# Question #a4806

May 18, 2016

$\frac{\left[N {O}_{3}^{-}\right]}{\left[N {H}_{4}^{+}\right]} = 1.17 \times {10}^{- 27}$

#### Explanation:

This is a mis - leading question as the -ve ${E}_{c e l l}^{\circ}$ value tells us that the reaction cannot happen as written left to right under standard conditions.

For a reaction to occur the free energy change must be -ve:

$\Delta G = - n F {E}_{c e l l}$

This means ${E}_{c e l l}$ must be positive for this to occur. In this case we are told ${E}_{c e l l}^{\circ} = - 0.271 \text{V}$.

However, we are not under standard conditions since $p H = 5.6$ and under standard conditions $p H = 0$.

So we need to use The Nernst Equation:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{n} \log Q \text{ } \textcolor{red}{\left(1\right)}$

(At ${25}^{\circ} \text{C}$)

$N {H}_{4 \left(a q\right)}^{+} + 2 {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(l\right)} \rightarrow N {O}_{3 \left(a q\right)}^{-} + 2 {H}_{\left(a q\right)}^{+}$

We are told the reaction has reached equilibrium. At this point the potential difference between the 2 half - cells has fallen to zero.

So $\textcolor{red}{\left(1\right)}$ becomes:

$0 = {E}_{c e l l}^{\circ} - \frac{0.05916}{n} \log K$

Note we have replaced $Q$ with $K$.

$\therefore 0 = - 0.271 - \frac{0.05916}{8} \log K$

$n = 8$ which is the number of moles of electrons transferred.

$\therefore - \frac{0.05916}{8} \log K = 0.271$

$\therefore \log K = - \frac{0.271 \times 8}{0.05916}$

$\log K = - 36.64$

$\therefore K = 2.29 \times {10}^{- 37}$

$K = \frac{\left[N {O}_{3}^{-}\right] {\left[{H}^{+}\right]}^{2}}{\left[N {H}_{4}^{+}\right] p {O}_{2}^{2}} \text{ } \textcolor{red}{\left(2\right)}$

We can find $\left[{H}^{+}\right]$ since $p H = 5.6$

$\therefore - \log \left[{H}^{+}\right] = 5.6$

$\therefore \left[{H}^{+}\right] = 2.512 \times {10}^{- 6} \text{mol/l}$

We are told $p {O}_{2} = 0.18 \text{Atm}$. We can use this in the expression since the values used for $K$ are normalised against standard conditions for which $p = 1 \text{Atm}$ and $K$ is dimensionless.

Now we can put values into $\textcolor{red}{\left(2\right)} \Rightarrow$

$2.29 \times {10}^{- 37} = \frac{\left[N {O}_{3}^{-}\right]}{\left[N {H}_{4}^{+}\right]} \times \frac{{\left(2.512 \times {10}^{- 6}\right)}^{2}}{{0.18}^{2}}$

$\therefore \frac{\left[N {O}_{3}^{-}\right]}{\left[N {H}_{4}^{+}\right]} = 1.175 \times {10}^{- 27}$

This, and the tiny value of $K$, tells us that the equilibrium lies almost completely to the left and the equation should more correctly written right to left.

Another mis - leading aspect of what I consider to be a badly written question is that the ${E}^{\circ} = 1.5 \text{V}$ value given is not needed.