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Supercapacitor Electrolyte

  1. Nov 14, 2015 #1
    Hey guys,

    Currently I am trying to build a low tech super capacitor based on the working the principle. Right now I am using a single beaker with two aluminium foil positive and negative electrodes with an electrolyte of 6M NaCl solution. The power source is a new-ish triple-A battery running at 1.4 volts. What happens is gas evolves on the electrodes. How do I keep this from happening? I understand that I must change the electrolyte but for what reason? Also will charge be able to be stored on the electrodes with a voltage below 1.23v using the current electrolyte? Thanks and appreciate any input.
     
  2. jcsd
  3. Nov 14, 2015 #2
    Capacitors are formed when two conductors are separated by an insulator. NaCl is not an insulator.

    Super-capacitors (if I understand the theory; not sure about that) allow a very small current to flow through the insulator making each carbon particle to act as it's own conductor. This way the total area of the capacitor is the sum of all the millions of carbon particles, and the distance between them is nearly zero. This leads to large capacitance.

    The carbon dust should be as fine as possible. The insulator liquid should nearly, but not completely, block current flow. (If the current is completely blocked, a proper voltage gradient won't form between the particles and most of them won't be part of the capacitor.) Obviously more current flow means more losses. (It can also lead to chemical reactions on the electrodes. :oldwink: )
     
  4. Nov 15, 2015 #3
    Regular supercapacitors use KOH as their electrolyte. That has higher conductivity than NaCl solution. For a supercapacitor, the electrolyte must be ionically conductive and not conduct the electrons from the anode to the cathode and thus maintaining charge separation. I think really my question is how do conventional supercapacitors achieve voltages of 2.35 volts when the breakdown voltage of aqueous electrolytes are 1.23v?
     
    Last edited: Nov 15, 2015
  5. Nov 15, 2015 #4

    CWatters

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  6. Nov 15, 2015 #5
    Actually I was just coming across this: http://diygadgets.co.za/diy-supercapacitor-home-made-super-battery/

    Basically this is instructions to make a supercapacitor using an aqueous electrolyte and it too operates only below 1.2 volts. This bodes the question for how do commercially available supercapacitors produce over 1.2 volts with aqueous electrolytes? Is there something I am missing here? If anyone trusts it here's a tid bit about electrolytes from Wikipedia's supercapacitor page for someone who can maybe explain? :


    Aqueous[edit]
    Water is a relatively good solvent for inorganic chemicals. Treated with acids such as sulfuric acid (H
    2SO
    4), alkalis such as potassium hydroxide (KOH), or salts such as quaternary phosphonium salts, sodium perchlorate (NaClO
    4), lithium perchlorate (LiClO
    4) or lithium hexafluoride arsenate (LiAsF
    6), water offers relatively high conductivity values of about 100 to 1000 mS/cm. Aqueous electrolytes have a dissociation voltage of 1.15 V per electrode (2,3 V capacitor voltage) and a relatively low operating temperature range. They are used in supercapacitors with low energy density and high power density.

    Organic[edit]
    Electrolytes with organic solvents such as acetonitrile, propylene carbonate, tetrahydrofuran, diethyl carbonate, γ-butyrolactone and solutions with quaternary ammonium salts or alkyl ammonium salts such as tetraethylammonium tetrafluoroborate (N(Et)
    4BF
    4[68]) or triethyl (metyl) tetrafluoroborate (NMe(Et)
    3BF
    4) are more expensive than aqueous electrolytes, but they have a higher dissociation voltage of typically 1.35 V per electrode (2.7 V capacitor voltage), and a higher temperature range. The lower electrical conductivity of organic solvents (10 to 60 mS/cm) leads to a lower power density, but since the energy density increases with the square of the voltage, a higher energy density.
     
  7. Nov 15, 2015 #6

    anorlunda

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    The Wikipedia text you quoted says 1.2v per electrode with 2 electrodes. It forms two capacitors in series. Go back and read the whole Wikipedia article on supercapacitors.
     
  8. Nov 15, 2015 #7
    I understand. Now how do make that work in a supercapacitor? If you read the instructions to build one I've provided, you have the basic design of a supercapacitor with two electrodes. According two what your picking from the wiki passage, that designed capacitor should also function at 2.3v and not breakdown above 1.15v. Why are commercial supercapacitors rated at 2.3v in this case? What am I missing here?
     
  9. Nov 15, 2015 #8

    CWatters

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    Two capacitors in series have twice the breakdown voltage of one. eg 1.2 * 2 is about 2.4V.
     
  10. Nov 15, 2015 #9
    Right, but conventional 2.3v supercapacitors are single cells rather than two in series as I understand. I don't get how that works. I also understand how this can easily be explained by adding the voltages per electrode, but even then the potential difference would be greater than 1.48v causing electrolysis within an aqueous electrolyte.
     
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