Problem 56 - Chemistry: An Atoms-Focused Approach

Question

Chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) is produced by the following reaction of chlorate \(\left(\mathrm{ClO}_{3}^{-}\right)\) with \(\mathrm{Cl}^{-}\) in acid solution: \(2 \mathrm{ClO}_{3}^{-}(a q)+2 \mathrm{Cl}^{-}(a q)+4 \mathrm{H}^{+}(a q) \rightarrow\) $$ 2 \mathrm{ClO}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ a. Determine \(E^{\circ}\) for the reaction. b. The reaction at \(298 \mathrm{K}\) produces a mixture of gases in the reaction vessel in which \(P_{\mathrm{ClO}_{2}}=2.0\) atm and \(P_{\mathrm{C}_{1}}=1.00\) atm. Calculate \(\left[\mathrm{ClO}_{3}^{-}\right]\) if, at equilibrium, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{Cl}^{-}\right]=10.0 \mathrm{M}.\)

Step-by-Step Solution

Answer

Solution: The standard cell potential (E°) for the reaction is 4.19 V. The concentration of chlorate ions at equilibrium is 3.57 x 10^-23 M.

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

To determine the cell potential, we first need to identify the half-reactions and find their standard reduction potentials (\(E^{\circ}\)) from a table of standard reduction potentials. The half-reactions are: 1. \(\mathrm{ClO}_{3}^{-}\left(aq\right)+6\mathrm{H}^+\left(aq\right)+6\mathrm{e}^-\rightarrow3\mathrm{H}_2\mathrm{O}\left(l\right)+\mathrm{Cl}^{-}\left(aq\right)\). (Chlorate Reduction) 2. \(\mathrm{Cl}^{-}\left(aq\right)+\mathrm{e}^-\rightarrow\frac{1}{2}\mathrm{Cl}_{2}\left(g\right)\). (Chlorine Reduction) Then, find the standard reduction potentials from a table: \(E_1^{\circ} = -1.47\ \text{V}\) (Chlorate Reduction) \(E_2^{\circ} = +1.36\ \text{V}\) (Chlorine Reduction)

Step 2: Calculate the standard cell potential

To determine the standard cell potential, we need to subtract the reduction potential of the reducing half-reaction from the reduction potential of the oxidizing half-reaction. Since we want chlorine to be reduced, we need to multiply the second equation by 2 and then reverse the first equation: \(3\mathrm{H}_2\mathrm{O}\left(l\right)+\mathrm{Cl}^{-}\left(aq\right) \rightarrow \mathrm{ClO}_{3}^{-}\left(aq\right)+6\mathrm{H}^+\left(aq\right)+6\mathrm{e}^-\) \(2(\mathrm{Cl}^{-}\left(aq\right)+\mathrm{e}^-\rightarrow\frac{1}{2}\mathrm{Cl}_{2}\left(g\right))\) \(E_{cell}^\circ = -(E_1^\circ) + 2E_2^\circ = -(-1.47) + 2(1.36) = 1.47 + 2.72 = 4.19\ \text{V}\)

Step 3: Write the Nernst equation and set up the equilibrium expression

The Nernst equation relates the standard cell potential, E_cell, and the concentration of the species in the reaction: \(E_{cell} = E^\circ - \frac{0.0592}{n}\log{Q}\) where \(n\) is the number of moles of electrons exchanged and \(Q\) is the reaction quotient. To solve for the chlorate concentration at equilibrium, set up the equilibrium expression of reaction quotient: \(Q = \frac{[\mathrm{ClO}_{2}]^2[\mathrm{Cl}_2]}{[\mathrm{ClO}_{3}^{-}]^2[\mathrm{H}^{+}]^4[\mathrm{Cl}^{-}]^2}\)

Step 4: Plug in given values and solve for chlorate concentration

At equilibrium, we are given the following values: \(P_{\mathrm{ClO}_{2}}=2.0\) atm, \(P_{\mathrm{C}_{1}}=1.00\) atm, \([\mathrm{H}^{+}]=[\mathrm{Cl}^{-}]=10.0\ \mathrm{M}\) Assuming ideal gas behavior, we can calculate the molar concentrations of \(\mathrm{ClO}_{2}\) and \(\mathrm{Cl}_{2}\) using the ideal gas law: \([\mathrm{ClO}_{2}]=\frac{P_{\mathrm{ClO}_{2}}}{RT}=\frac{2.0\ \text{atm}}{0.0821\frac{\text{L.atm}}{\text{mol.K}}\cdot298\ \text{K}}=0.0813\ \text{M}\) \([\mathrm{Cl}_{2}]=\frac{P_{\mathrm{C}_{1}}}{RT}=\frac{1.00\ \text{atm}}{0.0821\frac{\text{L.atm}}{\text{mol.K}}\cdot298\ \text{K}}=0.0406\ \text{M}\) Since the system is at equilibrium and the reaction has occurred, the cell potential at equilibrium is zero: \(0 = 4.19 - \frac{0.0592}{6}\log{\left(\frac{0.0813^2 \cdot 0.0406}{[\mathrm{ClO}_{3}^{-}]^2 \cdot 10^4 \cdot 100}\right)}\) Now, solve for \([\mathrm{ClO}_{3}^{-}]\): \(\log{\left(\frac{0.0813^2 \cdot 0.0406}{[\mathrm{ClO}_{3}^{-}]^2 \cdot 10^4 \cdot 100}\right)} = \frac{6 \cdot 4.19}{0.0592} = 42.65\) \(\frac{0.0813^2 \cdot 0.0406}{[\mathrm{ClO}_{3}^{-}]^2 \cdot 10^4 \cdot 100} = 10^{42.65}\) \([\mathrm{ClO}_{3}^{-}]^2 = \frac{0.0813^2 \cdot 0.0406}{10^{42.65} \cdot 10^4 \cdot 100} = 1.27 \times 10^{-46}\) Thus, the concentration of chlorate ions at equilibrium is: \([\mathrm{ClO}_{3}^{-}] = \sqrt{1.27 \times 10^{-46}} = 3.57 \times 10^{-23}\ \mathrm{M}\).

Key Concepts

Standard reduction potential

In electrochemistry, the standard reduction potential, denoted as \( E^{\circ} \), is a crucial concept used to predict the direction of electron flow in an electrochemical cell. It represents the tendency of a chemical species to gain electrons, which is also known as reduction. Each half-reaction has its own standard reduction potential, which can be found in a table of standard reduction potentials.

To determine the standard cell potential for a reaction, the standard reduction potentials of the substances involved are considered. For example, if we have a reduction reaction involving the conversion of \( \mathrm{ClO}_{3}^{-} \) to \( \mathrm{Cl}^{-} \), the standard reduction potential for that reaction is \( -1.47 \ \text{V} \). In contrast, when \( \mathrm{Cl}^{-} \) is reduced to \( \mathrm{Cl}_2 \), the standard reduction potential is \( +1.36 \ \text{V} \). These values allow us to calculate the overall standard cell potential by combining the respective half-reaction potentials.

• If the standard reduction potential is positive, the species is more likely to be reduced.
• If negative, the species is less likely to gain electrons.

Knowing \( E^{\circ} \) aids in determining the feasibility and direction of any redox reaction under standard conditions.

Nernst equation

The Nernst equation is a powerful formula in electrochemistry that relates the cell potential to the concentrations (or pressures) of the reactants and products involved in the reaction. It is given by the formula:\[E_{cell} = E^{\circ} - \frac{0.0592}{n}\log{Q}\]

Where \( E_{cell} \) is the cell potential at any given moment, \( E^{\circ} \) is the standard cell potential, \( n \) is the number of moles of electrons exchanged, and \( Q \) is the reaction quotient. The Nernst equation helps to calculate cell potentials under non-standard conditions, as it incorporates the concentrations and pressures of reactants and products.

In practice, the Nernst equation is used to determine how the cell potential changes as the reaction progresses. The logarithmic term adjusts the standard potential based on the reaction's progress, providing a real-time measure of its voltage output. This is particularly useful in understanding how changing concentrations or pressures affect the potential of an electrochemical cell.

• The equation emphasizes that cell potential decreases as equilibrium is approached.
• Therefore, as reactants are consumed and products formed, \( Q \) increases, leading to a drop in cell potential.

Reaction quotient

The reaction quotient, denoted as \( Q \), is a dimensionless number that provides a snapshot of the relative concentrations of reactants and products during a reaction. It is calculated similarly to the equilibrium constant, \( K \), but unlike \( K \), \( Q \) can be determined at any point in the reaction, not just at equilibrium.

For the reaction of \( \mathrm{ClO}_{3}^{-} \), \( \mathrm{Cl}^{-} \), and \( \mathrm{H}^{+} \) ions to produce \( \mathrm{ClO}_{2} \) and \( \mathrm{Cl}_2 \), the reaction quotient is expressed as:\[Q = \frac{[\mathrm{ClO}_{2}]^2[\mathrm{Cl}_2]}{[\mathrm{ClO}_{3}^{-}]^2[\mathrm{H}^{+}]^4[\mathrm{Cl}^{-}]^2}\]

By comparing \( Q \) with \( K \), the direction of the reaction can be predicted:

• If \( Q < K \), the reaction will proceed forward, producing more products.
• If \( Q > K \), the reaction will go backward, forming more reactants.
• If \( Q = K \), the reaction is at equilibrium.

This concept plays a key role in calculating the equilibrium states of reactions and is integral to utilizing the Nernst equation in real-life applications.

Equilibrium concentration

Equilibrium concentration involves the concentrations of all reactants and products in a chemical reaction when it has reached a state of balance, meaning no further changes occur in their amounts. At this point, the forward and reverse reactions occur at the same rate, thus maintaining constant concentrations.

For the reaction to form \( \mathrm{ClO}_{2} \) and \( \mathrm{Cl}_2 \) from \( \mathrm{ClO}_{3}^{-} \), \( \mathrm{Cl}^{-} \), and \( \mathrm{H}^{+} \), determining the equilibrium concentration of a component like \( \mathrm{ClO}_{3}^{-} \) involves balancing the pressures and concentrations using the reaction quotient \( Q \) and the ideal gas law, alongside the Nernst equation.

In cases where pressures of gaseous products are given, they can be converted into molar concentrations using the ideal gas law formula:\[[\mathrm{Gas}] = \frac{P_{\mathrm{Gas}}}{RT}\]This gives the equilibrium concentration of gas components in a reaction.

• Equilibrium concentrations can be drastically different from initial ones depending on reaction conditions and stoichiometry.
• Understanding this allows chemists to manipulate conditions to favor the formation of desired products.