Electrolysis with inert electrodes (1 Viewer)

[DAAVE]

How many of us know how to predict what will happen at the anode and the cathode during electrolysis with inert electrodes?

Is there any easy way of remembering it?

 

[smallcattle]

i think all you need to know is that inert electrodes will allow for slower reactions than the reactive metals

 

[DAAVE]

Nah not what I mean...

Say u electrolysise CuSO4.

U get copper on the cathode but u get oxygen gas at the anode because sulfate is too stable to be oxidised.

and theres more that we have to remember, I think the syllabus calls them 'selected solutions' but it doesnt actually tell us what they are... well yeah...

 

[d_elmo]

when u get a question, look at all the different species that could possibly be oxidised or reduced. eg. in a NaCl solution with inert electrodes, the Na+ ions cannot be oxidised and lose another electron to become Na2+, but it can be reduced. go through all the species like that, and then using the standards table thinggo find out the emf each half reaction produces - the one with the most positive value out of the oxidation equations u have written will be oxidised, and the same for reduction.

 

[rumour]

I think the water will undergo electrolysis where there is an inert electrode, use the table to predict this.

 

[LSP]

selected solution can be in:
molten - which do not have water. so dont think anything about water
e.g. NaCl
Anode: Cl- ------> 1/2 Cl2 + e-
Cathode: Na+ + e- -----> Na (s)

dilute solution - this case you need to consider all possible solution that can be present in that electrolyte
e.g. NaCl possible ions present Na+, Cl-, H2O
Anode: Cl- ------> 1/2 Cl2 + e- E = -1.36V
H2O ------> 1/2 O2 + 2H+ + 2e- E = -1.23 V
Cathode: Na+ + e- -------> Na(s) E = -2.71 V
H2O + e- -------> 1/2 H2 (g) + OH- E = -0.83V
Look at the E value and use the one that requires less energy for the reaction to take place. this is the value closest to zero
:. Anode: H2O ------> 1/2 O2 + 2H+ + 2e- E = -1.23 V
Cathode: H2O + e- -------> 1/2 H2 (g) + OH- E = -0.83V

concentrated solution
taking same example. in this case there are more Na+ and Cl- ions than water
so once again look at the possible reaction above and determine which reaction is more likely to occur. This case
Anode: Cl- ------> 1/2 Cl2 + e- E = -1.36V
Cathode: H2O + e- -------> 1/2 H2 (g) + OH- E = -0.83V

 

[DAAVE]

OK basically I've come up with some rules.

It is impossible to reduce aqueous solutions of things with potentials <-1V

It is really hard to oxidise things as the number increases (fluoride is hard to oxidise), however you can oxidise things like Cl if it has a high enough concentration.

It is really hard to oxidise ions like SO4 etc.

U can predict most of if by knowing them.