What is the emf of this cell when:

A voltaic cell utilizes the following reaction:
4Fe^2+(aq) + O2(g) + 4H^+(aq) --> 4Fe^3+(aq) + 2H2O(l)

What is the emf of this cell when:
Fe^2+ = 2.0 M,
Fe^3+ = 1.1×10−2 M,
Po2 = 0.42 atm and
pH of the solution in the cathode compartment is 4.1?

Express your answer using two significant figures.
I came up with .36 and was incorrect??

4Fe^2+(aq) + O2(g) + 4H^+(aq) --> 4Fe^3+(aq) + 2H2O(l)

The two half-reactions are:

4Fe2+(aq) ==> 4Fe3+(aq) + 4e- . . . .Eo = -0.771 V
O2(g) + 4H+(aq) + 4e- ==> 2H2O(l) ..Eo = +1.229 V
======================================…
4Fe2+(aq) + O2(g) + 4H+(aq) ==> 4Fe3+(aq) + 2H2O(l) . .Eo = +0.458 V

If pH = 4.1, then [H+] = 10^-pH = 10^-4.1 = 0.000079 M
Note that [H2O] does not appear in the Q expression since H2O is a pure liquid.

Ecell = Eo - (0.059/n) log Q = 0.458 - (0.059/4) log ([Fe3+(aq)]^4] / ([Fe2+(aq)]^4)(P O2)([H+(aq)]^4))
= 0.458 - (0.0148) log ((0.011)^4 / (2.0)^4 (0.42) (0.000079)^4)) = 0.458 - (0.0148)(-7.75) = 0.573 V

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