Problem 55 - Chemistry: An Atoms-Focused Approach

Question

A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides. The following reaction describes the formation of \(\mathrm{NO},\) one of the products: \(3 \mathrm{Cu}(s)+8 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \rightarrow\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(\ell)\) a. Starting with the appropriate standard potentials from Appendix \(6,\) calculate \(E_{\text {ren }}^{\circ}\) for this reaction. b. Calculate \(E_{\text {rxn }}\) at \(298 \mathrm{K}\) when \(\left[\mathrm{H}^{+}\right]=0.100 \mathrm{M}\) \(\left[\mathrm{NO}_{3}^{-}\right]=0.0250 \mathrm{M},\left[\mathrm{Cu}^{2+}\right]=0.0375 M,\) and the partial pressure of \(\mathrm{NO}=0.00150\) atm.

Step-by-Step Solution

Answer

Answer: The cell potential for this redox reaction under the given non-standard conditions is approximately 1.28 V.

Step 1: Identify the half-reactions and their standard reduction potentials

In this redox reaction, copper reacts with nitric acid to produce nitrogen monoxide gas, copper ions, and water. First, let's write the half-reactions for each substance involved: Copper half-reaction: \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q)+2 e^{-}\) Nitrogen monoxide half-reaction: \(2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+3 e^{-} \rightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) Use the standard reduction potentials from Appendix 6: \(E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=0.34 \, \text{V}\) \(E^{\circ}_{\mathrm{NO}_{3}^{-}/\mathrm{NO}}=0.96 \, \text{V}\) Note that the copper half-reaction needs to be reversed because it is oxidized, and the nitrogen monoxide half-reaction can stay as is because it is reduced.

Step 2: Calculate \(E_{\text{rxn}}^{\circ}\) and find the cell potential under non-standard conditions using the Nernst equation

The standard cell potential is the difference between the standard reduction potential of the cathode (reduction) reaction and the standard reduction potential of the anode (oxidation) reaction: \(E_{\text{rxn}}^{\circ} = E^{\circ}_{\mathrm{cathode}}-E^{\circ}_{\mathrm{anode}}\) Since copper is being oxidized, we will reverse its half-reaction and change the sign of its standard reduction potential. Therefore, \(E_{\text{rxn}}^{\circ}= E^{\circ}_{\mathrm{NO}_{3}^{-}/\mathrm{NO}}-( - E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}})=0.96 \, \text{V} + 0.34 \, \text{V}= 1.30 \, \text{V}\) Now, we'll apply the Nernst equation to find the cell potential under non-standard conditions: \(E_{\text{rxn}} = E_{\text{rxn}}^{\circ} - \dfrac{RT}{nF}\ln Q\) \(E_{\text{rxn}} = E_{\text{rxn}}^{\circ} - \dfrac{0.0592}{n}\log Q\) \(n=6\), since there are 6 electrons exchanged Using the given concentrations and partial pressure, we build the reaction quotient \(Q\): \(Q = \dfrac{\left[ \mathrm{Cu}^{2+} \right]^3\left[\mathrm{NO} \right]^2}{\left[\mathrm{H}^{+}\right]^8\left[\mathrm{NO}_{3}^{-}\right]^2} = \dfrac{(0.0375)^3(0.00150)^2}{(0.100)^8(0.0250)^2}\) Now we substitute the values and find \(E_{\text{rxn}}\): \(E_{\text{rxn}} = 1.30 \, \text{V} - \dfrac{0.0592}{6}\log \left( \dfrac{(0.0375)^3(0.00150)^2}{(0.100)^8(0.0250)^2} \right) \approx 1.28 \, \text{V}\) So, under the given non-standard conditions, the cell potential, \(E_{\text{rxn}}\), for this reaction is approximately 1.28 V.

Key Concepts

Redox Reaction

Redox reactions, short for reduction-oxidation reactions, are essential in electrochemistry. In these reactions, electrons are transferred from one substance to another. This transfer results in one substance being oxidized (losing electrons) and another being reduced (gaining electrons).

In the provided copper and nitric acid reaction, copper (\(\text{Cu(s)}\)) acts as the reducing agent and undergoes oxidation, turning into copper ions (\( ext{Cu}^{2+}\)). Meanwhile, the nitrate ions (\( ext{NO}_3^-\)) are reduced to form nitrogen monoxide (\( ext{NO}\)). These transformations are broken down further into half-reactions representing the oxidation and reduction processes.

• Oxidation Half-Reaction: Involving the loss of electrons by copper and resulting in the formation of copper ions.
• Reduction Half-Reaction: Where the nitrate ions gain electrons to produce nitrogen monoxide.

Understanding redox reactions involves recognizing the gain and loss of electrons, which is fundamental to electrochemical cell potentials.

Standard Reduction Potential

The standard reduction potential (\(E^ ext{o}\)) is a measure of the tendency of a chemical species to be reduced, expressed in volts. It's vital to predicting the direction of electron flow in a redox reaction. Referenced against the standard hydrogen electrode, these potentials provide a universal scale.

In the given reaction, we have two half-reactions:

• Copper Oxidation: \(\text{E}^{ ext{o}}_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V}\), indicating a lower tendency for copper to be reduced.
• Nitrate Reduction: \(\text{E}^{ ext{o}}_{\text{NO}_3^-/\text{NO}} = 0.96 \, \text{V}\), showing a higher tendency for nitrate ions to gain electrons.

By subtracting the oxidation potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction, we find the overall cell potential. This calculation confirms if the reaction is spontaneous and in which direction it occurs.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry. It helps us determine the cell potential (\(E\)) of a redox reaction under non-standard conditions by considering temperature and concentration effects.

The equation is expressed as:
\[E = E^ ext{o} - \dfrac{RT}{nF}\ln Q\]
Where:

• \(E^ ext{o}\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged in the reaction.
• \(F\) is the Faraday's constant.
• \(Q\) is the reaction quotient.

The simplified version used in our exercise reflects constant temperature conditions, and it takes concentration and pressure data to find the non-standard cell potential. The Nernst equation shows how real-world conditions can shift potential from the standard state, enabling the calculation of cell potential at any given moment.

Reaction Quotient

The reaction quotient (\(Q\)) is an important factor in electrochemical calculations. It provides a snapshot of the current concentrations or pressures of the reactants and products involved at any given time in a reaction.

To calculate \(Q\), we construct it similarly to an equilibrium constant expression: it's the ratio of the concentrations (or partial pressures) of the products to the reactants, each raised to the power of their coefficients in the balanced equation.

In our example, for the reaction quotient:\[Q = \dfrac{\left[ \text{Cu}^{2+} \right]^3\left[\text{NO} \right]^2}{\left[\text{H}^{+}\right]^8\left[\text{NO}_3^-\right]^2}\]
This formula includes concentrations of ions and partial pressures of gases reflecting the current state of the redox reaction. The value of \(Q\) indicates how far the reaction has proceeded compared to equilibrium, and this impacts the cell potential calculation via the Nernst equation.