Author Topic: Considering the Formation of Water Molecule  (Read 2103 times)

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Re: Considering the Formation of Water Molecule
« Reply #8 on: April 02, 2016, 10:05:31 am »
In the water molecule the oxygen is the most electro negative atom that atracts to it the electrons of the hydrogen to the point it form a covalent strong bond.

when we drop a metal in water it will interact with it because a metal also have an electronegativity,,

lf it has a lower electronegativity than hydrogen that forms the water molcule it will form a hydroxyde such as barium sodium or potassium...

the question is the lowest reactive metals will have a electronegativity of the order of hydrogen 2.2 like platinum iridium paladium ... nickel chromium and iron is around 1.8

oxygen is much  more electronegative is 3,44 so what happen is that it wants more electrons than metals want for example....

when we drop alcaly metals on water the metal will react by substituting the hydrogen of the molecules with an electron... causing the molecules to partial break into H neutral atom and OH- ion, because it will give up also its electron to form a positive ion since its lowe electronegativity than hydrogen being more or less bellow 1...... i mean the metal.. so the point really is that is a substitution reaction... when we drop the metal straight on water...

misteriously if we have 3,44-2,2 we get 1,24v the minimum voltage to start a substitution reaction by electrolysis where the substitutin occur at the electrode surface...

so the importance for the reaction to occur is to have this 1,24v to allow it going...

this if we have platinum electrodes on both sides...

some metals will go into solution at a voltage lower than this aplied like silver

the question is that stainless steel takes over 1,44v to get decent current level...

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Re: Considering the Formation of Water Molecule
« Reply #9 on: April 02, 2016, 10:31:51 am »
so basically what i mean is that if we could positively charge an electrode it will have a greater electronegativity...

since it will want more electrons it will act like as an oxygen to the water molecule perspective so thats where the oxygen will be generated...

if we apply a negative voltage similarly it will act as electron that will substitute the hydrogen electron on the molecule liberating hydrogen..

this easier occurs by half reactions than the water molecule as a whole meaning that is easier to ionize two molecules and take a hydrogen atom from each than is to break apart the whole water molecule...

this mean that the generation of radicals is significant during aply of high voltage

platinum works as a electrocatalist meaning it helps the ions to contact the electrode thru the dual layer barrier. reducing the potential required to start the reaction...

but it also promote spontaneous recombination when hydrogen and oxygen touches it at the same time..

the dual layer is a high capacitance

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Re: Considering the Formation of Water Molecule
« Reply #10 on: April 02, 2016, 12:30:21 pm »
1+1×2-4=0

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Re: Considering the Formation of Water Molecule
« Reply #11 on: April 02, 2016, 16:22:07 pm »
? what you mean/?