electrolysis dilemma

I tried to set up a electrolysis cell to produce hydrogen and oxygen. I used 2 stainless steel plates in KOH it worked fine. The problem came about when I tried to replace the 2 plates with a stack of 5 or 6 pairs of plates wired + - + - + etc. All the positive plates were wired together and all the negative plates were wired together.

I expected this arrangement to draw more current and produce more gas but it did not. Yet it is used in lead acid batteries and works fine. I did manage to improve the output by using two long parallel plates wound together in a spiral fashion.

Why is it that stacks of alternating plates do not improve the hydrogen and oxygen produced?
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 Admin This is not arrangement that is used in lead batteries, they are made of identical cells connected in series. Perhaps your current source is already working at maximum output?
 Each cell in a lead acid battery is made up of parallel plates a lead compound in a + - + - configuration. Then the cells are connected in series.

electrolysis dilemma

OK, small misunderstanding.
 What are you using for your power source?
 12 volt car battery.
 R = $$\rho$$ (l/A) So the resistance in your cell drops as you increase your electrode area (A). The resistance however also increases as you decrease the electrode spacing (l). So in effect what you may have done is kept the overall cell resistance constant - so current shouldn't change. rho is the electrical resistivity of the system - related to your electrolyte If you wish to increase the current I would increase the electrode AREA but keep the electrode spacing constant. That is enlarge the cell volume to fit more electrodes but kep the electrode gap the same. The electrical resistance in this case would drop and the current will increase - MORE gas evolution at the electrode surface Conversely, you can decrease the electrode spacing but keep the electrode area the same.

 Quote by Driftwood1 The resistance however also increases as you decrease the electrode spacing (l).
Quite the opposite.

Which you admitted in the last phrase BTW.

 Quote by Borek Quite the opposite. Which you admitted in the last phrase BTW.
typo - of course R is directly proportional to the inter-electrode distance and inversely proportional to the Area of the electrodes.

The point was to highlight the fact that you can double the electrode area, but if you half the electrode gap at the same time the electrical resistance remains constant so the current does not change.

If we can get the electrode area and spacing for both configuations I suspect that the ratio (l/A) has remained constant so the resistance hasnt changed - explains why there is no increase in current and no increase in gas generation at the electrode surface.

cheers

 Quote by Driftwood1 The point was to highlight the fact that you can double the electrode area, but if you half the electrode gap at the same time the electrical resistance remains constant so the current does not change.
If you double area you have also to double the distance to not change resistance. If you half the electrode gap while doubling electrode surface, resulting resistance will be four times lower.
 Very basic high school level science generally includes Ohms law and how it applies to various applications. In a battery cell for example The Electrical resistance R is directly proportional to the inter-electrode spacing and inversely proportional to the area of the electrode (this is for uniform constant electric fields - for other geometries this simple equaton does not apply) The proportionallity constant (rho) is the resistivity of the electrolyte or substance between the anaode and cathode. So, if you double the electrode area by introducing more electrodes, but at the same time you also double the inter-electrode spacing then the electrical resistance remains the same - current is constant - therefore gas production doesnt change Amazing how Ohms law creaps into things Where has Arydberg vanished to?

 Quote by Borek Perhaps your current source is already working at maximum output?
interesting.....

The electrode gap is obviously NOT equal zero and the electrode area isnt infinite.

Power = I^2{rho(l/A)}
 Admin You are forgetting about internal resistance of the current source, it limits maximum current in the system. You can't get more juice from the power supply than it was designed for, even if you short circuit it.

 Quote by Borek You are forgetting about internal resistance of the current source, it limits maximum current in the system. You can't get more juice from the power supply than it was designed for, even if you short circuit it.
true...

The electrode plates however behave like a linear Ohmic resistor (if we assume that the electrode gap is constant producing a uniform electric field and current density)

Why doesnt Arydberg provide the dimensions of the initial and modified electrode set up?
 I did not change the electrode spacing in all cases it was about 1/4 inch. You are trying to make a simple problem out of this. It is not simple.