/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 149 A galvanic cell is based on the ... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 149

A galvanic cell is based on the following half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{array}$$ In this cell, the silver compartment contains a silver electrode and excess \(\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. Assuming \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\space \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

Short Answer

Expert verified
In this galvanic cell with specified half-reactions, the potential at 25掳C is 0.46 V. To obtain a cell potential of 0.52 V, approximately 0.017 moles of NH鈧� should be added to the copper compartment.

Step by step solution

01

Determine the balanced overall cell reaction

First, we need to balance the two half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & (1) \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & (2) \end{array}$$ Multiply reaction (1) by 2, and then subtract the resulting reaction from reaction (2) to get the overall balanced reaction: 2Ag鈦� + 2e鈦� 鈫� 2Ag(s) (multiply by 2) Cu虏鈦� + 2e鈦� 鈫� Cu(s) --------------------------------------- 2Ag鈦� + Cu 鈫� 2Ag(s) + Cu虏鈦�
02

Calculate cell potential using standard electrode potentials

We can now calculate the cell potential (E_cell) using the standard electrode potentials: E_cell = E潞_reduced - E潞_oxidized In this case, Ag鈦� + e鈦� 鈫� Ag(s) is the reduction reaction and Cu虏鈦� + 2e鈦� 鈫� Cu(s) is the oxidation reaction. So, we have: E_cell = 0.80 V - 0.34 V = 0.46 V The potential for this cell at 25掳C is 0.46 V. #b. Moles of NH鈧� needed for specific cell potential#
03

Calculate the [Cu(NH鈧�)鈧劼测伜] needed for the desired cell potential

We have to use the Nernst equation to find the required concentration of Cu(NH鈧�)鈧劼测伜: E_cell = E潞_cell - \(\frac{RT}{nF}\) ln(Q) Here, E_cell is 0.52 V (given), E潞_cell is 0.46 V (from previous step), R is the gas constant (8.314 J/mol路K), T is the temperature (25掳C = 298 K), n is the number of electrons transferred (2), F is Faraday's constant (96485 C/mol), and Q is the reaction quotient. 0.52 V = 0.46 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(Q) Now, let's find Q. We can write it as: Q = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}] \times [\mathrm{Ag}^+]^2}{[\mathrm{Cu}^{2+}]}\) The standard Gibbs free energy change for dissolution of AgCl is: 螖G潞 = -RT ln(K_sp) = -nFE潞_AgCl From the information given in the problem, we can find E潞_AgCl: E潞_AgCl = -\(\frac{螖G潞}{nF}\) = -\(\frac{RT \times ln(K_{sp})}{nF}\) = -\(\frac{8.314 \times 298 \times ln(1.6 \times 10^{-10})}{96485}\) 鈮� 0.225 V We also know that E潞_cell = E潞_Ag - E潞_AgCl, so: E潞_Ag = E潞_cell + E潞_AgCl = 0.46 V + 0.225 V 鈮� 0.685 V The concentration of Ag鈦� ions is: [Ag鈦篯 = \(\sqrt{\frac{K_{sp}}{[\mathrm{Ag}]}} = \sqrt{\frac{1.6 \times 10^{-10}}{1}}\) 鈮� 4.0 脳 10鈦烩伓 M Now we can find the required [Cu(NH鈧�)鈧劼测伜] to achieve 0.52 V: 0.52 V = 0.46 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(\(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}] \times (4.0 \times 10^{-6})^2}{2.0}\)) On solving the above equation, we get: [Cu(NH鈧�)鈧劼测伜] 鈮� 1.91 M
04

Calculate the moles of NH鈧� required

To find the moles of NH鈧� required, we'll use the given expression for K: K = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}]}{[\mathrm{Cu}^{2+}] \times [\mathrm{NH}_3]^{4}}\) Rearranging the equation and substituting the values: [\mathrm{NH}_3]鈦� = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}]}{K \times [\mathrm{Cu}^{2+}]}\) = \(\frac{1.91}{1.0 \times 10^{13} \times 2.0}\) [\mathrm{NH}_3] 鈮� 0.017 M Now, considering this problem asked for the moles of NH鈧� needed and a 2.0 M solution of Cu虏鈦� in 1.0 L: Moles of NH鈧� required = (0.017 mol/L) 脳 1.0 L 鈮� 0.017 mol The moles of NH鈧� that would have to be added to give a cell potential of 0.52 V at 25掳C are approximately 0.017 mol.

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Key Concepts

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

Understanding Standard Electrode Potentials

Standard electrode potentials, expressed as E掳 values, are measures of the inherent tendency of redox species to gain or lose electrons. These values serve as a benchmark telling us how likely an element is to be reduced under standard conditions, which are typically 25掳C (298 K), 1 atm pressure, and 1M concentration for each ion. When comparing two half-reactions in a galvanic cell, the reaction with the higher E掳 acts as the cathode (reduction), while the one with the lower E掳 acts as the anode (oxidation).

In our exercise, the standard electrode potentials of silver and copper are given as +0.80V and +0.34V, respectively. This implies that under standard conditions, silver ions are more likely to gain electrons and form solid silver compared to copper ions, which makes silver the cathode and copper the anode.

Interpretation of E掳 values:

  • A higher E掳 value indicates a greater likelihood of reduction.
  • A negative E掳 value signifies a tendency to undergo oxidation.
  • Standard electrode potentials are crucial for calculating the overall cell voltage by subtracting the anode E掳 from the cathode E掳.

Understanding these potentials is essential as they are foundational elements of electrochemistry and crucial to determining the direction in which an electrochemical reaction will proceed.

The Nernst Equation and Its Application

The Nernst equation is a relationship that allows us to calculate the actual cell potential under non-standard conditions, taking into account the concentration (or activities) of the reactants and products. It's a crucial formula that bridges the gap between real-world scenarios and standard conditions.

General Form of the Nernst Equation:

E = E掳 - (RT/nF) ln(Q)

Here, E is the cell potential, E掳 is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons exchanged, F is Faraday's constant, and Q is the reaction quotient reflecting the concentrations of the reactants and products.

Adjusting Cell Potential:

  • For reaction spontaneity, E needs to be positive. This, at times, requires modifying reactant or product concentrations.
  • The logarithmic relationship in the Nernst equation means that even small changes in concentration can have a significant impact on cell potential.

In our exercise, the Nernst equation is used to calculate the required concentration of complex ions to adjust the cell potential to a specified value, which reflects the versatility of this equation in practical electrochemistry.

Chemical Equilibrium In A Galvanic Cell

Chemical equilibrium is a state where the rates of the forward and reverse reactions are equal, resulting in no net change of the reactants and products over time. In a galvanic cell, establishing equilibrium involves balancing redox reactions, ensuring charge neutrality, and often involves complex ion formation and dissolution processes influencing the cell's dynamics.

How Equilibrium Affects Cell Potential:
  • As a reaction approaches equilibrium, the cell potential moves towards zero because there's less driving force for the reaction to proceed in either direction.
  • Variations in reactant or product concentrations shift the equilibrium position and consequently alter the cell potential.

In the context of our exercise, the chemical equilibrium of the silver chloride dissolution and copper-ammonia complex formation are pivotal to calculating the final cell potential and the amount of ammonia needed to achieve a specific cell potential.

Complex Ion Formation and Its Impact on Galvanic Cells

Complex ion formation is an equilibrium process where central metal ions bond with surrounding ligands to form charged species. This process can significantly alter the concentrations of participating ions in a solution, thus impacting the reaction quotient, Q, in the Nernst equation and ultimately affecting the cell potential.

Role in Electrochemistry:

  • Complexation can change the solubility of salts (like AgCl), affecting ion availability for the redox reactions.
  • Formation constants (Kf) of complex ions indicate their stability and help in calculating equilibrium concentrations.

The exercise guides us through the formation of a copper-ammonia complex, utilizing the formation constant (K) to calculate the moles of ammonia needed to alter the cell potential. Complexation reactions are thus integral to the understanding of electrochemical systems.

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Most popular questions from this chapter

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} $$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\ \mathrm{OH}^{-}(a q) &=-157 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0 \end{aligned} $$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1.

When aluminum foil is placed in hydrochloric acid, nothing happens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.

Consider the following half-reactions: \(\begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & & \mathscr{E}^{\circ}=1.188 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{E}^{\circ}=0.755 \mathrm{V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} & \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=0.96 \mathrm{V} \end{aligned}\) Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M},\) the cell potential is observed to be 0.31 V. Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).
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