[lrw1793]

Cu(s) | Cu²⁺(aq), SO₄^2‒(aq) || SO₄^2‒(aq), H⁺(aq), O₂(g, 1 atm.) | Pt(s)

The half reactions:
Cu(s) → Cu²⁺(aq) +2e-
SO₄^2-(aq) + 4H⁺(aq) + 2e- → 2H₂O(l) + SO₂(g)
Cu(s) + 2H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) + SO₂(g)

E° (Cu / Cu²⁺) = +0.34 V and E° (O₂, H⁺ | H₂O) = +1.23 V, calculate the standard free energy change per mole of copper for the formal cell reaction.

Cu(s) → Cu2+(aq) + 2e-  +0.34V 
4H+(g) + 2O2(g) + 4e- → 2H2O(g) = +1.23V
 
E^o cell = E^o reduction -  E^o oxidation
+1.23V – 0.34V = 0.89V

For Cu(s) + 2H⁺(g) + O₂(g)  → Cu²⁺(aq) + 1/2H₂O(g)
Standard free energy change deltaG = -nFE = -1.78F = -171743.3 Jmol^-1

Write out the Nernst equation for the reaction. Calculate the measured cell potential difference at 298 K if the pH is 5 and the oxygen is replaced with air. Is copper more or less likely to corrode under these conditions than under standard conditions?

Okay here's where I'm drawing a blank. How do I work out the activities of the ions?
I know pH = -log10[H⁺]
10^-5 = [H⁺]
How do I find the activity coefficient for this?
Cu(s) +  2H⁺(aq) + O₂(g) → Cu²⁺(aq) + 1/2H₂O(l)
Omitt solid and liquids
Air so pressure = 1 atm and can be omitted

E cell = 0.89 – (8.314 x 298/2F) x (ln[Cu²⁺] /[H⁺]²)


 

[Borek]

O₂(g, 1 atm.)

SO₂(g)

These parts contradict each other.

[lrw1793]

O₂(g, 1 atm.)

SO₂(g)

These parts contradict each other.

Oh yes, you're right. So the half reactions are:
Cu(s) → Cu²⁺(aq) +2e-
4H+(aq) + O₂(g) + 4e- → 2H₂O(l)
Overall reaction:
2Cu(s) + H₂SO₄(aq) + O₂(g) → 2CuSO₄(aq) + 2H₂O(l) 

So I think from the pH I can work out the activity of H+, but then what about for Cu2+?

[lrw1793]

Wait I think it would just be:
E cell = 0.89 – (8.314 x 298/2F) x (ln[H+]²)
E cell potential = 0.89 – (-0.29) = 1.18V
This is higher and therefore reaction has a greater tendency to take place when pH = 5