/*! 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 87 Consider the following galvanic ... [FREE SOLUTION] | 91影视

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Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Short Answer

Expert verified
The cell potential (E) for this galvanic cell is approximately 0.851 V, and the Gibbs free energy change (螖G) for the given reaction at these conditions is approximately -164000 J/mol.

Step by step solution

01

Identify the half-reactions

Since we have the overall reaction, let's determine the half-reactions as reduction: (i) For Cr: \(Cr^{3+}(aq) + e^{-} \rightarrow Cr^{2+}(aq)\) (ii) For Co: \(Co^{2+}(aq) + 2e^{-}\rightarrow Co(s)\)
02

Write and calculate the Nernst equation for the cell potential

The Nernst equation is: \(\mathscr{E} = \mathscr{E}^{\circ} - \frac{0.05916}{n} \log Q\) Here, we need to find the standard cell potential, E掳, which could be calculated by combining the standard reduction potentials for each half-cell. But since we are given the equilibrium constant, we can use the relationship between E掳 and K instead: \(\mathscr{E}^{\circ} = \frac{0.05916}{n} \log K\) We know that n = the number of electrons transferred, which is 2 for this reaction. And K = 2.79 脳 10^7, so we can now find E掳: \(\mathscr{E}^{\circ} = \frac{0.05916}{2} \log (2.79 \times 10^{7}) \approx 1.026\, V\) Now, we need to find the reaction quotient, Q. The reaction formula for Q is: \(Q = \frac{[\mathrm{Cr}^{3+}]^{2}}{[\mathrm{Cr}^{2+}]^{2}[\mathrm{Co}^{2+}}\) Given the concentrations: [Cr鲁鈦篯 = 2.0 M, [Cr虏鈦篯 = 0.30 M, [Co虏鈦篯 = 0.20 M, we calculate Q: \(Q = \frac{(2.0)^2}{(0.30)^2(0.20)} = 444.44\) Now, we can calculate E using the Nernst equation: \(\mathscr{E} = 1.026 - \frac{0.05916}{2} \log 444.44 \approx 0.851\, V\)
03

Calculate 螖G using the cell potential

We can find the Gibbs free energy change (螖G) using the formula: \(\Delta G = -nFE\) Where n is the number of electrons transferred (n = 2), F is the Faraday constant (F 鈮 96485 C/mol), and E is the cell potential we calculated in the previous step (E 鈮 0.851 V). Now, we can calculate 螖G: \(\Delta G = - (2)(96485 C/mol)(0.851 V) \approx -164000\, J\, mol^{-1}\) So, the cell potential (E) for this galvanic cell is approximately 0.851 V, and the Gibbs free energy change (螖G) for the given reaction at these conditions is approximately -164000 J/mol.

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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 fundamental concept in electrochemistry that provides a way to calculate the electric potential of a galvanic cell under non-standard conditions. It accounts for the effect of ion concentration on the cell potential. This makes the Nernst equation essential for understanding how batteries work, how cells sustain their membrane potential, and other applications where the ion concentration is crucial.

The general form of the Nernst equation is as follows:
\[\mathscr{E} = \mathscr{E}^{\circ} - \left(\frac{0.05916}{n}\right) \log Q\]
where \(\mathscr{E}\) is the cell potential under non-standard conditions, \(\mathscr{E}^{\circ}\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the electrochemical reaction, and \(Q\) is the reaction quotient reflecting the ratio of product and reactant activities.

In practice, the Nernst equation allows us to assess how cell potential will change as the concentration of reactants and products shifts, which helps us understand the behavior of electrochemical cells in a dynamic environment.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationships between electrical energy and chemical reactions. This field of study is crucial for the development of technology such as batteries, fuel cells, corrosion prevention, and electroplating. In the exercise involving the galvanic cell, we delve into the heart of electrochemistry, exploring how chemical energy is converted into electrical energy and vice versa.

The galvanic cell is a classic electrochemical setup in which a spontaneous redox reaction generates an electrical current. This process entails the flow of electrons from the anode (where oxidation occurs) to the cathode (where reduction takes place) through an external circuit. Understanding galvanic cells reveals the practical application of redox reactions and provides a real-world context for the thermodynamic principles at play in electrochemistry.
Gibbs Free Energy
Gibbs free energy (\(\Delta G\)) is a thermodynamic quantity that measures the maximum amount of work that can be done by a thermodynamic process at constant temperature and pressure. In the context of electrochemistry, the change in Gibbs free energy is directly related to the electrical work that can be harnessed from a galvanic cell.

The relationship between the Gibbs free energy change and cell potential is given by the equation:
\[\Delta G = -nFE\]
where \(\Delta G\) is the change in Gibbs free energy, \(n\) is the amount of substance (measured in moles of electrons transferred), \(F\) is the Faraday constant, and \(E\) is the cell potential. A negative value for \(\Delta G\) suggests that the reaction is spontaneous, driving the electrons through the circuit, while a positive value indicates a non-spontaneous reaction requiring an external source of energy to proceed.
Equilibrium Constant
The equilibrium constant (\(K\)) expresses the extent to which a chemical reaction occurs until reaching a state of equilibrium. In the context of the electrochemical cell, the equilibrium constant is related to the standard cell potential (\(\mathscr{E}^{\circ}\)) by the equation:
\[\mathscr{E}^{\circ} = \left(\frac{0.05916}{n}\right) \log K\]
In the given galvanic cell, we deal with a high equilibrium constant value, which indicates that the products of the redox reaction are heavily favored at equilibrium. This is characteristic of a reaction that will proceed forward to a significant extent under standard conditions, producing a strong driving force for the flow of electrons, and thus a higher cell potential.

Understanding the equilibrium constant in relation to electrochemical cells helps students predict the direction and extent of reactions, the likelihood of a cell to perform work, and how changes in concentration affect cell voltage and free energy.

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Most popular questions from this chapter

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{lr} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-} & 80^{\circ} \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow_{2 \mathrm{H}_{2} \mathrm{O}} & 2.01 \mathrm{V} \\ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} & -0.83 \mathrm{V} \\ \mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na} & -2.71 \mathrm{V} \end{array}$$

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 \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

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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