Question

Consider the reduction of nitrate ion in acidic solution to nitrogen oxide \(\left(E_{\mathrm{red}}^{\circ}=0.964 \mathrm{~V}\right)\) by sulfur dioxide that is oxidized to sulfate ion \(\left(E_{\text {red }}^{o}=0.155 \mathrm{~V}\right)\). Calculate the voltage of a cell involving this reaction in which all the gases have pressures of \(1.00 \mathrm{~atm}\), all the ionic species (except \(\left.\mathrm{H}^{+}\right)\) are at \(0.100 M\), and the \(\mathrm{pH}\) is \(4.30 .\)

Short answer

Question: Calculate the voltage of a cell involving the reduction of nitrate ion to nitrogen oxide and the oxidation of sulfur dioxide to sulfate ion in acidic solution, given the following information: Standard reduction potential of nitrate ion (NO3-) to nitrogen oxide (NO): 0.964 V Standard reduction potential of sulfur dioxide (SO2) to sulfate ion (SO4^2-): 0.155 V pH of the solution: 4.30 Concentrations: [NO3-] = 0.100 M, [SO2] = 0.100 M, and [SO4^2-] = 0.100 M Pressures: P(NO) = 1.00 atm, P(SO2) = 1.00 atm Answer: The voltage of the cell is approximately 0.712 volts.

Step-by-step solution

Write the half-reactions

Here are the given half-reactions involved in this cell: 1. Reduction of nitrate ion (NO3-) to nitrogen oxide (NO): \(\mathrm{NO}_3^- + to \ \mathrm{NO}\) 2. Oxidation of sulfur dioxide (SO2) to sulfate ion (SO4^2-): \(\mathrm{SO}_2 \to \mathrm{SO}_4^{2-}\)

Balance the half-reactions

We need to balance the half-reactions by adding appropriate number of electrons and water/hydrogen ions. The balanced half-reactions are: 1. Reduction of nitrate ion: \(\mathrm{NO}_3^- + 4\mathrm{H^+} + 3\mathrm{e^-} \to \mathrm{NO} + 2\mathrm{H_2O}\) 2. Oxidation of sulfur dioxide: \(\mathrm{SO}_2 + 2\mathrm{H_2O} \to \mathrm{SO}_4^{2-} + 4\mathrm{H^+} + 2\mathrm{e^-}\)

Find the overall cell reaction

Add the balanced half-reactions to find the overall cell reaction: \(\mathrm{NO}_3^- + 4\mathrm{H^+} + 3\mathrm{e^-} + \mathrm{SO}_2 + 2\mathrm{H_2O} \to \mathrm{NO} + 2\mathrm{H_2O} + \mathrm{SO}_4^{2-} + 4\mathrm{H^+} + 2\mathrm{e^-}\) Simplify the reaction: \(\mathrm{NO}_3^- + \mathrm{SO}_2 + 2\mathrm{H^+} + \mathrm{e^-} \to \mathrm{NO} + \mathrm{SO}_4^{2-}\)

Calculate the standard cell potential

Use the given standard reduction potentials to calculate the standard cell potential: \(E_{\text {cell}}^{\circ} = E_{\mathrm{red}}^{\circ} - E_{\mathrm{oxd}}^{\circ} = 0.964 \mathrm{~V} - 0.155 \mathrm{~V} = 0.809 \mathrm{~V}\)

Determine the concentration of hydrogen ions

Find the concentration of hydrogen ions (H+) using the given pH value: \(\mathrm{pH} = -\log_{10}[\mathrm{H^+}]\) \([\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-4.30} = 5.01\times10^{-5}\ M\)

Use the Nernst equation to find the cell voltage

The Nernst equation relates the cell voltage to the standard cell potential and the concentrations of the species involved: \(E_{\text {cell}} = E_{\text {cell}}^{\circ} - \frac{RT}{nF} \ln Q\) where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday's constant, and \(Q\) is the reaction quotient. Assuming the temperature is \(298\ \mathrm{K}\), and using the values of \(R\) and \(F\), the Nernst equation becomes: \(E_{\text {cell}} = E_{\text {cell}}^{\circ} - \frac{0.0592}{n} \log Q\) Here, \(n = 1\) since there's one electron involved, and \(Q = \frac{[\mathrm{NO}][\mathrm{SO}_4^{2-}]}{[\mathrm{NO}_3^-][\mathrm{SO}_2] [\mathrm{H^+}]^2} = \frac{(0.100)(0.100)}{(0.100)(0.100)(5.01\times10^{-5})^2} = 3.99\times10^6\) Now, plug in the values into the Nernst equation: \(E_{\text {cell}} = 0.809 - \frac{0.0592}{1} \log (3.99\times10^6) \approx 0.712\ \mathrm{V}\) Therefore, the voltage of the cell is approximately \(0.712\) volts.

Key concepts

Nernst Equation

The Nernst equation plays a fundamental role in understanding the behavior of electrochemical cells under non-standard conditions. It allows us to calculate the electric potential of a half-cell or full cell under different temperatures, pressures, and concentration conditions. It's expressed as:

\[\begin{equation}E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln Q\end{equation}\]
where:

• \[\begin{equation}E_{\text{cell}}\end{equation}\] is the cell potential under non-standard conditions,
• \[\begin{equation}E_{\text{cell}}^{\circ}\end{equation}\] is the standard cell potential,
• \[\begin{equation}R\end{equation}\] is the universal gas constant (8.314 J/(mol·K)),
• \[\begin{equation}T\end{equation}\] is the temperature in Kelvin,
• \[\begin{equation}n\end{equation}\] is the number of moles of electrons transferred in the electrochemical reaction,
• \[\begin{equation}F\end{equation}\] is the Faraday's constant (96485 C/mol), and
• \[\begin{equation}Q\end{equation}\] is the reaction quotient, which represents the ratio of the product of concentrations of the reaction products to the reactants, each raised to the power of their stoichiometric coefficients.

For the reaction with nitrate ions and sulfur dioxide, the Nernst equation is particularly critical when dealing with changes in concentration and differing pH levels, affecting the overall cell potential.

Standard Cell Potential

The standard cell potential, often denoted as \[\begin{equation}E_{\text{cell}}^{\circ}\end{equation}\], is a measurement that indicates how capable a cell is at generating an electric potential under standard conditions. These conditions are typically 1 molar concentration for all solutions, a pressure of 1 atmosphere for any gases, and a temperature of 298 K (25°C).

The standard cell potential can be calculated using standard reduction potentials for the half-reactions involved. For an overall reaction, it's found by subtracting the standard reduction potential of the anode (oxidation half-reaction) from the standard reduction potential of the cathode (reduction half-reaction). In our exercise, the calculation was:\[\begin{equation}E_{\text{cell}}^{\circ} = E_{\text{red}}^{\circ}_{\text{cathode}} - E_{\text{red}}^{\circ}_{\text{anode}}\end{equation}\].

The value gives us an insight into the driving force behind the electrochemical reaction when no external voltage is applied. A positive \[\begin{equation}E_{\text{cell}}^{\circ}\end{equation}\] indicates a spontaneous reaction under standard conditions.

Redox Reactions

Redox reactions are the foundation of electrochemical cell processes. They encompass all chemical reactions in which atoms have their oxidation number (oxidation state) changed.

Redox reactions can be broken down into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained. The substance that loses electrons is called the reducing agent, whereas the substance that gains electrons is termed the oxidizing agent.

In the given exercise, sulfur dioxide (SO2) is oxidized to sulfate (SO4^2-), giving up electrons, while nitrate ion (NO3^-) is reduced to nitrogen oxide (NO), accepting electrons. In the cell, these processes occur simultaneously and are spatially separated into two half-cells linked by a salt bridge or membrane, which allows for the flow of ions to maintain charge balance during the electrochemical reaction.