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An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Short Answer

Expert verified
The potential of the electrochemical cell at 25掳C is 1.40 V. The concentration of Al鲁鈦 in the unknown solution is approximately 5.95 脳 10鈦宦 M.

Step by step solution

01

Identify relevant formulas

We will use the Nernst equation to find the potential of the electrochemical cell: \(E = E^\circ - \frac{0.0592}{n} \log Q\) Where \(E\) is the cell potential, \(E^\circ\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient. In this problem, we are given the concentrations of Ni虏鈦 and Al鲁鈦, and we need to find the standard cell potentials for the nickel and aluminum half-reactions. We can find this information in a standard reduction potential table.
02

Find standard cell potentials

Using a standard reduction potential table, we find that: \(E_{Ni^{2+}/Ni}^\circ = -0.25 \, V\) \(E_{Al^{3+}/Al}^\circ = -1.66 \, V\) As the cell potential is calculated as the difference between the potentials of each half-reaction, the standard cell potential will be: \(E_{cell}^\circ = E_{Ni^{2+}/Ni}^\circ - E_{Al^{3+}/Al}^\circ\)
03

Calculate the standard cell potential

Using the standard cell potentials found in step 2: \(E_{cell}^\circ = (-0.25) - (-1.66) = 1.41 \, V\)
04

Calculate the reaction quotient

The reaction quotient \(Q\) for this cell is given by: \(Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Al}^{3+}]}\) Using the given concentrations of both ions: \(Q = \frac{1.0}{7.2 \times 10^{-3}} = 138.89\)
05

Calculate the cell potential

Now we can use the Nernst equation to find the cell potential at 25掳C: \(E = E^\circ - \frac{0.0592}{n} \log Q\) As aluminum loses 3 electrons during the redox reaction, \(n = 3\). Substitute the values obtained so far: \(E = 1.41 - \frac{0.0592}{3} \log 138.89 = 1.40 \, V\) So the potential of the cell at 25掳C is 1.40 V. Now for part b:
06

Set up the Nernst equation with the unknown concentration of Al鲁鈦

We are given the measured cell potential and asked to find the unknown concentration of Al鲁鈦 in the solution. We will use the Nernst equation with unknown \(Q\): \(E = E^\circ - \frac{0.0592}{n} \log Q\) Here, \(E\) is given as 1.62 V, and \(Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Al}^{3+}]}\).
07

Solve for the unknown concentration of Al鲁鈦

Rearrange the Nernst equation to solve for \(Q\): \(\log Q = \frac{n}{0.0592} (E^\circ - E) = \frac{3}{0.0592} (1.41 - 1.62)\) Calculate \(Q\): \(Q = 10^{-3.773} = 1.68 \times 10^{-4}\) Now use \(Q\) to find the unknown concentration of Al鲁鈦: \([\mathrm{Al}^{3+}] = \frac{1.0}{Q} = \frac{1.0}{1.68 \times 10^{-4}} = 5.95 \times 10^{-3} M\) The concentration of Al鲁鈦 in the unknown solution is approximately 5.95 脳 10鈦宦 M.

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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 plays a key role in electrochemistry. It allows us to calculate the potential of an electrochemical cell under non-standard conditions. It is represented mathematically as:\[E = E^\circ - \frac{RT}{nF} \ln Q\]Where:
  • \( E \) is the cell potential under non-standard conditions.
  • \( E^\circ \) is the standard cell potential.
  • \( R \) is the universal gas constant \(8.314 \, \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}\).
  • \( T \) is the temperature in Kelvin.
  • \( n \) is the number of moles of electrons exchanged in the reaction.
  • \( F \) is Faraday鈥檚 constant, approximately \(96485 \, \text{C}\,\text{mol}^{-1}\).
  • \( Q \) is the reaction quotient.
At standard temperature \(25^{\circ} \text{C}\) or \(298 \text{K}\), the equation simplifies to:\[E = E^\circ - \frac{0.0592}{n} \log Q\]This simplified version is used frequently for calculations at room temperature, as shown in the original exercise. The Nernst Equation shows how the cell potential decreases with increasing product concentration \([\text{Al}^{3+}]\) relative to the reactant concentration \([\text{Ni}^{2+}]\).
This equation is indispensable for predicting how changes in concentration affect the electrochemical cell potential.
Standard Cell Potential
The standard cell potential, \( E^\circ \), is a measure of the driving force behind an electrochemical reaction. It is determined under standard conditions: \(1 \, \text{M}\) concentration, \(1 \, \text{atm}\) pressure, and a temperature of \(25^{\circ} \text{C}\). The standard cell potential is calculated from the standard reduction potentials of the half-reactions involved.

Determining Standard Cell Potential

For a typical electrochemical cell:
  • Identify the half-reactions and their respective standard reduction potentials from standard tables.
  • Calculate the standard cell potential by taking the difference between the reduction potential of the cathode and the anode.
  • In the provided exercise:
    • The standard reduction potential for \(\text{Ni}^{2+}/\text{Ni}\) is \(-0.25 \text{ V}\).
    • For \(\text{Al}^{3+}/\text{Al}\), it's \(-1.66 \text{ V}\).
    • Hence, \(E_{cell}^\circ = E_{\text{Ni}^{2+}/\text{Ni}}^\circ - E_{\text{Al}^{3+}/\text{Al}}^\circ = 1.41 \text{ V}\).
This potential indicates the voltage that the cell generates spontaneously, under standard conditions, and serves as a baseline for further calculations with the Nernst Equation.
Knowing the standard cell potential helps in understanding the energy efficiency and feasibility of electrochemical cells in practical applications.
Reaction Quotient
The Reaction Quotient, \( Q \), is a valuable concept when examining the state and direction of a chemical reaction. It provides insight into whether a reaction is at equilibrium, or which way it needs to shift to reach it.

Understanding the Reaction Quotient

In the context of the Nernst Equation:
  • \( Q \) is calculated as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.
  • For the exercise mentioned, the reaction quotient \( Q \) is defined as:\[Q = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\text{Ni}^{2+}]}{[\text{Al}^{3+}]}\]
  • This equation signifies that as the product concentration increases or the reactant concentration decreases, \( Q \) increases, consequently reducing the cell potential \( E \).

Importance of Reaction Quotient

The value of \( Q \) helps predict the direction of the reaction:
  • If \( Q < K_{eq} \), the reaction will proceed forward to reach equilibrium.
  • If \( Q > K_{eq} \), the reaction will shift backward.
  • When \( Q = K_{eq} \), the system is at equilibrium, and there is no change in concentrations over time.
Determining \( Q \) in real-time conditions, as shown in the exercise, allows for the adjustment and control of electrochemical processes, ensuring optimal functioning of cells.

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Most popular questions from this chapter

Consider the following half-reactions: $$\begin{array}{ll}{\mathrm{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.77 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.73 \mathrm{V}} \\\ {\mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.62 \mathrm{V}}\end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0\(M\) in chloride ion and 0.020\(M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that 99\(\%\) of a metal must be plated out before another metal begins to plate out.)

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C} )\) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\) . b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\). c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

What mass of each of the following substances can be produced in 1.0 h with a current of 15 A? a. \(\mathrm{Co}\) from aqueous \(\mathrm{Co}^{2+}\) b. \(\mathrm{Hf}\) from aqueous \(\mathrm{Hf}^{4+}\) c. \(\mathrm{I}_{2}\) from aqueous \(\mathrm{KI}\) d. \(\mathrm{Cr}\) from molten \(\mathrm{CrO}_{3}\)

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

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