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?
what the difference between hydration and hydrogenation
Yes, your approach is correct, but there is a slight error in your calculation. Let's go through the steps again to ensure accuracy.
First of all, you correctly identified the oxidation and reduction half-reactions:
Oxidation Reaction: Fe -> Fe2+ + 2e-
Reduction Reaction: 2e- + 2H2O + 1/2O2 -> 4OH-
Now, let's calculate the standard cell potential, E°, by subtracting the reduction potential from the oxidation potential:
E° = E°cathode - E°anode = -0.44 V - (-0.83 V) = 0.39 V
Since the partial pressures of oxygen are not at standard conditions, we need to use the Nernst equation to determine the actual cell potential, E:
E = E° - (0.0592 V/n) * log(Q)
Where:
E = cell potential under non-standard conditions
E° = standard cell potential
0.0592 V/n = the Nernst constant at room temperature (25°C)
Q = reaction quotient
However, it seems that there's an error in your calculation of the reaction quotient, Q. The reaction quotient is calculated using the partial pressures of the species involved in the reaction:
Q = [OH-]^4 / [Fe2+]
Now, let's calculate the reaction quotient using the given partial pressures of oxygen:
Q = (OH-)4 / [Fe2+]
At the point with 0.20 atm of oxygen, there will be 2 OH- ions produced for every Fe2+ ion formed. Therefore, [OH-] = 2[Fe2+].
At the point with 0.0010 atm of oxygen, there will be 4 OH- ions produced for every Fe2+ ion formed. Therefore, [OH-] = 4[Fe2+].
Now, let's calculate Q1 using the partial pressures at the point with 0.20 atm of oxygen:
Q1 = (2[Fe2+])^4 / [Fe2+] = 16[Fe2+]^3
And let's calculate Q2 using the partial pressures at the point with 0.0010 atm of oxygen:
Q2 = (4[Fe2+])^4 / [Fe2+] = 256[Fe2+]^3
Now, substitute these values into the Nernst equation:
E = E° - (0.0592 V/2) * log(Q2 / Q1)
= 0.39 V - (0.0296) * log(256[Fe2+]^3 / 16[Fe2+]^3)
= 0.39 V - (0.0296) * log(16)
≈ 0.39 V - (0.0296) * 1.2041
≈ 0.39 V - 0.0357
≈ 0.354 V
So, the correct voltage difference between the two points of corroding iron with different oxygen partial pressures is approximately 0.354 volts.