1. Reduction of O2 in acidic and neutral environment is given together with standard free
energy of the components involved as follows .
O2  4H   4e   2H 2O
a. Acidic:
∆G0 0
0
-56700 kal/mol
b. Neutral:
O 2  2H 2 O  4e   4OH 
∆G0 0 -56700 kal/mol
-37600 kal/mol
Prove that the reduction potential of oxygen in both environments can be calculated by same
equation, assuming that oxygen is dissolved in each solution.
a) Oxygen reduction potential in acidic environment:
O2  4H   4e   2H 2O
∆G0 0
0
-56700 cal/mol
Standard electrode potential of this reaction; Assuming that 1 cal = 4,1868 Joul and
ΔG 0   zFE 0
E0  
 Pr oducts
0,059
ΔG 0
 2 x56700 x 4,18

 1,23 Volt ; E  E 0 
log
zF
4 x96500
 Re ac tan ts
z
When corresponding
values of the terms are substituted in Nernst equation; the electrode potential depending on
ion concentrations of the solution;
4
0,059
0,059

E  1.23 
log 1 H   .PO2  1,23 
 4. log (H  ) 0,015 log PO2
4
4
E 1.23  0.059pH  0,015log PO2


b) Oxygen reduction potential in nuetral environment:
O 2  2H 2 O  4e 
∆G0

0 -56700 cal/mol
4OH
37600 cal/mol
Standard equilibrium potential of this reaction:
E0  
 4.37600   2.56700.4,18  0,4 Volt
ΔG 0

zF
4.96500
Electrode potential for this reaction with changing ion concentration according to the Nernst
equation: E  0,4  0,059. log OH   0,015. log PO Remembering the following definitions;
[H+] [OH-] = 10 -14 and pH= -log [H+]
Equation can be formulated as
E  0,4  0,059(pH  14 )  0,015 log PO2 or E  1.23  0.059pH  0,015log PO2
Finally it is seen that electrode potential of oxygen reduction reaction in both environment is
the same.
2
2. Standard electrode potentials of the following reduction reactions presenting two half-cell
are give as follows:
𝐴𝑙 3+ + 3𝑒 − = 𝐴𝑙
𝐸 𝑜 = −1,662 𝑉𝑜𝑙𝑡
𝑃𝑏 2+ + 2𝑒 − = 𝑃𝑏 𝐸 𝑜 = −0,126 𝑉𝑜𝑙𝑡
a. Find out standard electrode potential of the total reaction for corrosion cell, 3𝑃𝑏 +
2𝐴𝑙 3+ → 3𝑃𝑏 2+ + 2𝐴𝑙, and indicate if cell reaction would continue in the direction as
written.
b. Calculate the amount of potential shift of the corrosion cell in the solution where
concentration of 𝐴𝑙 3+ and 𝑃𝑏 2+ ions are 10−3 and
10−1 Mol respectively and clarify
whether this shift would change the direction the reaction proceeds?
Make the use of the Nernst equation when calculating electrode potentials.
a. Standard electrod potential of the corrosion cell:
𝐴𝑙 3+ + 3𝑒 − = 𝐴𝑙
𝐸 𝑜 = −1,662 𝑉𝑜𝑙𝑡
+ 𝑃𝑏 = 𝑃𝑏 2+ + 2𝑒 −
+ 𝐸 𝑜 = +0,126 𝑉𝑜𝑙𝑡
3𝑃𝑏 + 2𝐴𝑙 3+ → 3𝑃𝑏 2+ + 2𝐴𝑙
𝐸 𝑜 = −1,536 𝑉𝑜𝑙𝑡
Reaction must proceed in the opposite direction from that written in the reaction, since ∆𝐺 0 =
−𝑧𝐹𝐸 0 > 0:
3𝑃𝑏 + 2𝐴𝑙 3+ ← 3𝑃𝑏 2+ + 2𝐴𝑙
b. According to Nernst equation, cell potential in presence of 𝐴𝑙 3+ and 𝑃𝑏 2+ ions at given
concentration;
E = −1.536 −
0.0592
6
0.0592
(3𝑙𝑜𝑔10−1
6
3
[Pb2+ ]
log [Al3+]2 = E= − 1.536 −
−3 )
0.0592
6
3
[10−1 ]
log [10−3 ]2 =−1,536 −
− 2𝑙𝑜𝑔10
= −1,536 − 0,0296(−3 + 6) = −1,6248
Amount of potential shift = -1,6248 - (-1,536)= - 0,0888 Volt in negative direction.
Therefore reaction will continue to proceed in opposite direction as indicated in (a). Another
word potential shit due to ion concentration causes no change in direction of the reaction
indiccated in (a).