[NewtoAtoms]

Hello Chemists,

I have stumbled across a ridiculous question, which is plaguing me. I have done some leg work so I will post it below, however if someone could give me some light, I would deeply appreciate the assistance to the end of the tunnel.

Q:  The corrosion of iron can be thought of as an electrochemical cell reaction.  Calculate the voltage difference between two points of corroding iron differing only in their partial pressures of oxygen : 0.20 atm of oxygen at one point and 0.0010 atm of oxygen at another.  The reaction is:

H₂0 + Fe (s) + 1/2 O₂ (g) ----------> Fe(OH)₂

I have figured out the 1/2 reactions which are:

Oxidation Reaction:  Fe ---------> Fe²⁺ + 2^e-
Reduction Reaction: 2e^- + H₂0 + 1/2 O₂ ----------> 2OH^-

I have also come to learn that this is NOT a standard conditions, therefore I must use the Nernst Equation.

E = E^o - 0.0257 V/n (1n Q)

Now where on earth do the pressures fit into this equation?
I have searched many examples in my text and on the internet and this equation usually considers different molar concentrations, however not pressures.

Is it still possible however to say:  E = (O) - 0.0257 V/n (1n Fe²⁺/O₂)

Can someone please, I say OH PLEASE, help me with where the partial pressures go in this question?

I would be so grateful for your time.

Newtoatoms

[Borek]

Hint:

http://en.wikipedia.org/wiki/Standard_hydrogen_electrode

[NewtoAtoms]

Thank you Borek,

I went and read all up on SHE's.  Now, thanks to your hint, I can apply the SHE as the oxidation reaction and my other oxygen 1/2 reaction equation as the reduction reaction (2 different time with two different atm)

STEP #1 - For pp 0.20 atm:

Reduction Reaction: 2e^- + H₂0 + 1/2 O₂ (0.20 atm) ----------> 2OH-
oxidation Reaction:  2H⁺ + 2e^- ----------> H₂

E = RT/F (1n aH⁺/pH₂/P^o)^1/2

NOW MY ISSUE ARISES here:

R = 8.314 472(15) J K−1 mol−1
T = 25^oC
F = 96 485.339 9(24) C/mol
pH2 = 1 atm
p0 (the special atm) = 0.20 atm

However what on earth is aH+??
I realize that it is is the activity of the hydrogen ions, aH+=fH+ CH+ /C0 but what on earth is that equal to in my original question??  Because obviously as soon as I have that value, I can calculate E for the 0.20 atm, and then do the same thing for 0.0010 atm. 

Your help is once again greatly appreciated!

Newtoatoms

[Borek]

I wasn't hinting at SHE being one of your electrodes. Try to understand how does the pressure of hydrogen enters reaction quotient and how it is related to your problem with oxygen pressure.

[NewtoAtoms]

Hello Borek,

Okay then if the SHE is not an electrode then if I apply the nernst equation found on the 'hint' page to this situation perhaps I can calculate V.

For example:

Reduction Reaction: 2e^- + H₂0 + 1/2 O₂ (0.20 atm) ----------> 2OH^-                                        -0.83 V
oxidation Reaction:  2H+ + 2e- ----------> H₂                                                                       - 0.44 V

E^ocell= Eocathode - Eoanode
E^ocell= -0.83 -(-0.44)
E^ocell = -0.39 V

THEREFORE

E = E^o - RT (1n Q)
            --
            nF

E = -0.39 - 0.0257 V 1n (1)
                --------       --
                    2            1/2(O₂)

Q is really derived via aA + bB ----> cC + dD therefore Q= cC + dD / aA + bB
In this situation - the H₂0 is a liquid, the Fe is a solid, and the Fe(OH)₂ is a solid therefore it wouldn't be considered.

E = -0.39 - 0.01V (1 )
                         ---
                        1/2(0.20atm)

E = -0.39 - 0.01 (2.30)
E = -0.39 - 0.02
E = -0.41 V

There in this situation under non-standard conditions with a pressure of 0.20 atm, there would be -0.41 V

Am I correct now Borek?

Thank you for your time!

Newtoatoms

[Borek]

Seems to me you have no idea how the reaction quotient should look like for the reaction where some of the reagents are gasous and others are dissolved.

You have to mix both - concentrations and pressures.

For SHE it was

E = E₀ + RT/nF ln ([H⁺]/p_H2^1/2)

You have to write similar equation for oxygen half cell. It must contain both OH^- and p_O2.

[NewtoAtoms]

Hello Borek,

You are totally right I really have no idea in regards to electrochemistry, that's what I said off the start!
I do keep plugging away and continuously keep trying....never to give up. 

So my mistake was the fact that I calculated Q in regards to the overall reaction and not the oxygen half cell.

Therefore now that I have re-directed my view to ONLY the oxygen half cell I can create the Q via:

2e- + H₂0 (l) + 1/2  O₂(g) ----------> 2OH^- (aq)

Q = 2(OH^-)
      ------
      1/2(O₂)

However my original question doesn't mention anything about concentrations, but only pressures.
Therefore am I correct to assume that the concentrations of OH- would therefore be standard and 1 M??

Q = 2(1 M)
     ------ 
     1/2(0.20atm)

Q = 2 / 0.10
Q = 20

[Borek]

You are totally right I really have no idea in regards to electrochemistry, that's what I said off the start!

Reaction quotient is not electrochemistry, it is a concept used in many other places as well.
 

However my original question doesn't mention anything about concentrations, but only pressures.
Therefore am I correct to assume that the concentrations of OH- would therefore be standard and 1 M??

Concentration of OH^- doesn't matter - it is the same in both parts of the system, so in the final answer it will cancel out. You may assume it is 1M, or you may derive formula for the potential difference and it will not contain OH^-.