain this to me briefly? I'd really appreciate it! Does it just have to do with *wanting* the reaction to proceed?

 

[circulus vitios]

Could somebody explain this to me briefly? I'd really appreciate it! Does it just have to do with *wanting* the reaction to proceed?

Yeah, I don't get this either. Why are you guys calculating voltages on the assumption that the concentrations will equilibrate, and not calculating voltages on the given concentrations without equilibration?

 

[V5RED]

Yeah, I don't get this either. Why are you guys calculating voltages on the assumption that the concentrations will equilibrate, and not calculating voltages on the given concentrations without equilibration?

You can't calculate voltage for a cell without taking equilibration into account.

If Q does not equal 1, then you need to use the Nernst equation to calculate voltage. The Nernst equation is based on the idea that a system in equilibrium has no voltage because the reaction has gone as far as it can go.

If you want the proof math wise, look at the Nernst equation.
Eactual=Estd-RT/nF*lnQ
Estd=RT/nF*lnK

If Q=K (ie the system is at equilibrium), then we get:
Eactual=Estd-RT/nF*lnK=Estd-Estd=0

edit:Think of it like any other reaction. As a reaction proceeds, reactants become products. In a cell, this reaction involves the transfer of electrons from the anode to the cathode. Once the reaction is at equilibrium, there will be no more net reaction and thus no more net transfer of electrons and thus no voltage.

 

[sciencebooks]

edit:Think of it like any other reaction. As a reaction proceeds, reactants become products. In a cell, this reaction involves the transfer of electrons from the anode to the cathode. Once the reaction is at equilibrium, there will be no more net reaction and thus no more net transfer of electrons and thus no voltage.

This helped! Basically having more reactants pushing the reaction forward and since this reaction involves the flow of electrons, increased reactants = increased electron flow = increased voltage.

From my knowledge, it seems like the connection is that current = charge per time... So increased reactions = increased current (and by V=IR) = increased voltage as well. Is this correct?