The figure of 1.2v seems to be use as the max an aqueous double-layer capacitor can operate. Is this voltage electrode material dependent or is it mainly based on the electrolyte e.g. water? If it is electrode dependent then I would expect this voltage to also vary. Or is this the activation energy require to decompose water?
Any ref. to a paper kindly welcome.
WayneL
You don't need a reference. Look up the delta_G for the decomposition of water, and then translate that into delta-E. When electrolysing water, this is roughly the voltage needed to break up the water, and any voltage over that then goes into ohmic heating.
Thanks Dieter for the help, again.
E= delta_G/-nF
from CRC delta_G = -237,000Jmol-1
:. -237,000/-2x96458
= 1.2285v
WayneL
WayneL schrieb:
This is a correct calculation for the theoretical value for the decomposion of water. But you have to take into consideration the overvoltage which is material dependent. Guenter
The original question was about the maximum voltage the capacitor formed by the double layer can take. Up to 1.2 V, there is hardly any current, but above that, water will be electrolysed, and the capacitor will be leaking. How strongly, depends on the electrode material (Butler-Volmer relations). Pt would presumably be worst, and Hg best. But in all cases, more than 1.2 V will make it leaky. Obviously, in another solvent that voltage will be different.
...In this case then the electrolysis voltage will have to be added to the cell overvoltage. ie. Cell voltage in a charged state say 2 volts. 2v +
1.2285 = 3.2285V will be the necessary input voltage. We are getting slightly off the orginal question though.See an article on "electrochemical capacitors" in the Electrochemistry Encyclopedia
Good luck: Z.N.
WayneL wrote:
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