Electrochem Q

[chiddler]

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Say we have a galvanic cell. How do we increase the rate of reaction? How do we decrease rate of reaction?

thanks.

 

[pfaction]

This is exactly why I asked my "Keq>1 exception ratio" question.

So TBR says, in their last chapter of GChem, you increase the voltage by increasing the cathode and decreasing the anode.

A|A+|B+|B
Anode react|Anod prod|Cathode react|Cathode prod
We want to make B+ large as hell and A+ small as hell.

For example their 10.13
ZN|0.10|1.00|Cu
has greater voltage than
ZN|1.0|0.10|CU

tldr: decrease anode prod, increase cathode reactant.

 

[milski]

I sort of suck at electro chem but I think the only way to control it by controlling the load that connects the electrodes. The cell can move a fixed amount electrones before reaching equilibrium and being completely discharged. There is probably some sort of maximum rate when the resistance between the electrodes is zero and if I had to guess, I would say that that's controlled by the surface area of the electrodes and maybe the concentration of the electrolytes.

 

[pfaction]

Oh also you can alter the cathodes and anodes themselves, IIRC. But not sure about that. TBR says this is one way to increase the voltage.

There is probably some sort of maximum rate when the resistance between the electrodes is zero and if I had to guess, I would say that that's controlled by the surface area of the electrodes and maybe the concentration of the electrolytes

TBR says at equilibrium, voltage will be 0 so you have to always keep Q<Keq.

 

[milski]

Oh also you can alter the cathodes and anodes themselves, IIRC. But not sure about that. TBR says this is one way to increase the voltage.



TBR says at equilibrium, voltage will be 0 so you have to always keep Q<Keq.

Zero resistance as a load between the electrodes - you need some sort of electrical connection between them on the "dry side" to have the reaction going.

 

[chiddler]

Cathode voltage must increase? Why not the anode?

I mean the way i'm seeing it is the more positive the E value is for a given redox reaction, the more voltage it generates be it at cathode or anode.

I sort of suck at electro chem but I think the only way to control it by controlling the load that connects the electrodes. The cell can move a fixed amount electrones before reaching equilibrium and being completely discharged. There is probably some sort of maximum rate when the resistance between the electrodes is zero and if I had to guess, I would say that that's controlled by the surface area of the electrodes and maybe the concentration of the electrolytes.

I was thinking about changing surface area very much. But more surface area of either node would mean less potential per surface area? Does this logic even apply to electrodes?

 

[chiddler]

Zero resistance as a load between the electrodes - you need some sort of electrical connection between them on the "dry side" to have the reaction going.

what now?

 

[milski]

Cathode voltage must increase? Why not the anode?

I mean the way i'm seeing it is the more positive the E value is for a given redox reaction, the more voltage it generates be it at cathode or anode.



I was thinking about changing surface area very much. But more surface area of either node would mean less potential per surface area? Does this logic even apply to electrodes?

I don't think the potential depends on the area? It's just a difference of concentrations. More area would give you more possibilities for electrons to move and will certainly help you with high currents. I don't know how much of a limiting factor that would be and when it would matter. 😕

 

[milski]

what now?

Just trying to say that shorting the two electrodes will result in a much higher rate than connecting them through a 5 megaohm resistor.

I won't have a writing section on my test and as a result I can let my writing be sloppier than ever. 😛

 

[chiddler]

screw this. i think i'm overthinking a question i answered.

 

[chicagoryan]

Doesn't this involve E = Estd - (0.0592/n)log(Q)? Therefore it depends on Q. Increasing the concentration of the cathode lowers Q and raises E, opposite result for increasing the concentration of the Anode. I may have misinterpreted the question.

 

[pfaction]

Doesn't this involve E = Estd - (0.0592/n)log(Q)? Therefore it depends on Q. Increasing the concentration of the cathode lowers Q and raises E, opposite result for increasing the concentration of the Anode. I may have misinterpreted the question.

Is that what I said roughly? Wanted to make sure.

 

[chiddler]

i don't think more E entails a faster reaction. just a bigger K constant.

deltaG = -nFE = -RTln(Keq)

 

[MedPR]

Pretty sure galvanic depends on the solvent and the half-reaction potentials of the two electrodes. The reaction slows as one electrode degrades and the other plates out, but I don't know how you would go about actively controlling it at various points.

 

[DrRichand1]

doesn't an increase in Temp increase the voltage

 

[rjosh33]

doesn't an increase in Temp increase the voltage

Yes, though to varying degrees depending on the situation.

milski had the right idea (per usual) -- it's easiest (and most reliable) to increase or decrease the current by changing the resistance of the wire that connects the electrodes while holding the voltage constant, in accordance with Ohm's law.

Increasing/decreasing the reactants and/or products may or may not change the current, depending on the reaction involved.