Question

The oxidation of NH4+ to NO3– in acid solution (pH = 5.600) is described by the following equation:

NH+4​(aq)+2O2​(g)=NO−3​(aq)+2H+(aq)+H2​O(l)
The overall reaction for the oxidation has an E°cell = –0.2710 V.
E° = 1.50 V for the half-reaction
NO-3​(aq)+10H+(aq)+8e−=NH+4​(aq)+3H2​O(l)
Be sure to solve the equation using a base-10 logarithm (log) rather than the natural logarithm (ln).
What is the ratio of [NO3–] to [NH4+] at 298 K if PO2 = 0.180 atm? Assume that the reaction is at equilibrium.

Answer 1

The ratio of [NO3–] to [NH4+] at 298 K can be determined using the Nernst equation. The Nernst equation is given by: E = E° – (0.0592/n) log(Q). Where E is the cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient.

Step-by-step explanation:

The ratio of [NO3–] to [NH4+] at 298 K can be determined using the Nernst equation. The Nernst equation is given by:

E = E° – (0.0592/n) log(Q)

Where E is the cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient.

In this case, the reaction quotient Q can be expressed as:

Q = [NO3–] / [NH4+]

We know that E = 0 because the reaction is at equilibrium. Substituting E = 0 and E° = –0.2710 V into the Nernst equation, we can solve for the ratio [NO3–] / [NH4+].

Answer 2

The Nernst equation allows us to determine the equilibrium constant
`(\(K_{\text{eq}}\))` for a redox reaction using the given standard cell potential
`(\(E^\circ_{\text{cell}}\))` and the actual cell potential
`(\(E_{\text{cell}}\)):`

`\[E_{\text{cell}} = E^\circ_{\text{cell}} - (RT)/(nF) \cdot \log K_{\text{eq}}\]`

Step-by-step explanation:

Given \(E^\circ_{\text{cell}} = -0.2710\) V, at equilibrium,
`\(E_{\text{cell}} = 0\)` (since the reaction is at equilibrium). With temperature \(T = 298\) K, the Faraday constant
`\(F = 96485\) C/mol, and \(n = 8\) (from the balanced half-reaction)`, the Nernst equation becomes:

`\[0 = -0.2710 - \frac{(8.314 \, \text{J/mol} \cdot \text{K})(298 \, \text{K})}{8(96485 \, \text{C/mol})} \cdot \log K_{\text{eq}}\]`

Solving for
`\(\log K_{\text{eq}}\):`

`\[\log K_{\text{eq}} = (-0.2710 * 8(96485))/(298 * 8.314)\]`

`\[K_{\text{eq}} = 10^{(-0.2710 * 8(96485))/(298 * 8.314)}\]`

This equilibrium constant is related to the concentrations of the species involved in the redox reaction. For the reaction:

`\[NO_3^-(aq) + 10H^+(aq) + 8e^- \rightleftharpoons NH_4^+(aq) + 3H_2O(l)\]`

The equilibrium constant expression
`(\(K_{\text{eq}}\))` in terms of the ratio of products to reactants
`(\([NO_3^-]\)` to
`\([NH_4^+]\))` is used to find this ratio. Given the
`\(K_{\text{eq}}\)` value from the Nernst equation, you can set up an expression involving the equilibrium concentrations of
`\(NO_3^-\)` and
`\(NH_4^+\)` and solve for their ratio to determine
`\([NO_3^-]/[NH_4^+]\)` at 298 K and
`\(PO_2 = 0.180\)` atm.