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miniradman
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Homework Statement
A voltaic cell utilises the following reaction:
4 Fe2+(aq) + O2(g) + 4 H+(aq) ---> 4 Fe3+(aq) + 2 H2O(l)
What is the emf of this cell when [Fe2+] = 1.3 M, [Fe3+] = 0.010 M, PO2 = 0.51 bar and the pH of the solution in the cathode is 3.5.
The standard reduction potentials for each half-reaction.
Fe3+(aq)+ e- ---> Fe2+(aq) Eored = +0.77 V
O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) Eored = +1.23 V
Homework Equations
## E = E^o - \frac{RT}{nF}ln|Q| ##
Q=[Products][Reactants]
The Attempt at a Solution
I'm not exactly sure how to incorporate the pH and Partial pressure of oxygen into this equation, but I had a go.
## \frac{P}{RT} = C ##
## \frac{51}{(8.314)(298)} = C ##
## {0.027M = C} ##
C being the concentration of Oxygen:
Then proceeded to find the concentration of hydrogen
## pH = -log_{10}|[H^+]| ##
## 3.5 = -log_{10}|[H^+]| ##
## 10^{-3.5} = [H^+] ##
## 0.00032M = [H^+] ##
Then I found the half reactions for the redox reaction:
## 4Fe^{2+} \rightarrow 4Fe^{3+} + 4e^- ## ---- Oxidation (+0.77V)
## 4e^- + O_2 + 4H^+ \rightarrow 2H_2O ## ---- Reduction (+1.23V)
Hence,
## E^o = E_{reduction} - E_{oxidation} ##
## E^o = (1.23) - (0.77) ##
## E^o = 0.46V ##
Now to find Q (I think this is where my error could be, I'm not sure what to do with the water, but I figured the concentration may be 1?).
## Q = \frac{[products]}{[reactants]} ##
## Q = \frac{[Fe^{3+}]}{[H^+]^4[O_2][Fe^{2+}]^4} ##
## Q = \frac{(1*10^{-8})}{(0.00032)^4(0.027)(1.3)^4} ##
## Q = 1250 ## (rounded, obviously)
Now using the equation:
## E = E^o - \frac{RT}{nF}ln|Q| ##
## E = 0.46 - \frac{(8.314)(298)}{(4)(96485)}ln|1250| ##
## E = 0.46 - 0.0457 ##
## E = 0.41V ##
however, this is incorrect. The correct answer is:
0.37V