$10$ Question $(1$ point$)$

See page $917$

The oxidation of $N{H}_4^{+}$to $N{O}_3^{-}$in acid solution $)=(5\mathrm{.}600$ is described by the following equation:

$N{H}_4^{+}(aq)+2O_2(g)longrightarrowN{O}_3^{-}(aq)+2H^{+}(aq)+H_{2}O(l)$

The overall reaction for the oxidation has an $E{?}_{\text{c}\text{e}\text{l}\text{l}}=-0\mathrm{.}2710V.$

$E=1\mathrm{.}50V$ for the half$-$reaction

$N{O}_3^{-}(aq)+10H^{+}(aq)+8e^{-}longrightarrowN{H}_4^{+}(aq)+3H_{2}O(l)$

$2$nd attempt

See Hint

Be sure to solve the equation using a base$-10$ logarithm $(log)$ rather than the natural logarithm $($In$).$

What is the ratio of $N{O}_3^{-}$to $N{H}_4^{+}$at $298K$ if $P_{O2}=0\mathrm{.}180$ atm? Assume that the reaction is at equilibrium.