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Electrolysis help


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in my chemistry class we were recently given a electrolysis experiment to do with three variables to chose from (our group chose voltage), after the experiments were complete ourteacher told the class that there were 5 possible reactions (personally i think there are more) that could have occured and that we were supposed to figure out what they were and include them in our reports. (just so you know this is in final year of high school) I have so far only been able to figure out two of them which are

2H2O -> O2 + 4H+ + 4e- = -1.23 V

2H2O + 2e- -> H2 + 2OH- = -0.83


the experiment itself consists of two copper electrodes ina an aqueous copper nitrate solution we ran it at 4, 6, 8 and 10 volts on a power pack about which i can infer from the fact that as the voltage was increased more products were formed that when the voltage is increased the amperage also increases.


so could someone please help out a stressed school goer and inform me what reactions could be occuring in this experiment.


P.S. from 6 volts upwards a blue solid formed on the cathode along with the dark brown cupric oxide and at 10 volts after the power had been turned off most of the cupric oxide fell off the cathode (gradually not all at once) and when it did a bunch of bubbles came from the cupric oxide (both when it initially came off and when it was lying on the bottom of the beaker)

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blue compound is probably a copper oxide, are you familiar with writing half equations for the electon transfer (oxidation and reduction) reactions, then you have to multiply the half equations by a number that will make the amount of electons on either side balance. Then you add the two together and balance.


The reations taking place at differing voltages will make different reactions take place. Lookup the electro-chemical series on google if you dont already have one, this will give you the potential differences that are needed for different reactions, so then you know what to expect. If you need more helpo just ask.

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i thought the brown/black stuff was copper oxide and i think i have figured out what the gas would be (NO3-)


2H2O -> 4e-+ 4H+ + O2 = -1.23V

Cu+ -> Cu2+ + e- = -0.8V

Cu -> Cu2+ + 2e- = -0.34V

NO3- + 4H+ + 3e- -> NO + 2H2O = +0.95V

NO3- + 2H+ + e- -> NO2 + H2O = +0.81V

are these the 5 half reactions?

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