But I don't know how that comes along. Can you explain please,
It's 4OH- ---> O2 + 2H2O + 4e-
Write electrons on right hand side - you can't really minus minus in reactions so just move it to the other side The charge is - and there are 4 hydroxide ions In order to become neutral, 1 hydroxide ion loses 1 electron. There are 4 hydroxide ions (to balance the equation) and consequently 4 electrons.
But I don't know how that comes along. Can you explain please,
As explained by the other poster, but here is my take on trying to explain. Not sure if you have learnt redox chemistry, but say you have. Most chemical reactions are redox reactions - involving both reduction and oxidation - these are complementary
In a balanced redox equation, you don't see electrons being added on left hand side(LHS) or right hand side(RHS), but actually complete redox equations are made up of usually two half equations.
Now, the key here is half- ; this half-equation shows the flow/movement of electrons on paper to clarify things. the equation you posted is one such half equation - you need to be able to know is that happening at cathode or anode (i.e. is it reduction/oxidation happening there, if so, you need to know what reduction/oxidation means in terms of electron loss/gain)
it might seem daunting, but with practice, it'd get easier! trust me, that is how we all chemists learn too (perhaps i am generalising too much, )
Write electrons on right hand side - you can't really minus minus in reactions so just move it to the other side The charge is - and there are 4 hydroxide ions In order to become neutral, 1 hydroxide ion loses 1 electron. There are 4 hydroxide ions (to balance the equation) and consequently 4 electrons.
Okay so if OH is neutral then how does that form water and oxygen?
OH- is easier to discharge than halide ions, BUT in the concentrations found usually the halide ions vastly outnumber the hydroxide ions (1 x 10-7M) and are preferentially discharged.
When the halide ions are very dilute (0.01M and lower), then oxygen is discharged.
There is also a concentration ratio at which they are both simultaneously discharged.
I'm not actually sure. May go back into solution but I assume it depends on the set up - whether it's in a lab or on an industrial scale.
So you know when you were talking about the polyatomic and mono atomic, what if we had to electrolyse copper sulphate solution. Then we've got SO42- and OH- which one would discharge?
So you know when you were talking about the polyatomic and mono atomic, what if we had to electrolyse copper sulphate solution. Then we've got SO42- and OH- which one would discharge?
Read through the thread again!
You have already been told that OH- is always discharged (as far as school chemistry is concerned) UNLESS there are halide ions at highish concentration.