Question

Challenge Write the balanced equation for the cell reaction and calculate the standard cell potential for the reaction that occurs when these half-cells are connected. Describe the reaction using cell notation. $$ \begin{array}{l}{\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}} \\ {\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}}\end{array} $$

Short answer

The balanced cell equation can be written as: \(4\mathrm{NO}_{3}^{-}+16 \mathrm{H}^{+} + 3\mathrm{O}_{2}+6\mathrm{H}_{2} \mathrm{O} \rightarrow 4\mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O} + 12\mathrm{OH}^{-}\). The standard cell potential (\(E^\circ_\text{cell}\)) is calculated to be 0.56 V. The cell notation for this reaction is: \(|\, \mathrm{NO}_{3}^{-}/\mathrm{NO} \, |\, \mathrm{H}^{+} \, ||\, \mathrm{OH}^{-} \, |\, \mathrm{O}_{2} \, |\).

Step-by-step solution

Identify the oxidation and reduction half-reactions

In the given half-cell reactions, we can see that: 1. Nitrate ion (NO3-) is reduced to nitric oxide (NO) by gaining electrons. 2. Oxygen (O2) is reduced to hydroxide ions (OH-) by gaining electrons.

Balance the half-cell reactions

Since the first half-reaction has three electrons and the second one has four, we need to multiply both half-cell reactions so that they have the same number of electrons. Multiplying the first half-reaction by 4 and the second half-reaction by 3, we get: \(4 \times (\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-}\rightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O})\) \(3 \times (\mathrm{O}_{2}+2\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-}\rightarrow 4\mathrm{OH}^{-})\) Which results in: \(4\mathrm{NO}_{3}^{-}+16 \mathrm{H}^{+}+12 \mathrm{e}^{-}\rightarrow 4\mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O}\) \(3\mathrm{O}_{2}+6\mathrm{H}_{2} \mathrm{O}+12 \mathrm{e}^{-}\rightarrow 12\mathrm{OH}^{-}\)

Add the balanced half-cell reactions

Now that the half-cell reactions have the same number of electrons, we can add them to obtain the balanced cell reaction: \(4\mathrm{NO}_{3}^{-}+16 \mathrm{H}^{+}+12 \mathrm{e}^{-} + 3\mathrm{O}_{2}+6\mathrm{H}_{2} \mathrm{O}+12 \mathrm{e}^{-} \rightarrow 4\mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O} + 12\mathrm{OH}^{-}\) The final balanced reaction is: \(4\mathrm{NO}_{3}^{-}+16 \mathrm{H}^{+} + 3\mathrm{O}_{2}+6\mathrm{H}_{2} \mathrm{O} \rightarrow 4\mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O} + 12\mathrm{OH}^{-}\)

Calculate the standard cell potential

From standard potential tables, we can find the standard potentials for the original unbalanced half-cell reactions: 1. \(E^\circ_{\mathrm{NO}_{3}^{-}/\mathrm{NO}} = +0.96\,\text{V}\) (for the reduction of nitrate ions) 2. \(E^\circ_{\mathrm{O}_{2}/\mathrm{OH}^{-}} = +0.40\,\text{V}\) (for the reduction of oxygen) To calculate the standard cell potential, we only need to subtract the second half-reaction's standard potential value from the first: \(E^\circ_\text{cell} = E_1^\circ - E_2^\circ = 0.96\,\text{V} - 0.40\,\text{V} = 0.56\,\text{V}\)

Write the cell notation

We can now write the cell notation for the reaction: \(\mathrm{NO}_{3}^{-}/\mathrm{NO} \, |\, \mathrm{H}^{+} \, ||\, \mathrm{OH}^{-} \, |\, \mathrm{O}_{2}\) So, the cell notation for this reaction is: \(|\, \mathrm{NO}_{3}^{-}/\mathrm{NO} \, |\, \mathrm{H}^{+} \, ||\, \mathrm{OH}^{-} \, |\, \mathrm{O}_{2} \, |\)

Key concepts

Redox Reactions

Redox reactions are fundamental to electrochemistry. They involve two simultaneous reactions: oxidation and reduction. Oxidation is the loss of electrons, whereas reduction is the gain of electrons. Together, these processes account for the transfer of electrons that powers electrochemical cells. In the original exercise, we had two reduction reactions. The nitrate ion (\(\text{NO}_3^-\)) was reduced to nitric oxide (\(\text{NO}\)), and oxygen (\(\text{O}_2\)) was reduced to hydroxide ions (\(\text{OH}^-\)).
Understanding which species undergo oxidation or reduction is crucial for balancing redox reactions. The electrons lost by one species must be equal to those gained by another. This balance often requires multiplying the number of electrons in each half-reaction to make the overall electron exchange equal.
Identifying and balancing such redox reactions is crucial, as it determines how efficiently an electrochemical cell can produce electrical energy. Always ensure that both the mass and charge are balanced in these reactions.

Standard Cell Potential

The standard cell potential, denoted as \( E^\circ_{\text{cell}} \), is a crucial measure in electrochemistry. It tells us the voltage or potential difference a cell can produce when the system is at standard conditions. These conditions include a temperature of 298 K, a pressure of 1 atmosphere, and concentrations of 1 M for all reactants.
To find the \( E^\circ_{\text{cell}} \), follow these steps:

• Identify the standard potentials (\( E^\circ \)) for each half-reaction from a potential table. These are measured in volts (V).
• Subtract the \( E^\circ \) of the oxidation reaction from the \( E^\circ \) of the reduction reaction.

In the exercise, \( E^\circ_{\text{cell}} \) was calculated as \( 0.96 \, \text{V} - 0.40 \, \text{V} = 0.56 \, \text{V} \). A positive \( E^\circ_{\text{cell}} \) indicates a spontaneous reaction under standard conditions. This means the reaction will proceed without external work, effectively producing electricity.

Half-Cell Reactions

Half-cell reactions are the individual oxidation or reduction reactions that occur in separate compartments of an electrochemical cell. These reactions contain either the transfer of electrons or the change in oxidation states that make up the whole cell process.
The concept can be broken down into:

• Anode Half-Cell: The site of oxidation, where electron loss occurs.
• Cathode Half-Cell: The site of reduction, where electron gain occurs.

In practice, these half-cells are physically separated, often by a salt bridge or membrane, that allows ion exchange to maintain charge neutrality. Each half-cell has its own electrode and participates in only one half of the redox process.
In our case, we looked at the reduction of nitrate and oxygen in their respective half-cell reactions:

• \( \mathrm{NO}_3^- + 4 \mathrm{H}^+ + 3 \mathrm{e}^- \rightarrow \mathrm{NO} + 2 \mathrm{H}_2 \mathrm{O} \)
• \( \mathrm{O}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{e}^- \rightarrow 4 \mathrm{OH}^- \)

Balancing these half-reactions is crucial before adding them to form a full redox reaction. The exact number of electrons exchanged in each half-reaction must match to combine them effectively without any excess charges.