Text iron fe metal is corroded in an aerated dilute acid...

Text: Iron (Fe) metal is corroded in an aerated dilute acid solution with a pH of 7. Since the corrosion potential (Ecorr) of this corrosion cell is -0.45 Volt (SHE) at pH=7, calculate the cathodic current density required for oxygen reduction using the following equation: i = io * exp((b/2.3)*(Ecorr - EoFe)) According to your results, explain which cathodic reaction is dominant in the corrosion of iron for this corrosion cell? (Take RT/F= 0.0591) Given: EoFe = -0.44 V (SHE) aFe2+ = 10^-6 Tafel constant, b (2.3βa) = 0.06 V/decade io = 10^-11 A.cm^-2 EoH = 0.00 V (SHE) Tafel constant, b (2.3βa) = 0.12 V/decade io = 10^-10 A.cm^-2 PH2 = 1 atm. (This question is a corrosion question)

Transcript

00:01 From here, if we see, we have to find the rate of corrosion of the iron.

00:06 So iron has a problem that it undergo corrosion whenever it is exposed to the moist climate.

00:14 So here it is given that ph, it is 2.

00:18 So from here, h positive concentration, it is minus, so from here ph, it is minus low of h positive concentration.

00:28 So from here, h positive concentration would be 10 raise to the power minus 2.

00:32 So from here what we can say anodic polarization.

00:38 So anodic polarization will occur.

00:43 And due to the anodic polarization, electrode potential for the corrosion, it is equal to electron potential, in note 10 log, i corrosion divided by the i, which is here, operational.

01:01 So from here, also the cathode, catholic polarization can be written both anodic and the catholic polarization are written for this particular case so from here a node corrosion it can be written as e h2 plus bc log i corrosion and along with this i h2 so from here what is happening we should see this reaction f2 positive it will be gaining to electron and it would be for forming f f e note corrosion it can be written as e note minus r t divided by the nf l n one divided by the f2 positive concentration so from here if we see this value of the e note which is here for this particular case this e note for this case here it is minus 0 .44 and along with this it is written as 8 .314 multiplied with temperature in this particular case the temperature which is here it can be written as 298 kelvin divided by the 2 multiplied with this 965 00 ln 1 divided by the 8 10 to the power minus 8 so from here e corrosion it would be be for the case of the fv minus 0 .67 65 volts so this is a corrosive electron volt so if we see what is happening two moles of the hydrogen it will be taking two electrons it will be giving h2 so from here if we see the e for this h2 it can be written as we can say e note it will be written as e note it will be written as e note minus rt ln k minus rt not ln ln should be here and here it should be 2 so from here it is ln divided by the 10 to the power we can say the concentration of the h positive so this is like this so from here if we see e for the h2 so it would be e note for the h2 it is 0 so from here, e h2, it would be coming out to be minus 0 .1174 volts...

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