Current increase electrolytic cell

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I made a cell using copper and aluminum with liquid bleach electrolyte. I used two different meters to measure current. The analog meter highest setting is 250 ma and DMM has 10 amp setting. When using analog meter the current slowly climbs, but the highest I've seen is 200 ma. When I use DMM the current slowly climbs but may go as high as 3 amps. Any ideas why the big difference in measurements.
 

Analog meter has a current shunt. Often the movement is a 10K/V sensitivity and the shunt is limiting current. Anyway, you need to read the fine manual that came with it. DMM also has a current shunt but the value is much lower.

You did not say how much voltage you applied. If the applied voltage is like 5-10V, the case is fully understandable.

Liquid bleach is an undefined material. Better to use solutions of known chemicals.
 
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My apologies, I'm confusing electrolytic with galvanic cell. I thought maybe the meter resistance had something to do with it but I can't understand why. If the cell can deliver x amount of current it seem to me they would both measure the same or relatively close. I don't get it. This is just something I scrapped together with what I have at hand. I realize bleach isn't optimal but it's all I have at the moment.

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This is the contraption.
 

I thought maybe the meter resistance had something to do with it but I can't understand why...
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Because a galvanic cell or an electrochemical cell, both have considerable internal resistance. Commercial cells (say a Pb acid cell or a Li based one) are designed to minimize this internal resistance.

Perhaps now you understand why two measurements with two different meters can give two different results.

But I asked about the voltage: you did not specify the applied voltage (electrolytic cell) or the EMF produced (galvanic cell).

These are modelled as a voltage source in series with a resistance. In reality it is more complex because of internal polarization.

The slow rise of current is because of Al (it has a passivating oxide layer that must get eaten up by the electrolyte).

Unfortunately you will not be able to reproduce the same results after some time.
 
The EMF (galvanic) varies considerably but on average about 1.5 volts open circuit. I have seen as high as 1.95 volts. I measured roughly .8 volts at 2.8 amps on one test. You are 100 % correct about not reproducing same results. Once you stop drawing current if you wait more than a few seconds it won't return to that level. I have to clean plates with muriatic acid and change electrolyte to get similar current measurements.
 

First you must try to get reproducible results.

By design, your setup is irreproducible. Why?

To get a stable potential at the Cu plate, you must use Cu(II) at the interface. Use a dilute solution (say 10%) of copper sulphate, pour on a filter paper and stick on the copper plate.

Same way, for the Al plate, use a dilute solution of aluminium alum, soak on a filter paper and stick on the Al plate.

The filter papers must stay wet but not dripping.

Use a damp separator (a wet filter paper will do) and stack these two plates.

Now you have a setup something like this:

Cu-Cu(II) (on filter paper)- separator (damp filter paper)- Al(III) (alum solution on filter paper) -Al

The electrodes must be conditioned before use. But that is a different story.

But this set up will be far more reproducible. Expected potential will be around 1.8V
 
The major effect is warming - as the liquids heat up -the conductivity increases and the current goes up - and it heats more ...! - it is a bit of a positive feedback spiral - a lot of battery researchers have encountered this sort of thing too - the final limit is the number of ions in solution ...
 
I found a cheap drain cleaner that contained water , sodium hypochlorite and sodium hydroxide. With this product I get very different results. The voltage came up to 1.74 immediately, as compared to the bleach which increased slowly. The amperage was there also. Started at 5.5 amps and gradually fell off. But here is my question. When I place cell in this liquid there is a strong reaction at aluminum plate which forms gas bubbles. The odd thing is when I start drawing current the reaction slows down drastically. Anyone have and explanation for this.
 

When I place cell in this liquid there is a strong reaction at aluminum plate which forms gas bubbles. The odd thing is when I start drawing current the reaction slows down drastically. Anyone have and explanation for this.
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The alkali (sodium hydroxide) removes the protective oxide layer from the Al surface. It also attacks the metal with production of Hydrogen. The gas bubbles you see are hydrogen formed by oxidation of the Al.

When you start drawing current, the reaction does not slow down. But it may appear (incorrectly) that the hydrogen evolution has considerably reduced. The fact is that the hydrogen ions are now pushed toward the copper plate.
 

What other gas could be produced in this reaction. I don't think that hydrogen gas would make you cough and burn your nose.

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Would the sodium hypochlorite release some other gas.
 

What other gas could be produced in this reaction. I don't think that hydrogen gas would make you cough and burn your nose....
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Hydrogen evolution is the basic reaction. The hypochlorite acts a depolariser and you will get HCl produced as a result. HCl is an acid and will get neutralised by the alkali (NaOH). Therefore there will be no gas produced in the first few minutes of the electricity production.

If the alkali (NaOH) gets over first (while NaOCl is still left) you will get HCl gas which can cause irritation (make your cough and burn your eyes and nose).

If the NaOCl gets consumed first (while NaOH is still left), you will get only H2 gas released from the Cu surface.

If you see lots of gas evolution from the Al surface, the reaction is not contributing to the electricity generation. The side reaction will consume the electrode (Al; Cu electrode is not consumed at all) without supplying current.
 
If the NaOCl gets consumed first (while NaOH is still left), you will get only H2 gas released from the Cu surface.

In post #9 you stated that hydrogen is released from Aluminum plate. I must be misunderstanding something.
 

In post #9 you stated that hydrogen is released from Aluminum plate. I must be misunderstanding something.
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Hydrogen gas will appear at the Al electrode when there is no current. As soon as you start drawing current, hydrogen ions will move towards the Cu electrode and get discharged there.

The internal circuit (current path) is completed by hydrogen ions moving from the Al electrode to the Cu electrode. The external circuit is completed by electrons going from the Al electrode to the Cu electrode

If you use Al and Cu wires as the electrodes dipped in a electrolyte in a glass vessel, you will be able to make out more clearly.
 
What effect does potassium permangenate have on sulfuric acid. My electrodes are graphite sheet with sulfuric acid , water and potassium permangenate solution. The aluminum side has naoh solution and both solutions are on paper towel. With sulfuric acid only the initial voltage is around 2 volts. With permangenate added the initial voltage is much higher. The sample I'm using at the moment started at 2.8 volts.
 

I am not sure whether you are conducting these experiments with sufficient care and safety protection. Potassium permanganate mixed with sulphuric acid is a very powerful oxidant and can cause fire or explosion.

The success of a electrochemical cell depends on the separator membrane or film that keeps chemicals from reacting from either side with each other. But it must also allow easy passage of ions, most importantly in this case of Hydrogen ions.

Acid (sulphuric acid used on one side) and alkali (NaOH is used on the other side) will neutralize each other.

Common cells use Zn instead of Al. It uses MnO2 instead of KMnO4 as the depolarizer. Instead of H2SO4, it uses NH4Cl, a rather weak acid (it is salt of NH3 and HCl).

In this case KMnO4 acts as an oxidising agent and converts the H2 gas being produced to water.

Please give due respect of chemicals.
 
OK, firstly, working with such materials are not safe and cause in worst condition cancer, so be aware of what you are doing.
when you google the standard potential of metals or elements you will see that Al has standard potential of -1.66 V and Cu has standard potential of +0.34 V. it means when you use them in an electrolyte in a cell, it must have the theoretical EMF of +2.0 V. but what you see in practice is lower than that because the cell you made is not ideal. making an ideal cell needs some extra parts and reagents like the ionic bridge or membrane to separate electrolytes from each other and using same ions than your electrode around each electrode.
bleachers and cleansers has chemicals that will result in producing Cl2 gas (chlorine) which is the first chemical warfare used in WW II, so obviously DO NOT USE them. the best electrolyte for the galvanic cell is ions of your electrodes but if you don't have them, use KOH (or potash) because it will only produce H2 and O2 in water and attack to your electrodes less than others. aslo be aware of explosive potential of H2!!!
in electrolysis always Anions reach Anode and get Oxidized. remember it as AAO.
and on the other side always Cations reach Cathode and get Reduced. remember it as CCR
in Galvanic cells it is vice versa, so you can say that Anode and Cathode has changed.
in galvanic cells when the standard potential of an element is negative like Al in your cell, so it gives electrons to cations and Reduce them freely. like 2H+, 2e -> H2. also the higher potential electrode will take electrons from anions and oxide them, like the Cu electrode in your cell. 2OH- -> 4e , O2 , 2H+
in electrolysis it just changes, so the anions will go to the Al side and cations will go to the Cu side. as simple as that.
if there is any Cl- anions in the electrolyte, it will compete with other ions and therefore get oxidize to Cl2 gas.
 


I have been attempting to make a rechargeable cell using aluminum and graphite material. The cell appears to take a charge but I'm not sure if it's just charging like a capacitor would. In the attached is a plot of one discharge cycle. Is there any indicator in the plot to whether or not it's a capacitive discharge or not. When I initially formed the cell I discharged it down to 1.5 volts. I then allowed the cell to recover and it settled at around 1.72 volts. I then applied 2.5 volts for 2 minutes. After I disconnected power the voltage on the cell slowly dropped till it was somewhat stable at 1.85 volts. I then started the discharge as seen on the plot which is in 30 second intervals. The load was 235 ohm resistor.
 

Congrats either a 32 Farad capacitor ( 1.80 - 1.75V for 210 sec on a 7.65mA discharge ) - or a battery ( actually a battery ) - ( 446uA-hour to 1.75V )
 
I have been attempting to make a rechargeable cell using aluminum and graphite material.
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It is not possible to use Al and C electrodes to make a rechargeable cell using water as an electrolyte. You need to use non-aquous electrolyte for recharging to be effective.

When you are charging such a system, you are basically causing electrolysis of water. This energy is lost for ever. Al is not being regenerated.

I cannot predict the output if the processes are not clear.
 
I am using using aluminum sulphate only on paper separator.

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At least I think it is. I formed it myself so I'm not positive.
 

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