[minjie]

Calculate the cell potential (Ecell, V) of a galvanic cell with the following electrodes: Oxygen electrode (pure oxygen, partial pressure 1 atm, in acidic media pH = 3.0) and Zn electrode in ZnSO₄ solution (activity of Zn(2+)-ions: a_Zn²⁺ = 0.1). Standard electrode potentials:
O₂(g) + 4H⁺(aq) + 4e^–  →  2H2O(l)       E0 =    1.229 V
Zn₂+(aq) + 2e^–  →  Zn(s)                                  E0 = – 0.763 V
pH = – log a_H+
Write out equation of the net reaction in this cell.


It's a very difficult question to me, can anyone help me to finish it? Thank you very much

[Borek]

We can help you finish it - once you start.

[minjie]

It's my understanding for cell potential:  E=-0.763-1.229=-1.992 V, is it correct

[Borek]

1.992 would be a standard potential. Not a bad start, but you are not there yet - now you need to write Nernst equation for the cell. Or two equations for half cells.

[okryan]

O2(g) + 4H+(aq) + 4e–  →  2H2O(l)       E0 =    1.229 V (this is the most likely reduction (largest positive E0) . So other reaction is reversed)
Zn(s)  → Zn2+(aq) + 2e–   - E0 = + 0.763 V
(Zn(s)  → Zn2+(aq) + 2e–) x 2   - E0 = + 0.763 V      ;

The equations are then added 

O2(g) + 4H+(aq) + 4e–  + 2Zn(s)  →  2H2O(l) + 2Zn2+(aq) + 4e–

O2(g) + 4H+(aq)  + 2Zn(s)  →  2H2O(l) + 2Zn2+(aq)   emf = 1.229 V + + 0.763 V = 1.992 V

But I still don't understand why it has PH related.

[okryan]

here is another way,
Do you know which is correct? Borek
Zinc
E = E0 –           Q =  =  = 10
E = - 0,763 –  ln 10 = - 0,79 V
Oxygen
E = E0 –           Q =  =  = 107   a(H+) = 10-pH = 10 -3
E = - 1,229 –  ln 107 = 1,13 V

Zinc has the lower potential so it is the anode and Oxygen the cathode.

E° cell= E cathode – E anode
E° cell = 1,13 + 0,79
E° cell = 1,92 V

[Borek]

Nernst equation. Write it for both half cells.