Homework Help: Electrochemical Cells (involving partial pressures and pH)

1. Nov 4, 2014

1. The problem statement, all variables and given/known data
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

2. Relevant equations
$E = E^o - \frac{RT}{nF}ln|Q|$

Q=[Products][Reactants]

3. 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

2. Nov 4, 2014

Staff: Mentor

What is the Nernst equation for the formal potential of this half cell?

3. Nov 4, 2014

Ok...

$E = E^o - \frac{RT}{nF} ln| Q |$

$Q = \frac{[H_2O]}{[O_2][H^+]^4}$

$Q = \frac{1^2}{(0.00032)^4(0.027)}$

$ln|Q| = 35.8$

so:

$E = (1.23) - \frac{(8.314)(298)}{(4)(96485)} (35.8)$

$E = (1.23) - 0.230 = 1$

Hmm, and If I did the same with the other half cell?

I get $E = 0.801$, but I digress...

Last edited: Nov 4, 2014
4. Nov 4, 2014

Staff: Mentor

Q should use partial pressure of the gas, not the concentration.

5. Nov 4, 2014

Righto, I converted it to a Kpa, I just has a feeling that Bar is not a the correct unit of measurement.

$Q = \frac{1}{(51)^4(0.027)}$

$ln|Q| = -11.9$

$E = 1.23 - \frac{(8.314)(298)}{(4)(96485)}(-11.9)$

$E = 1.306V$

Quick question, why do I use the partial pressure of Oxygen as opposed to concentration ($M$), doesn't the $Q$ have to be dimensionless?

6. Nov 5, 2014

Staff: Mentor

To confuse you further, is Q for

HA <-> H+ + A-

dimensionless?

Q should be constructed using not concentrations, but dimensionless activities. Activity of a substance - for a diluted solution - is approximately equal its concentration, so we typically omit the activity coefficients assuming it equals 1 - but in effect we also introduce dimensionality into the equation.

7. Nov 5, 2014

I see, I always assumed it to be dimensionless (as a ratio of concentration), since it's analogous to equilibrium constant.

I'm still not sure exactly sure how to calculate $Q$

8. Nov 5, 2014

Staff: Mentor

Standard state of a gas is 1 bar, so just putting 0.51 as a partial pressure of oxygen should work.