gas yield from bubble volume
DESCRIPTION
This show the direct relationship between electrochemical bubble formation and gas yield in respect to electrode geometry and how to calculate the yield from the prior.TRANSCRIPT
Bubble nucleation and gas yield
We used video to record the bubble nucleation on the cathodes and on the water surface ofH3PO4 and H2SO4, due to opacity. Due to perspective distortion in the video we used theactual cathode dimensions for scaling for example the a 1.6667 deviation in width, 10 mmin the video = d and 6 mm actual size = a, d/a = 1.6667 which was used for gas yield as in 2-D approximation.5 Gas yields from bubble nucleation were calculated with the use equation1 and 2:
V=(π(r2
2−r1
2)h)
3(1)
V= volume of gas in cm3; r2= radius of cathode with bubble film in mm; r1=initial radius of cathode inmm; h= height of bubble formation in mm; π=3.14159. By use of volume of a hollow cone or eq. 1allowed us to determine the volume of bubble nucleation on cathode 4. The value (r 2
2−r 1
2) from
eq.1 and eq.2 negated the cathode radius giving only the bubble nucleation volume which correspondedto gas yield and active sites on the cathodes.
π(r22−r1
2)h=V (2)
V= volume of gas in cm3; r2= radius of cathode with bubble film in mm; r2=initial radius of cathode inmm; h=height of bubble nucleation in mm; π=3.14159. Eq. 2 similar to eq. 1 as the volume of a pipe,
0.5(A
96,500)(24,466cm3
/mol)=cm3 (3)
Since Faraday's law showed the relationship between deposition is propotional to amperes eq. 3 wasused as the theoretical yield. 0.5 = cathhode reaction stoichiometry for 2H, 1 mol e-/2 mol e- ; A=corresponding amps used for each cathode; 96,500= 1 mole of coulombs; 24,466 cm3/mol= volume of1 mole of ideal gas in cubic centimeters at 298 K; cm3=volume of gas yield.