Consider the following reaction:

$4F{e}^{2+}(aq)+O_2(g)+4H^{+}(aq)4F{e}^{3+}(aq)+2H_{2}O(l)$

under the following conditions at $298K$ :

$[F{e}^{2+}]=1\mathrm{.}3M$

$[F{e}^{3+}]=0\mathrm{.}010M$

$P_{O_2}=0\mathrm{.}50$atm

$pH=3\mathrm{.}5($cathode half$-$cell$)$

Part A: Calculate the three variables needed for the Nernst Equation in Part B: standard cell potential, $E{?}_{\text{c}\text{e}\text{l}\text{l}\text{,}}($in V$),$ moles of electrons transferred $($in mol$),$ and Q$.$

Hint: It is okay to mix pressure values with aqueous concentration values in the $Q$ expression.

Round your answer for $E$ cell to TWO places past the decimal. If your answer is negative, include the sign.

Part B: Use the rounded values of $E{?}_{\text{c}\text{e}\text{l}\text{l}},$ moles, and $Q$ from Part A to calculate the cell potential, $E_{\text{c}\text{e}\text{l}\text{l}},($in V$)$ of the reaction.

Round your answer to TWO places past the decimal. If your answer is negative, include the sign.

Part C: Use the rounded value of $E_{\text{c}\text{e}\text{l}\text{l}}$ from Part $B$ to calculate the free energy, $G,($in $kJ)$ of the reaction.

Round your answer to the nearest whole number. If your answer is negative, include the sign.

$G=$ kJ

Part D: Under standard conditions, this reaction is $$ choose your answer... Under the nonstandard conditions given, this reaction is choose your answer... $$ Compared to standard conditions, the reaction under nonstandard conditions was