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Topic: Finding K of a redox reaction  (Read 6567 times)

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Offline atakmanga

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Finding K of a redox reaction
« on: April 19, 2013, 06:53:40 PM »
Ok this problem is driving me insane.
The equation is:
4OH-+2Cr(OH)3+3ClO- :rarrow: 2CrO42-+3Cl-+5H2O

E° is 1.02 v and ΔG° is −5.91×105 J.

My question is what is K??? I've tried log K=(E°*n/.0592) but the answer is not coming out right.

Please help

Offline UG

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Re: Finding K of a redox reaction
« Reply #1 on: April 19, 2013, 07:22:05 PM »
There is an equation that relates ΔG° to K, please see:

http://en.wikipedia.org/wiki/Gibbs_free_energy#Useful_identities

Offline atakmanga

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Re: Finding K of a redox reaction
« Reply #2 on: April 19, 2013, 07:56:25 PM »
ok i used that too.  but it just errors out on my calculator. this isn't the only problem i have worked on this subject, but it is the only one i'm having trouble with.

Offline UG

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Re: Finding K of a redox reaction
« Reply #3 on: April 19, 2013, 07:59:30 PM »
Can you show us the calculations and numbers you used?
Edit: I see what you mean by error on calculator!

Offline UG

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Re: Finding K of a redox reaction
« Reply #4 on: April 19, 2013, 08:08:42 PM »
Were you given the E° value and ΔG° or were they calculated from a previous part of the question?

Offline atakmanga

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Re: Finding K of a redox reaction
« Reply #5 on: April 19, 2013, 09:27:23 PM »
calculated from a previous question. I know they are right though, because mastering chemistry said so...

Offline Borek

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Re: Finding K of a redox reaction
« Reply #6 on: April 20, 2013, 04:01:00 AM »
Please show exactly what numbers do you use and how. My calculator survived and displayed a result.
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Offline Big-Daddy

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Re: Finding K of a redox reaction
« Reply #7 on: April 21, 2013, 11:37:46 AM »
E° is not needed at all. Converting from ΔG° to Kc isn't even an electrochemistry problem, it's basic thermodynamics: ΔG°=-RT*loge(Kc). Kc=e^(-ΔG°/(RT)). Why should you need E°? (I'm assuming by K you mean the equilibrium constant).

Edit: I also see what they mean. Borek I am running calculation Kc=e^(-ΔG°/(RT))=e^(-((-5.91*(10^5))/(8.314*298.15))). The value is over 10100. Surely we will not get an equilibrium constant that high ...

Offline Borek

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Re: Finding K of a redox reaction
« Reply #8 on: April 21, 2013, 04:39:23 PM »
Why should you need E°?

Where do you get ΔG° from? My bet is that in this particular problem the value OP gave was calculated from E°.

Quote
Edit: I also see what they mean. Borek I am running calculation Kc=e^(-ΔG°/(RT))=e^(-((-5.91*(10^5))/(8.314*298.15))). The value is over 10100. Surely we will not get an equilibrium constant that high ...

Actually I have ignored ΔG° given and calculated K directly from E° - and I got a result below 10100 (not that K over 10100 is unreasonable! Try to estimate K for oxidation of sodium by elemental chlorine).

Not to mention the fact that if someone is unable to calculate 10x just because x is too large for their calculator, it means they have no idea what they are doing.
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Offline Big-Daddy

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Re: Finding K of a redox reaction
« Reply #9 on: April 21, 2013, 04:55:25 PM »

Where do you get ΔG° from? My bet is that in this particular problem the value OP gave was calculated from E°.

That's fine. I was just pointing out that the OP's attempt to calculate Kc then does not require E° once we've got ΔG° as the OP has - we don't need nFE=RTloge(Kc), as we've already got -nFE in the form of ΔG° summed up.

Actually I have ignored ΔG° given and calculated K directly from E° - and I got a result below 10100 (not that K over 10100 is unreasonable! Try to estimate K for oxidation of sodium by elemental chlorine).

I have no idea how to estimate Kc from a reaction but I agree you could get values well over 10100, but these would generally not be considered reversible reactions at all. But ok, I take your point.

The OP's calculation of ΔG° may be faulty then? I get a value of 3.5·10103 moldm-3 with the numbers I wrote in my above post.

Not to mention the fact that if someone is unable to calculate 10x just because x is too large for their calculator, it means they have no idea what they are doing.

Some people may not have learnt change of base in their maths class yet? :p In any case we (at least I) can work out the number pretty easily, it just seems slightly unreasonable for this reaction.

Offline Borek

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Re: Finding K of a redox reaction
« Reply #10 on: April 21, 2013, 05:36:07 PM »
Some people may not have learnt change of base in their maths class yet?

It is not about changing base, it is about understanding what the logarithm is.
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Offline Big-Daddy

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Re: Finding K of a redox reaction
« Reply #11 on: April 21, 2013, 05:44:30 PM »
Some people may not have learnt change of base in their maths class yet?

It is not about changing base, it is about understanding what the logarithm is.

All this abstract talk can't be helpful to someone who doesn't know this bit of maths yet. Let's just write it down: e^x=10^y, we want y, so take log10. y=log10(e^x)=x*log10(e). Applying this to our current problem, y=(-((-5.91*(10^5))/(8.314*298.15)))*log10(e)=103.5. So Kc=10103.5 which is converted shortly into 3.5*10103.

Offline Borek

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Re: Finding K of a redox reaction
« Reply #12 on: April 21, 2013, 06:43:56 PM »
Someone that is expected to use exponential function to calculate K should know properties of the exponential function, period. And the most basic property is

ax+y=axay

so

10103.5 = 10103×100.5

The only thing you need calculator for is 100.5.
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