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Calculate the \(\mathrm{pH}\) of the cathode compartment for the following reaction given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\) when \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), and \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
Given the provided information and by following the steps, we can calculate the pH of the cathode compartment using the Nernst equation, half-reactions, and cell potentials. First, find the standard reduction and oxidation potentials (\(E_{\text{red}}^{\circ}\) and \(E_{\text{ox}}^{\circ}\)) from a reference table. Then, substitute the given values for the concentrations of Chromium (III), Aluminum (III), and Dichromate ions in the equation from Step 4. Solve for the H鈦 concentration, \([\text{H}^{+}]\). Finally, use the pH formula (\(pH = -\log_{10} [\text{H}^{+}]\)) to calculate the pH of the cathode compartment.

Step by step solution

01

Identify the Half-Reactions

The overall redox reaction is given as: \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\) The half-reactions are: Reduction: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2}\mathrm{O}(l)\) Oxidation: \(2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+6 \mathrm{e}^{-}\)
02

Write the Nernst Equation for the Reduction Half-Reaction

The Nernst equation relates the half-cell potentials to the concentrations of the species involved in the reaction. For the reduction half-reaction, the Nernst equation is: \(E_{\text{red}}=E_{\text{red}}^{\circ}-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{[\text{Cr}^{3+}]^{2}}{[\text{Cr}_{2}\text{O}_{7}^{2-}][\text{H}^{+}]^{14}}\right)\)
03

Write the Nernst Equation for the Oxidation Half-Reaction

For the oxidation half-reaction, the Nernst equation is given by: \(E_{\text{ox}}=E_{\text{ox}}^{\circ}-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{1}{[\text{Al}^{3+}]^{2}}\right)\)
04

Use the Given Overall Cell Potential to Relate the Half-Cell Potentials

The overall cell potential can be found by subtracting the oxidation half-cell potential from the reduction half-cell potential: \(\mathscr{E}_{\text {cell }} = E_{\text{red}} - E_{\text{ox}}\) Substitute the Nernst equations for \(E_{\text{red}}\) and \(E_{\text{ox}}\): \(\mathscr{E}_{\text {cell}} =[E_{\text{red}}^\circ-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{[\text{Cr}^{3+}]^{2}}{[\text{Cr}_{2}\text{O}_{7}^{2-}][\text{H}^{+}]^{14}}\right)] - [E_{\text{ox}}^\circ-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{1}{[\text{Al}^{3+}]^{2}}\right)]\)
05

Solve for the H鈦 Concentration

Given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\), \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\), we will need to find the standard reduction and oxidation potentials (\(E_{\text{red}}^{\circ}\) and \(E_{\text{ox}}^{\circ}\)) from a reference table. Then, substitute the given values in the equation from Step 4 and solve for the H鈦 concentration, \([\text{H}^{+}]\).
06

Calculate the pH

Once the H鈦 concentration is found, use the pH formula to calculate the pH: \(pH = -\log_{10} [\text{H}^{+}]\) Substitute the H鈦 concentration into the formula and find the pH of the cathode compartment.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst equation
The Nernst equation is a vital tool in electrochemistry used to calculate the cell potential or electrode potential of a chemical reaction under non-standard conditions. It provides insight into how the concentrations of the reactants and products in a solution will affect the voltage of a cell. The general form of the Nernst equation is:\[E = E^0 - \frac{RT}{nF} \ln Q\]where:
  • \(E\) is the cell potential under non-standard conditions,
  • \(E^0\) is the standard cell potential,
  • \(R\) is the universal gas constant (8.314 J/mol路K),
  • \(T\) is the temperature in Kelvin,
  • \(n\) is the number of moles of electrons transferred in the reaction,
  • \(F\) is Faraday's constant (approximately 96485 C/mol),
  • \(Q\) is the reaction quotient.
In aqueous solutions at room temperature, the Nernst equation can be simplified when using a logarithmic form with base 10 for ease of use:\[E = E^0 - \frac{0.05916}{n} \log_{10} Q\]This equation shows how the cell potential decreases as the reaction reaches equilibrium, where \(Q\) equals the equilibrium constant.
Redox reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes where electrons are transferred between two species. Understanding redox reactions is crucial for grasping how battery cells and other electrochemical cells work. In a redox reaction:
  • One species donates electrons and becomes oxidized (oxidation),
  • Another species gains electrons and becomes reduced (reduction).
The substance that is oxidized loses electrons and its oxidation state increases, while the substance that is reduced gains electrons and its oxidation state decreases. Redox reactions can be separated into two half-reactions, one for reduction and one for oxidation. For example, in the given exercise:
  • Reduction half-reaction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14 \mathrm{H}^+ \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_2\mathrm{O}\)
  • Oxidation half-reaction: \(2 \mathrm{Al} \rightarrow 2 \mathrm{Al}^{3+} + 6 \mathrm{e}^-\)
Each half-reaction involves a specific potential, which is used in calculating the overall cell potential.
Standard reduction potential
Standard reduction potential, denoted as \(E^0\), is a measure of the tendency of a chemical species to be reduced, measured in volts. It is determined under standard conditions, which are 25掳C, 1M concentration for each ion participating in the reaction, and 1 atm pressure for any gases involved.A higher \(E^0\) value indicates a greater tendency to gain electrons and be reduced, while a lower or negative \(E^0\) value suggests a lesser tendency. Standard reduction potentials are essential when calculating the overall cell potential in a redox reaction because they can indicate which way electrons will flow between the species in the reaction.In the context of the Nernst equation, the standard reduction potential is used to calculate the cell potential when concentrations are at equilibrium. It represents the potential difference when the electrodes are at a standard state. When combined with the standard oxidation potential (which is the reduction potential of the reverse reaction with the opposite sign), these values help in determining whether a reaction is spontaneous under standard conditions (a positive overall cell potential suggests spontaneity).
Half-cell reactions
Half-cell reactions are a way to represent the oxidation or reduction processes separately in an electrochemical reaction. Each half-cell consists of an electrode in contact with ions in a solution, where either oxidation or reduction occurs.A redox reaction involves two half-cells:
  • The reduction half-cell, where a species gains electrons.
  • The oxidation half-cell, where a species loses electrons.
For example, using the provided exercise:
  • Reduction half-cell: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14 \mathrm{H}^+ + 6e^- \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\)
  • Oxidation half-cell: \(2 \mathrm{Al} \rightarrow 2 \mathrm{Al}^{3+} + 6e^-\)
Each half-reaction occurs at an electrode; the cathode for reduction and the anode for oxidation. The flow of electrons from the anode to the cathode through an external circuit forms the basis for electricity generation in electrochemical cells. This system allows for the determination of cell voltages and is critical for calculations involving the Nernst equation and cell potentials.

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Most popular questions from this chapter

Consider the standard galvanic cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{aligned} $$

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{~kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) \(\mathscr{C}^{\circ}=0.68 \mathrm{~V}\) \(\begin{array}{ll}\text { b. } \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} & \mathscr{E}^{\circ}=-1.18 \mathrm{~V} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.036 \mathrm{~V}\end{array}\)

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

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