Question

For a PE-pH diagram, derive the function for the boundary at which [NH*] = [NO2] for pH <7. The relevant half-reaction to consider is NO2 + NHA (not balanced) En° = 0.893 V

Answer 1

The boundary is given by Nernst's equation:

${\displaystyle E=E^{\circ }-{\frac {RT}{nF}}\ln(Q)}$

Where Q is the ratio between the concentration of products and of reactants of the involved half-reaction, elevated to thei corresponding stoichiometric coefficients, T is the temperature in K, R is the universal constant, n is the number of electrons involved in the half-reaction, F is Faraday's constant and E° is the standard potential- Taking into account the value of the constants and the transformation of ln to log₁₀, we have:

$E=E_{0}-\frac{0.05916V}{n}\cdot log(Q)$

Now, to know what Q stands for, we need to balance the half-reaction. Since we are working at pH below or equal to 7, we will consider acid conditions. The balanced half-reaction is:

$NO_{2}^{-}&plus;8H^{&plus;}&plus;6e^{-}\rightarrow NH_{4}^{&plus;}&plus;2H_{2}O$

Thus, considering that the concentration of water is constant (and thus, it does not appear in the expression of Q), our Nernst equation turns into:

$E=E_{0}-\frac{0.05916V}{6}\cdot log\left (\frac{[NH_{4}^{&plus;}]}{[NO_{2}^{-}][H^{&plus;}]^{8}} \right )$

Since we are caculating the equation in which [NH₄⁺] = [NO₂^-], they are cancelled in our equation, which results in:

$E=E_{0}-0.00986V\cdot log\left (\frac{1}{[H^{&plus;}]^{8}} \right )$

Applying the properties of log, particularly: log(xⁿ) = n*log(x), we have:

$E=E_{0}-0.00986V\cdot 8\cdot log\left (\frac{1}{[H^{&plus;}]} \right )=E_{0}-0.0788V\cdot log\left (\frac{1}{[H^{&plus;}]} \right )$

Which, since log(1/[H⁺]) = pH, is (adding the value for E°):

$E=0.893V-0.0788V\cdot pH$

And this is the function for the boundary.