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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: deafcaves on November 04, 2016, 05:06:24 AM

Title: Calculating Area of an Electrode
Post by: deafcaves on November 04, 2016, 05:06:24 AM

Hello,

I am attempting to calculate the area of my electrodes. I have carried out a scan rate study using cyclic voltammetry with Ferricyanide (5mM, PBS). I have my values, however, I´m having trouble deducing wherever the result I have are correct.

Scan Rate (mVs): 10, 25, 50, 100
ipa (uA): 12, 27, 39, 52

I have plotted ipa verses the square root of the scan rate and derived a relatively straight line. I used a diffusion coefficient of 7.2 x 10^-6 for Ferricyanide.

I calculate my electrode area to be...
=(2.69*10^5)*(1^(3/2))*SQRT(D)*SQRT(V)*0.05M*ipa
= 0.043cm^2

This seems particularly small. Any ideas?

Title: Re: Calculating Area of an Electrode
Post by: mjc123 on November 04, 2016, 05:30:36 AM

Well, it should be obvious that if ipa is proportional to sqrt(ν), you would expect either ipa/sqrt(v) or sqrt(v)/ipa in your equation, instead of which you have sqrt(v)*ipa, which should set alarm bells ringing. Now the equation I find is
ipa = Ipa/A = -2.69*10⁵*n^3/2*c_O^∞*D^1/2*v^1/2
Now rearrange that to get an equation for A.

Title: Re: Calculating Area of an Electrode
Post by: deafcaves on November 04, 2016, 05:38:58 AM

Thanks for your reply.

I think this is it?

A = ((-2.69*105*n3/2*cO∞*D1/2*v1/2)/Ipa)

Title: Re: Calculating Area of an Electrode
Post by: mjc123 on November 04, 2016, 05:43:59 AM

No - upside down.
Also be careful about units - does that factor 2.69*10⁵ assume that Ipa is in A, or mA,or µA? that ν is in V/s or mV/s? etc.

Title: Re: Calculating Area of an Electrode
Post by: deafcaves on November 04, 2016, 05:51:49 AM

Okay.

A = ipA/-2.69*105*n3/2*cO∞*D1/2*v1/2

Therefore:

A = 0.012 amps / -2.69*10^5 * 1 * 0.05M * SQRT(7.2*10^-6) *SQRT (0.01V)

= 0.0033cm^2

Title: Re: Calculating Area of an Electrode
Post by: mjc123 on November 04, 2016, 09:22:53 AM

You're being careless. 12 µA is not 0.012 A. 5 mM is not 0.05 M.
Instead of using 12 µA/sqrt(10 mV/s), why not use the slope of the straight line you plotted. That will probably be more accurate.
Are you using the right units? The reference where I got that equation said .."where Ipa is in A, D in cm²/s, v in V/s and c in mol cm^-3" (note!)

Title: Re: Calculating Area of an Electrode
Post by: deafcaves on November 04, 2016, 11:29:23 AM

My calculated slope is 0.00018 Amps/v^1/2.

Therefore:

= 0.00018 * 1 * sqrt (7.2*10^-6 cm^2 s1) * sqrt (0.01 V/s) * 0.000005 mol cm^3

= 2.44236E-13

Title: Re: Calculating Area of an Electrode
Post by: mjc123 on November 04, 2016, 12:32:40 PM

NOnononono. This is really, really sloppy. You divide by the rest of that stuff - have you forgotten already? And you don't need the factor sqrt(v) because that is accounted for in the slope. And you've forgotten the numerical factor.
A = slope/(2.69*10⁵*1*5*10^-6*sqrt(7.2*10^-6))
If you want to get anywhere in science you really MUST get the habit of taking care over things like copying numbers correctly, using the right units, converting between units (e.g. mM to M) correctly, rearranging equations correctly etc. If you're careless you will get things wrong all the time and not know why.