[Radu]

   Hi, I have a problem:
          In a saturated solution of Zn(OH)₂, a rectangular piece of metallic Zn is immersed.
   The pH of solution is 9 , ks Zn(OH)₂=10^-16 and E°_Zn²⁺/Zn=-0.763V.
        Calculate the pressure of H₂ at which the oxidation of Zn becomes possible in the above conditions.
   
       Here's what I did:
                     We need to calculate E_cell for : (-)Zn/Zn(OH)₂, OH^-_aq//H⁺_aq, H₂(x atm)(+).
                     E_cell=0.059lg[H⁺]-0.059/2 lg p_H2+1.235+0.059lg[HO^-].
                  The potential is positive and I find the maximum value of P_H2
        for which the oxidation is possible, by setting E_cell
=0; I obtain a ridiculous value of ≈10¹³atm. I should get a pressure lower than 5.56 mmHg.
          The problem sounded just like this and I don't know how I might get this answer.
   Any ideas where I'm wrong?
       

[Borek]

As far as I can tell 7.3×10¹³ is the equilibrium pressure, at which zinc will not react. Any lower pressure and it will be oxidized.

[Radu]

  Yes, it's true. But do we actually need so much force to stop Zn from being oxidated?
      That's the problem...I feel it's not in accordance with nature, Zn isn't such a reactive element.
       Anyway, numbers don't lie.

[Big-Daddy]

I got 6.2*10¹³ atm (probably just rounding differences), the pressure of H₂ has to be below this value for the oxidation to occur. Take home message (apparently): there is very little you can do pressure-wise to stop redox reactions from going forward, usually. (Since 0.763 is not a particularly large magnitude but still requires a huge pressure to oppose).

One interesting thing to note, is that the pH you gave in the beginning, 9, is not needed (except that, with the given fixed concentration of Zn²⁺ in the question, the pH could not be less than 9 with the solution still being at saturation, and if it were not at saturation we could not use the Ksp equilibrium to express [Zn²⁺] since it wouldn't hold unless it were saturated).

[Radu]

 Oh, I see, you're saying that at saturation [Zn²⁺][HO^-]²=Ks and that S*(2S)²=k_s. But I think it is just a lucky thing  that solubilty in distilled water matches the given pH( actually it doesn't match rigurously, but it is of the same order). Had it been saturated at pH-s greater than 9, the results would have been very different.

[Big-Daddy]

Oh, I see, you're saying that at saturation [Zn²⁺][HO^-]²=Ks and that S*(2S)²=k_s. But I think it is just a lucky thing  that solubilty in distilled water matches the given pH( actually it doesn't match rigurously, but it is of the same order). Had it been saturated at pH-s greater than 9, the results would have been very different.

There's nothing lucky about it. At saturation (by definition of K_sp), the K_sp equilibrium holds, so K_sp(Zn(OH)₂) = [Zn²⁺][OH^-]². Then [Zn²⁺]=K_sp(Zn(OH)₂) / [OH^-]² enables us to calculate [Zn²⁺] in the solution at this pH, given that we're at saturation. Now substitute [OH^-] = K_w / [H⁺] and then substitute for [Zn²⁺] into the expression for reaction quotient Q (of the redox reaction involving Zn/Zn²⁺ and H⁺/H₂ couples) and I find that it can be expressed in terms of just constants and the partial pressure of H₂. Of course [Zn²⁺] is dependent on [H⁺] but this dependence cancels through when finding Q.

[Radu]

   Oh, I see, my bad, very good remark!
    Actually, more straightforwardly,  no protons or hydroxides are involved in the overall cell reaction, so its potential is pH independent.
   

[Big-Daddy]

   Oh, I see, my bad, very good remark!
    Actually, more straightforwardly,  no protons or hydroxides are involved in the overall cell reaction, so its potential is pH independent.
   

Interesting comment. I checked out the related thermodynamics using that route and it all works.