New answers tagged electrolysis
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Whatever contamination in your tap water should be a great deal less than the concentration of CuSO4, but might be a factor in your experiment.
Conductivity is measured using a non-depositing AC current. The DC resistance or its inverse will depend on many factors, but all will require a measurement of voltage and current simultaneously.
Measuring the rate ...
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The solute must ionize, and as the concentration of ions in increased, further ionization is inhibited. The non-ionized molecules are inert, and interfere with reaction rate.
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Do I use the total surface area of both Cathode and Anode or just Anode?
You use the cross-section, just like it were a wire.
Cross-section of both Anode and Cathode or just Anode?
Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas.
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The rate of electrolysis (decomposition of water) is defined by the current that passes. If you apply a constant voltage, factors arise as suggested in the comment by MaxW.
At constant current, the rate of hydrogen (and oxygen) evolution would be somewhat different initially because on a microscopic level, hydrogen atoms have to cover the surface to some ...
1
For hydrogen to burn or explode in air, its concentration needs to be at least 4%.
Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing ...
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I asked a somewhat similar question a few weeks ago and Poutnik made a smart suggestion that I simply mix the H2 with nitrogen to keep it safely below explosive concentrations, which I set out to do prior to extracting it outside with a fan. So the nitrogen ensure that concentrations are low enough for the fan engine's sparks not to light it on, and the fan ...
5
First, sodium burns very nicely in air. However, it also burns in chlorine, returning back to the $\ce{NaCl}$ with which you started.
Most welders provide alternating current, meaning the the electrode becomes alternately positive and then negative (50 times per second [50 Hz] in much of the world, and 60 Hz in North America). Therefore, both $\ce{Cl2}$ gas ...
3
Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed.
6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small.
Consider also the graphite electrodes may get deteriorated by the ...
1
Absolutely a good experiment!
Diffusion of ions, propelled by the electric field (i.e., current) will occur at a given, calculable rate. Deposition of those ions (and conversion to $\ce{O2 + H2}$) will occur and the $\mathrm{pH}$ will rise at the cathode (negative electrode) and fall at the anode (positive).
Then back diffusion will occur as the $\ce{OH-}$ ...
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