- #1
TJT
Hi All,
I have a quick - probably quite basic - electrochemistry query I hope you can help with...
I have an anode and a cathode separated by water with current flowing between. There is some oxygen dissolved in the water so the reaction O2 + 2H20 + 4e -> 4OH occurs. Electrons flow out of the anode and into the cathode. At the anode, the electrons are used to convert O2 and H20 to OH (hydroxide). At the cathode electrons are removed to re-form O2 and H2) from the hydroxide?
My query is what proportion of the electrons are used in the anode/cathode reactions? If all the electrons are used for chemical reactions then is there no current flow?
Do you get zero current until you have converted all the O2 to OH? Doesn't seem likely to me but I'm sure I am missing something - I haven't done chemistry for a while...
Also, I know H2O is basically an insulator unless is has ionic substance in it so assume that there is some NaCl in the water and that this doesn't get involved in the reaction...
Would be grateful for any advice :-)
I have a quick - probably quite basic - electrochemistry query I hope you can help with...
I have an anode and a cathode separated by water with current flowing between. There is some oxygen dissolved in the water so the reaction O2 + 2H20 + 4e -> 4OH occurs. Electrons flow out of the anode and into the cathode. At the anode, the electrons are used to convert O2 and H20 to OH (hydroxide). At the cathode electrons are removed to re-form O2 and H2) from the hydroxide?
My query is what proportion of the electrons are used in the anode/cathode reactions? If all the electrons are used for chemical reactions then is there no current flow?
Do you get zero current until you have converted all the O2 to OH? Doesn't seem likely to me but I'm sure I am missing something - I haven't done chemistry for a while...
Also, I know H2O is basically an insulator unless is has ionic substance in it so assume that there is some NaCl in the water and that this doesn't get involved in the reaction...
Would be grateful for any advice :-)