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Electrolysis of water |
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Sep1-07, 11:01 AM
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#1
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chanller is
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Electrolysis of water
While doing electrolysis of water, I have put a rheostat and a battery in series with the apparatus. I found that when I increase the resistance of the rheostat, the number of bubbles produced per second decreases. How can you explain that?
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Sep1-07, 11:17 AM
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#2
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Gokul43201 is
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Before we can respond, we need to see what you think. Do you have any idea what other electrical quantities are affected by changing the resistance in a constant-voltage circuit? And the bubbles - what physical quantity do they relate to?
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Sep1-07, 11:32 AM
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Last edited by chanller; Sep1-07 at 11:39 AM..
Reason: Too impulsive!
#3
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chanller is
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Electrolysis of water
If the power withdrawn from a battery depends on the resistance of the circuit, then i guess current changes if P=IV. But, how can you keep the voltage from a battery constant while varying power withdrawn?
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Sep1-07, 11:35 AM
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#4
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Gokul43201 is
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Yes, that's correct, though it actually follows more directly from Ohm's Law, V=IR. So, does the current increase or decrease?
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Sep1-07, 11:49 AM
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Last edited by chanller; Sep1-07 at 11:56 AM..
#5
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chanller is
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I believe current for the electrolysis decreases if the voltage is constant and if current = electrons and electrons= gas bubbles, current decreases. Is it that when i increase resistance of the rheostat, power through it increases and that through the electrolysis decreases? This would explain why gas flow decreases but then it would mean power from the battery is constant! But, how can you keep the voltage from a battery constant while varying power withdrawn?
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Sep1-07, 01:35 PM
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#6
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Gokul43201 is
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There is no need to worry about power here. A bettery is, by definition, a device wich produces a constant voltage over a large range of currents. The power dropped (both at the rheostat and at the cell) decreases as you increase the resistance (because the current decreases, while the voltage is constant) but that's not the relevant quantity here. The relevant quantity is the current, which decreases.
As you mentioned, the current is simply the rate at which electrons are transfered between the electrodes. Each pair of electrons arriving at the cathode makes a molecule of hydrogen gas form there. The slower the rate of electrons, the slower will be the rate of hydrogen bubble formation.
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Sep1-07, 01:55 PM
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#7
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chanller is
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Well, what I did not know is that voltage from a battery is constant; now everything is clear! But can the voltage be changed and if yes, how?
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Sep1-07, 02:48 PM
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#8
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Astronuc is
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The battery represents the maximum allowable voltage, and then there is a voltage drop proportional to resisance and current.
Two resistance in series represent a voltage divider.
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Sep1-07, 05:03 PM
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#9
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chanller is
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I finally had to leaf through my physics book. Well I learned that physics was not just theory which I have to learn by heart!! Thanks everyone!
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Sep1-07, 06:34 PM
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#10
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chanller is
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Talking about electrolysis, what happens to the excess power input when gas production rate remains constant even though power is increased? Droplets of water settled at the top of my gas storage cylinder, does this mean excess power is converted into heat which vaporises water in the electrolyte?
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Sep2-07, 06:26 PM
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#11
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Astronuc is
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Originally Posted by chanller
Talking about electrolysis, what happens to the excess power input when gas production rate remains constant even though power is increased? Droplets of water settled at the top of my gas storage cylinder, does this mean excess power is converted into heat which vaporises water in the electrolyte?
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There is heat involved from resistance, and the heat will cause some liquid to evaporate which would condense on a cooler surface.
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Sep3-07, 07:55 AM
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#12
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chanller is
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If i use 1 molar sulphuric acid at rtp what voltage do you think i should use? 1.5V will be good?
Also, how can i keep the concentration of the acid constant?
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