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 the other. The reaction is: H2O + Fe (s) + 1/2 O2 (g) -- Fe(OH)2 (s)

Question

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 the other. The reaction is: H2O + Fe (s) + 1/2 O2 (g) --> Fe(OH)2 (s)  
H20 + Fe (s) + 1/2 O2 (g) ----------> Fe(OH)2

I figured out the 1/2 reactions which are:

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

this is NOT a standard conditions, therefore I know we need to use the Nernst Equation. 
 I figured out the Eo= ecathode-eanode = -.83-(-.44) = -0.39V
then use the nerst equation to find the voltage difference
E = -.39- (0.0592v/2)*log (0.001atm/0.20atm)
e = 0.054 volts

is this right?

lug · Answer

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