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\(\mathrm{Co}\left|\mathrm{Co}^{2+}\left(\mathrm{C}_{2}\right) \| \mathrm{Co}^{2+}\left(\mathrm{C}_{1}\right)\right| \mathrm{Co} ;\) for this cell, \(\Delta G\) is negative if : (a) \(\mathrm{C}_{2}>\mathrm{C}_{1}\) (b) \(\mathrm{C}_{1}>\mathrm{C}_{2}\) (c) \(\mathrm{C}_{1}=\mathrm{C}_{2}\) (d) unpredictable

Short Answer

Expert verified
The change in Gibbs free energy \(\Delta G\) is negative if \(C_2 > C_1\), which corresponds to option (a).

Step by step solution

01

Understand the Cell Notation

The given cell notation represents a galvanic cell (a type of electrochemical cell that generates electrical energy from spontaneous redox reactions). The cell is constructed with cobalt (Co) as the electrode material, Co(II) ions as the species in the electrolyte, and two different concentrations of Co(II) ions, labeled as \(C_1\) and \(C_2\). The anode (oxidation reaction) is on the left, and the cathode (reduction reaction) is on the right.
02

Relate Cell Potential to Gibbs Free Energy

The relationship between the Gibbs free energy \(\Delta G\) and the cell potential \(E_{cell}\) is given by the equation \(\Delta G = -nFE_{cell}\), where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, and \(E_{cell}\) is the cell potential. Since \(\Delta G\) is negative, the cell potential must be positive for the reaction to be spontaneous.
03

Apply Nernst Equation to Determine Cell Potential's Dependency on Concentration

The Nernst equation relates the cell potential \(E_{cell}\) to the standard cell potential \(E^\circ_{cell}\) and the reaction quotient \(Q\), which is dependent on the concentrations of the reacting species: \(E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q\). For this cell, \(Q\) is the ratio of concentrations \(\frac{C_2}{C_1}\).
04

Determine the Condition for a Positive Cell Potential

Since \(\Delta G\) is negative for a spontaneous reaction, \(E_{cell}\) must be positive. For \(E_{cell}\) to be positive, according to the Nernst equation, the log of \(Q (\frac{C_2}{C_1})\) must be positive. This occurs when the numerator of \(Q\), which is \(C_2\), is greater than the denominator, \(C_1\).

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, denoted as \(\Delta G\), is a thermodynamic quantity that measures the maximum amount of reversible work that can be performed by a system at constant temperature and pressure. It is a fundamental predictor of the spontaneity of a process. In the context of an electrochemical cell, a negative value of \(\Delta G\) indicates that a reaction can occur spontaneously. This aligns with the principle that nature favors processes which result in a decrease in free energy.

In an electrochemical cell, the change in Gibbs free energy is related to the electrical work that can be obtained from the chemical reaction occurring within the cell. The equation \(\Delta G = -nFE_{cell}\) connects \(\Delta G\) to the cell potential \(E_{cell}\), where \(n\) represents the number of moles of electrons exchanged in the reaction, and \(F\) is the Faraday constant, approximately 96,485 coulombs per mole of electrons. When \(\Delta G\) is negative, it implies that the cell potential is positive, and thus, the reaction can deliver electrical work.
Cell Potential
Cell potential, also known as electromotive force (EMF), represented as \(E_{cell}\), is a measure of how much voltage a cell can produce during a chemical reaction. It is the driving force that pushes electrons through an external circuit, creating electricity. The greater the cell potential, the more energy is available for electrical work.

The standardized conditions under which cell potentials are measured result in the standard cell potential, \(E^\theta_{cell}\). However, in practical scenarios, the actual cell potential will vary based on the concentrations of the reactants and products, as well as the temperature. The cell potential is always measured in volts (V). When dealing with a galvanic cell, like the one in our exercise with cobalt electrodes, a positive cell potential corresponds to a spontaneous reaction.
Nernst Equation
The Nernst equation, a central concept in electrochemistry, allows for the calculation of a cell's potential under non-standard conditions. The equation is expressed as \(E_{cell} = E^\theta_{cell} - \frac{0.0592}{n} \log Q\), where \(n\) is the number of moles of electrons transferred in the electrochemical reaction, \(Q\) is the reaction quotient, and \(E^\theta_{cell}\) is the standard cell potential.

The reaction quotient, \(Q\), reflects the ratio of the concentrations of the reaction products to the reactants raised to the power of their stoichiometric coefficients. In the context of our cobalt cell, \(Q\) is \(\frac{C_2}{C_1}\) as \(C_2\) and \(C_1\) represent the concentrations of cobalt ions at the cathode and anode respectively. The Nernst equation shows how the cell potential changes with the concentration of the electrolytes involved in the cell's reactions.
Electrochemical Cell
An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. There are two main types of electrochemical cells: galvanic cells, which convert chemical energy into electrical energy, and electrolytic cells, which do the opposite.

In galvanic cells, also called voltaic cells, spontaneous redox reactions occur that lead to the flow of electrons from the anode to the cathode through an external circuit, generating current. The anode is the electrode where oxidation takes place, while reduction occurs at the cathode. The exercise provided deals with a galvanic cell involving cobalt electrodes and ions at different concentrations, harnessing a spontaneous chemical reaction to produce electricity, characterized by a negative Gibbs free energy (\(\Delta G\)) and a positive cell potential (\(E_{cell}\)).
  • The anode is the site of oxidation: \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2e^-\).
  • The cathode is the site of reduction: \(\mathrm{Co}^{2+} + 2e^- \rightarrow \mathrm{Co}\).
  • The salt bridge allows for the flow of ions to maintain electrical neutrality within the internal circuit.

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Most popular questions from this chapter

Consider the following standard electrode potentials and calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the indicated disproportionation reaction : $$ \begin{aligned} 3 \mathrm{Mn}^{2+}(a q) & \longrightarrow \mathrm{Mn}(s)+2 \mathrm{Mn}^{3+}(a q) \\ \mathrm{Mn}^{3+}(a q)+e^{-} & \longrightarrow \mathrm{Mn}^{2+}(a q) ; \quad E^{\circ}=1.51 \mathrm{~V} \\ \mathrm{Mn}^{2+}(a q)+2 e^{\circ} \longrightarrow \mathrm{Mn}(s) ; & E^{\circ}=-1.185 \mathrm{~V} \end{aligned} $$ (a) \(1.2 \times 10^{-43}\) (b) \(2.4 \times 10^{-73}\) (c) \(6.3 \times 10^{-92}\) (d) \(1.5 \times 10^{-62}\)

The \(E^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction at the indicated concentrations is \(1.50 \mathrm{~V}\). Calculate the \(\Delta G\) in \(\mathrm{kJ}\) at \(25^{\circ} \mathrm{C}\) : $$ \operatorname{Cr}(s)+3 \mathrm{Ag}^{+}(a q, 0.1 \mathrm{M}) \longrightarrow \mathrm{Ag}(s)+\mathrm{Cr}^{3+}(a q, 0.1 \mathrm{M}) $$ (a) \(-140.94\) (b) \(-295\) (c) \(-212\) (d) \(-422.83\)

Equivalent conductivity of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is related to molar conductivity by the expression : (a) \(\Lambda_{\mathrm{eq}}=\Lambda_{m}\) (b) \(\Lambda_{\text {eq }}=\Lambda_{m} / 3\) (c) \(\Lambda_{\text {eq }}=3 \Lambda_{m}\) (d) \(\Lambda_{\text {eq }}=\Lambda_{m} / 6\)

Molar conductivity of a solution of an electrolyte \(A B_{3}\) is \(150 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}\). If it ionises as \(\mathrm{AB}_{3} \longrightarrow A^{3+}+3 B^{-}\), its equivalent conductivity will be : (a) \(150\left(\right.\) in \(\mathrm{Scm}^{2} \mathrm{eq}^{-1}\) ) (b) 75 (in \(\mathrm{Scm}^{2} \mathrm{eq}^{-1}\) ) (c) 50 (in \(\mathrm{Scm}^{2} \mathrm{eq}^{-1}\) ) (d) 80 (in \(\mathrm{Scm}^{2} \mathrm{eq}^{-1}\) )

A \(100.0 \mathrm{~mL}\) dilute solution of \(\mathrm{Ag}^{+}\) is electrolyzed for \(15.0\) minutes with a current of \(1.25 \mathrm{~mA}\) and the silver is removed completely. What was the initial \(\left[\mathrm{Ag}^{+}\right]\) ? (a) \(2.32 \times 10^{-1}\) (b) \(2.32 \times 10^{-4}\) (c) \(2.32 \times 10^{-3}\) (d) \(1.16 \times 10^{-4}\)

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