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Topic: Polarogram HW Question  (Read 708 times)

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Offline liverpool2406

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Polarogram HW Question
« on: March 19, 2022, 10:12:38 PM »
I am struggling with the question below. I have no idea where to begin.

Suppose that the diffusion current in a polarogram for reduction of Cd2+ at a mercury electrode is 14μA. If the solution contains 25 mL of 0.5 mM Cd2+, what percentage of Cd2+ is reduced in the 3.4 min required to scan from -0.6V to -1.2V?

Additional information that I know is that the reduction potential of the solution with mercury is -0.380V.

So there is not enough information for Ilkovic's. I know that the current is proportional to the rate of diffusion and also to the concentration difference. I also know the limiting current is equal to the diffusion current which is proportional to the concentration in bulk solution. I tried converting current to concentration through C=(I)(t)/(n)(F) and then using that with the concentration of original solution (C/(0.5/25)). That did not work (I got 0.000074%). I have no idea what to do now, but I think I need to do something with the given voltage change and known reduction potential.

The answer is 0.12%.

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Re: Polarogram HW Question
« Reply #1 on: March 20, 2022, 04:36:35 AM »
I am not sure what it means by the "diffusion current". Current is not constant, it is changing. Is it the maximum current registered? Some kind of an averaged current?

through C=(I)(t)/(n)(F) and then using that with the concentration of original solution (C/(0.5/25)).

If I understand correctly the logic you are trying to apply it looks OK, but formulas you wrote make no sense. What is 0.5/25? What units do you get from this division?
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Offline liverpool2406

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Re: Polarogram HW Question
« Reply #2 on: March 20, 2022, 11:04:59 AM »
I am not sure what it means by the "diffusion current". Current is not constant, it is changing. Is it the maximum current registered? Some kind of an averaged current?

Diffusion current is the current observed when the rate of electrolysis is limited by the rate of diffusion of analyte to the electrode. Diffusion current is the difference of limiting and residual current.


If I understand correctly the logic you are trying to apply it looks OK, but formulas you wrote make no sense. What is 0.5/25? What units do you get from this division?

Apologies, 0.5/25 is millimolar divided by milliliters to get concentration of the solution. I then divided the Cd2+ reduced by the concentration of solution to get the ratio. I then multiplied by 100% to get my percentage. But this is not correct.

C=(I)(t)/(n)(F) is the equation to find concentration. I is current. t is time. n is moles of electron. F is Faraday's constant.
« Last Edit: March 20, 2022, 11:28:30 AM by liverpool2406 »

Offline Hunter2

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Re: Polarogram HW Question
« Reply #3 on: March 20, 2022, 12:07:51 PM »
The mistake is 0,5 mM = 0,5 millimol/l what means 25 ml contain 0,5*10^-3 mol/l *0,025 l =0,0000125 mol = 12,5 μmol
Instead you used 0,5 mM = 0,5 mmol containing in 25 ml what gives wrong result of = 0,02 mol/l

Using c = I *t / nF gives

c= 14 *10^-6 A*3,4*60s/(2*96485 As/ mol) = 0,0148 μmol Cd

Percentage = 0,0148 μmol/12,5μmol*100% = 0,118%~ 0,12 %

Offline liverpool2406

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Re: Polarogram HW Question
« Reply #4 on: March 20, 2022, 12:15:20 PM »
Wow, I cannot believe it was that simple of a mistake. I was so determined on the idea that I misunderstood the content when I just did it incorrectly. Thank you so much.

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Re: Polarogram HW Question
« Reply #5 on: March 20, 2022, 03:02:11 PM »
I am not sure what it means by the "diffusion current". Current is not constant, it is changing. Is it the maximum current registered? Some kind of an averaged current?

Diffusion current is the current observed when the rate of electrolysis is limited by the rate of diffusion of analyte to the electrode. Diffusion current is the difference of limiting and residual current.

So it is much higher than the average current. In classical polarography current changes with the electrode size (mercury drop), Ilkovic equation describes the maximum current (AKA limiting current), but if you were to integrate to find the charge you will get less than I×t. That in turn means answer found using I×t is too high.

Quote
Apologies, 0.5/25 is millimolar divided by milliliters to get concentration of the solution.

Well, you already know it - but my question about units was intended to to point you in the right direction.

C=(I)(t)/(n)(F) is the equation to find concentration.

Again - units! That's not concentration, that's number of moles.
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