Give the cell reaction and $E_{cell}^o$ value for the cell constructed using the given standard electrode potentials: $E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ and $E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$.

(N/A) The standard electrode potentials are given as:
$E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ (Cathode reaction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$)
$E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^-$)
To construct the cell,the reaction with the higher reduction potential acts as the cathode and the one with the lower reduction potential acts as the anode.
Cell Reaction:
Anode: $2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^-$
Cathode: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$
Overall: $2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$
Calculation of $E_{cell}^o$:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = 1.23 \ V - (-0.44 \ V)$
$E_{cell}^o = 1.67 \ V$