Emily C.

asked • 04/30/20

Chem help please.1?

What is the emf of this cell when [Fe^ 2+ ]=1.1M [Fe^ 3+ ] =2.0*10^ -2 M,P O 2 = 0.45 atm and the pH of the solution in the cathode compartment is 3.2.


A voltaic cell utilizes the following reaction:

4Fe^2+ (aq)+ O2 (g) + 4H^+ (aq)—-> Fe^3+ (aq)+2H2O (l)


J.R. S.

tutor
The [H+] should be 10^-3.2 or 6.3x10^-4. And why didn't you raise [H+] to the fourth power? That seems like a rather small number, but shouldn't it be the case?
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04/30/20

Emily C.

When I try that, I get E=.46−0.5924log[2.0×10^-2]^4/[1.1]^4[10−3.2]^4[.45] = 3.436 which isn’t the right answer
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04/30/20

J.R. S.

tutor
What is the correct answer? If you have 10^-3.2 in the denominator and you raise that to the 4th power, you answer should be much larger than what you report. This may be more a problem in algebra than chemistry. LOL
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04/30/20

Emily C.

I don’t know what the correct answer would b, maybe I’m missing something else
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04/30/20

J.R. S.

tutor
I don't think so. Seems to me that you have the correct approach using Nernst and Q. I also found Eº to be 0.046 V so at least that part is correct. It seems that the correct answer might be slightly lower than Eº
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04/30/20

Emily C.

I’m so lost at how to solve this now
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04/30/20

1 Expert Answer

By:

J.R. S. answered • 05/01/20

Tutor
5.0 (145)

Ph.D. in Biochemistry--University Professor--Chemistry Tutor
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About this tutor ›

Here goes my feeble attempt, given that you already know how to find Eº and how to use the Nernst equation.

E = 0.46 - 0.0592/n log 2x10-2 / (1.1)4(0.45)(6.3x10-4)4

E = 0.46 - 0.0592/4 log 2x10-2 / (1.46)(0.45)(1.6x10-13)

E = 0.46 - 0.0148 log 2x10-2 / 1.05x10-13

E = 0.46 - 0.0148 log 1.9x1011

E = 0.46 - 0.0148 x 11.28

E = 0.46 - 0.17

E = 0.29 V


Check it carefully as I'm prone to making errors.

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Emily C.

Thank you so much for your help and patience. That looks right and it’s probably the right answer, maybe the system I’m imputing it in the computer with has the wrong answer. It’s happened before.
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05/01/20

J.R. S.

tutor
Glad I could help. The solution looks correct to me but one never knows. I did a fair amount of rounding at different stages of the problem which could easily throw the answer off a bit.
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05/01/20

Emily C.

E=.459-6.42X10^-3(ln(1.5x10^6))= E=.459-6.42x10^-3(14.42)=0.37 I was messing up the equation It was supposed to b E=E°-(RT/nF)ln(2.0x10^-2)^4/(1.1)^4(10^-3.2)^4(.45)
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05/01/20

J.R. S.

tutor
The 0.0592/n takes into account the F and converting ln to log @ 273K. So the equation I used is correct.
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05/01/20

Emily C.

Yes absolutely, thank u so much for ur help
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05/04/20

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