[shanmac]

a)  Give the cell diagram notation of the electrochemical cell that could be used to determine experimentally the dissociation constant (Kw) of water.

The standard reduction potential for:

O2 + 2H20 + 4e- ---------> 4 OH-    is 0.40 V

b)  Use the value from the above question to calculate the dissociation constant at 298 K.

Here's what I've come up with for (a):
The overall reaction for the electrolysis of water is:
 
2H2O (l)    2H2 (g) + O2 (g)

The oxidation half-reaction is 4OH- (aq):rarrow:O₂ (g) + 2H₂O (l) + 4e-     (0.40 V)
The reduction half-reaction is 2H₂O (l) + 2e- :rarrow:H₂2 (g) + 2OH- (aq)  (-0.83 V)

E〫= Ecathode - Eanode
          = -0.83 V - 0.40 V
       = -1.23 V
Cell diagram notation: Pt(s) | H2O (l) : H2SO4 (aq) | Pt(s)

Hoping that this looks good so far, I move on to the next part of the question:

E = E° -  (0.0257 V ÷ 4) ln Q

Because both cathode and anode are in the same solution, E° = 0
Therefore E = -0.006425 ln Q
. but I am lost from there.  Not sure what values I am working with the calculate Q.  Could anyone point me in the right direction?

[shanmac]

OK, so I have carried on grappling with this one on my own and decided to forgo the Nernst part entirely and work with the equation

ln K = -ΔG° ÷ RT

(Using -1.23 V as E° and the formula ΔG°=-nFE°, ΔG° came up as 474,780 J.)

Therefore, ln K = -191.63.... which leaves me with a K of 5.97 x 10^-84... clearly I have made a mistake somewhere and still have no idea.

Was I on the right track initially when I was going to use the Nernst equation?

[Borek]

Why don't you use H₂/H⁺ as the other half cell?

[shanmac]

OK, so now I've tried this:

Anode (oxidation): 2 H₂O(l) → O₂2(g) + 4 H⁺(aq) + 4e−     -1.23 V
Cathode (reduction): 2 H⁺(aq) + 2e− → H₂(g)    0.00 V

Which gives me an E°cell (E°cathode - E°anode) of 1.23 V.

Now using
ln K = -ΔG° ÷ RT

(With 1.23 V as E° and the formula ΔG°=-nFE°, ΔG° came up as -474,780 J.)

ln K = 474,780 J ÷ (8.314 J/K mol) (298 K)
ln K = 191.63
K = 1.7 x 10₈₃
...which of course is still wrong.
Where am I getting so far off-course?

[Borek]

The standard reduction potential for:

O2 + 2H20 + 4e- ---------> 4 OH-    is 0.40 V

b)  Use the value from the above question to calculate the dissociation constant at 298 K.

You have not used this value.

I would try to combine it with the H₂/H⁺.

Not that I am sure it will work, but you are told what to use, and H₂/H⁺ is a reference point with zero potential, so you can safely use it in your calculations (as opposed to some other potentials, randomly selected from the standard potentials table).