Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) \(2 \mathrm{Ce}^{++}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Ce}^{3+}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq})\) \(E_{\mathrm{cell}^{\circ}}=+1.08 \mathrm{~V}\) (b) \(6 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 6 \mathrm{Fe}^{2+}(\mathrm{aq})+\) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq}), E_{\text {cell }}{ }^{\circ}=-1.29 \mathrm{~V}\)

Short Answer

Expert verified
\( \Delta G^{\circ}_{(a)} = -207.648 \text{ kJ/mol}; \Delta G^{\circ}_{(b)} = 747.741 \text{ kJ/mol} \)

Step by step solution

01

Determine the number of moles of electrons exchanged in reaction (a)

For the given reaction (a), the cerium ions (Ce) are changing from a +2 to a +3 oxidation state, meaning each Ce atom is losing 1 electron. Since the reaction involves 2 moles of Ce, a total of 2 moles of electrons are transferred.
02

Calculate the standard reaction Gibbs free energy for reaction (a)

Use the formula \( \Delta G^{\circ} = -nFE^{\circ} \), where \( \Delta G^{\circ} \) is the standard Gibbs free energy change, \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant (approximately 96485 C/mol), and \( E^{\circ} \) is the standard cell potential. Plug in the values for the reaction (a) to get \( \Delta G^{\circ} = -(2 \text{ mol})(96485 \text{ C/mol})(1.08 \text{ V}) \).
03

Determine the number of moles of electrons exchanged in reaction (b)

For reaction (b), each chromium ion (Cr) changes from a +3 to a +6 oxidation state in the dichromate ion (\(\mathrm{Cr}_2\mathrm{O}_7^{2-}\)), resulting in the transfer of 6 moles of electrons for 2 Cr ions. Therefore, the reaction involves 6 moles of electrons being transferred.
04

Calculate the standard reaction Gibbs free energy for reaction (b)

Using the formula \( \Delta G^{\circ} = -nFE^{\circ} \) as before and plugging in the values for reaction (b), we find \( \Delta G^{\circ} = -(6 \text{ mol})(96485 \text{ C/mol})(-1.29 \text{ V}) \).

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Key Concepts

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

Electrochemistry
Electrochemistry is a branch of chemistry that deals with the interrelation of electrical currents and chemical reactions. It involves the study of the movement of electrons from one substance to another, a process fundamental to many types of electrochemical cells.

Electrochemical reactions can be used for a variety of applications, including energy storage in batteries, electroplating, and the industrial production of chemicals like chlorine and sodium hydroxide. The transfer of electrons during these reactions leads to changes in the oxidation state of the atoms and ions involved, which is key to understanding how electrochemical cells operate.
Galvanic Cells
Galvanic cells, also known as voltaic cells, are a type of electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. They consist of two half-cells, each containing an electrode in contact with an electrolyte.

Each half-cell has its own separate redox reaction, with the overall cell reaction being the sum of these half-reactions. Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction takes place. The potential difference between the two electrodes can be measured and is a source of electric power. Galvanic cells are the basis for common batteries.
Standard Cell Potential
The standard cell potential (\( E^{\text{\textdegree}} \)) is a crucial value in electrochemistry representing the potential difference between two electrodes of a galvanic cell when all components are at their standard states, usually 1 M concentration for solutes and 1 atm pressure for gases at 25°C. It is measured in volts (V).

The standard cell potential provides insight into the cell's ability to perform work when electrons transfer through the external circuit. A positive standard cell potential indicates that a reaction is spontaneous, while a negative value suggests a non-spontaneous reaction under standard conditions.
Gibbs Free Energy Equation
The Gibbs free energy equation connects thermodynamics and electrochemistry, allowing us to predict the spontaneity of a reaction. Given by the formula \( \Delta G^{\text{\textdegree}} = -nFE^{\text{\textdegree}} \), where \( \Delta G^{\text{\textdegree}} \) is the standard Gibbs free energy change, \( n \) is the number of moles of electrons transferred in the reaction, \( F \) is the Faraday constant (approximately 96485 C/mol), and \( E^{\text{\textdegree}} \) is the standard cell potential.

A negative value of \( \Delta G^{\text{\textdegree}} \) implies that the reaction is spontaneous under standard conditions. This fundamental equation bridges the gap between the electrochemical behavior of a cell and its thermodynamic properties and is key to solving problems related to galvanic cells and their energy changes.

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Most popular questions from this chapter

(a) What is the approximate chemical formula of rust? (b) What is the oxidizing agent in the formation of rust? (c) How does the presence of salt accelerate the rusting process?

When a pH meter was standardized with a boric acid-borate buffer with a pH of \(9.40\), the cell was \(+0.060 \mathrm{~V}\). When the buffer was replaced with a solution of unknown hydronium ion concentration, the cell potential was \(+0.22 \mathrm{~V}\). What is the \(\mathrm{pH}\) of the solution?

(a) Why are lead-antimony grids used as electrodes in the lead-acid battery rather than smooth plates? (b) What is the reducing agent in the lead-acid battery? (c) The lead-acid cell potential is about \(2 \mathrm{~V}\). How, then, does a car battery produce \(12 \mathrm{~V}\) for its electrical system?

The absolute magnitudes of the standard potentials of two metals \(\mathrm{M}\) and \(\mathrm{X}\) were determined to be (1) \(\mathrm{M}^{+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{M}(\mathrm{s}) \quad\left|E^{\circ}\right|=0.25 \mathrm{~V}\) (2) \(\mathrm{X}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{X}(\mathrm{s}) \quad\left|E^{\circ}\right|=0.65 \mathrm{~V}\) When the two electrodes are connected, current flows from \(M\) to \(\mathrm{X}\) in the external circuit. When the electrode corresponding to halfreaction 1 is connected to the standard hydrogen electrode (SHE), current flows from \(M\) to SHE. (a) What are the signs of \(E^{\circ}\) of the two half-reactions? (b) What is the standard cell potential for the cell constructed from these two electrodes?

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells: (a) \(\mathrm{Ni}(\mathrm{s})\left|\mathrm{Ni}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}(\mathrm{s})\) (b) \(\mathrm{C}(\mathrm{gr})\left|\mathrm{H}_{2}(\mathrm{~g}) / \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Cl}_{2}(\mathrm{~g}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ce}^{4+}(\mathrm{aq}), \mathrm{Ce}^{3+}(\mathrm{aq})\right| \operatorname{Pt}(\mathrm{s})\) (d) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{OH}^{-}(\mathrm{aq})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{Pt}(\mathrm{s})\) (c) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Hg}_{2} \mathrm{Cl}{ }_{2}\) (s) \(\mid \mathrm{Hg}(\mathrm{l})\)

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