HSC 2004 Chem Q 29 (a) (ii) - Predict Electrolysis rxns (1 Viewer)

[intuiit]

Ok - the question has a beaker of KBr solution and two Platinum electrodes in it.

What are the reactions at the anode and cathode?

Are they:

Anode: 2H2O + 2e- --> H2(g) + 2OH- [E = -0.83 V] Reduction

Cathode: 2 H2O --> O2 (g) + 4H+ + 4e- [E = -1.23 V] Oxidation

[E(total) = -2.06 V]

or are the rxns the other way round? If not, why do we get the H2 producing rxn only at the anode and the O2 producing rxn at the cathode?

I always get confused... the platinum is inert, right

 

[syriangabsta]

I can never do these types of questions!!

 

[intuiit]

Nevermind - that was an easy question:

solution --

These are the possible rxns.

Rxn 1: K + e- --> K E= -2.94V

Rxn 2: H2O + e- --> .5 H2 + OH- E= -0.83V

Rxn 3: .5 Br2 (aq) + e- --> Br- E= +1.1V

Rxn 4: .5 O2 + 2H+ --> H2O E = +1.23V


Compare the E(0)'s

E(0) Sign: - , -, +, +
Rxn: 1, 2, 3, 4

compare...

ox. red.
1 with 3

1 with 4

2 with 3

2 with 4

and the other way round for each of the above combos (but don't have to...1 is definitely our reduction - it has to be a positive)

Now... which one has the highest E(0)T?

It's rxns 4 and 1 as the oxidation and the reduction respectively!

Done - too easy.