During our experiments, Cyclic Potentiodynamic polarization curve is used to study corrosion behavior of materials. In this curve, Potential E (Volts) in ordinate is plotted against log Current density(I), where I in ampere/cm2. Area under this curve gives power density as I understood.
But some researchers reported that Area under the curve gives the unit of Charge i.e, Coulombs. Can anyone please bridge the gap between power density and charge?
Q = I dt
amp sec/ cm2 = ampere x sec
in polarisation experiment sweep rate is volt/sec
E axis is nothing but time axis at a given sweep rate
the area under the peak will give amp x sec / cm2
see the pdf
to caclulate charge density use dQ = jdt and integrate. On the other hand, power is calculated using P = UI.
I expect the Cyclic Potentiodynamic polarization to be cycling voltammetry. If so, check the textbooks for more theory (e.g. Bard/Faulkner or Hamann/Vielstich)
It's easiest when you plot your data as 'current vs time'. The data will then look similar to a 'current vs potential-plot' (since the potential is a linear function of time). Finding the charge is then easy by integrating the 'current vs time-plot'
Q = I dt
amp sec/ cm2 = ampere x sec
in polarisation experiment sweep rate is volt/sec
E axis is nothing but time axis at a given sweep rate
the area under the peak will give amp x sec / cm2
see the pdf
common unit for power is the Watt, which equals volts x amperes. power density refers to power obtained per unit of mass (grams) or volume (ml).
Charge is typically measure in coulombs (which is amperes x seconds). A related measure is the (charge) capacity which is typically measured in in amperes x hour.
so power is voltage x current while charge is the area under the current curve. You have to plot the current signal as is (not log current).
In your curve Potential E (Volts) in ordinate is plotted against log Current density (I), where I in ampere/cm2. The area under this curve has the meaning of “power per active surface of the working electrode”, but the unit of this area is not in Watts (per cm2) which is the common unit of Power per area units.
In order to obtain this area in Watts/cm2, you have to convert the data of [log Current density (I)] to Current density (A/cm2) using the equation logI = a → I=10a. If you plot a new curve of E(V) v.s. Current density (A/cm2) the area under the curve will have units of (Ampere/cm2)X(Volts) = Watts/cm2 and corresponds to the power per active surface area of your electrode.
If you want to determine the Charge, you have to convert your potential data to time dividing by the scan rate (V/V/s = s) and convert the [log Current density (I)] to Current density as given above. The area under the curve of Current density (A/cm2) vs. time (s) will give the passed charge per active surface area of the electrode in Coulombs (A/cm2)X(s) = C/cm2.
area under current curve is charge only in single case: when measure current is capacity current. When i=dq/dt.
Have a look at
http://www.bio-logic.info/assets/app%20notes/20101105%20-%20application%20note%2011.pdf
Dr. V. V. Elkin,
Sir, Kindly find attached the CV curves below. The values used here are Potential E in Volts and Current I in A/cm2. When I change the X-axis from linear to log10 scale, then I could find the curve which is shown in literature.
Then, How shall I proceed to obtain the area under the curve, so as to obtain in the unit of charge,i.e Coulombs??....
Looking forward your reply...
Suppose by that great rate (dE/dt) you measure only capacity. Current is capacitive. (It is doubtful since faradaic current equals zero at minus 0,5 V). By potential rate equalled 1V/s capacity current is equal numerically capacity I=CdE/dt=C*1=Qul/s*cm2. We integrate C on t. And we get Q equals roughly 0.2 Qul/cm2. OK.
let us suppose the aria form to be right-angled triangle with cathetuses 0.4 A/cm2 and 1s (v=1V/s). The product of cathetuses half-and-half is 0.2. Qul is illiteral Qoulombs. Sorry. Yours Yolkin.
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