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Chapter 18: Problem 114

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

Short Answer

Expert verified
The unknown metal M used to construct the standard galvanic cell is Copper (Cu). This is determined by calculating the standard electrode potential of the M虏鈦�/M half-cell using the given 螖G掳 value (-411 kJ) and the relationship \(\Delta G^\circ = -nFE^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant. The calculated standard electrode potential of 1.125 V is closest to the value for the Cu虏鈦�/Cu half-cell, indicating that the unknown metal is likely Copper.

Step by step solution

01

Calculate the number of moles of electrons transferred in the overall cell reaction

In the given overall cell reaction, we have 2 moles of Al鲁鈦� ions and 3 moles of metal ion M targeting 2 moles of Al atoms and 3 moles of M虏鈦� ions. We can balance the charges on both sides to find the number of electrons transferred. The charges on the reactants side: \( 2 \times 3 = 6 \) (from 2 moles of Al鲁鈦�) The charges on the products side: \( 3 \times 2 = 6 \) (from 3 moles of M虏鈦�) Both sides have a total charge of 6, so 6 moles of electrons are transferred in the overall cell reaction.
02

Use the relationship between 螖G掳 and the standard electrode potential

To find the standard electrode potential, E掳, of the M虏鈦�/M half-cell, we can use the following relationship that connects 螖G掳 with E掳 and the number of moles of electrons transferred (n) in the overall cell reaction: \( \Delta G^\circ = -nFE^\circ \) where F is the Faraday constant (approximately 96485 C/mol of electrons). Using the given 螖G掳 value of -411 kJ, convert it to J (since F is in C/mol): 螖G掳 = -411 kJ 脳 1000 J/kJ = -411,000 J Now, we can plug in the number of moles of electrons (n) and the Faraday constant (F) to find the standard electrode potential (E掳): \( E^\circ = \frac{-\Delta G^\circ}{nF} \)
03

Calculate the standard electrode potential of the M虏鈦�/M half-cell

Insert the values we found for 螖G掳, n, and F into the equation above: \( E^\circ = \frac{-(-411,000 \text{ J})}{6 \times 96485 \text{ C/mol}} \) \( E^\circ = \frac{411,000 \text{ J}}{6 \times 96485 \text{ C/mol}} = 1.125 \text{ V} \) Thus, the standard electrode potential of the M虏鈦�/M half-cell is 1.125 V.
04

Identify the unknown metal

Now that we have the standard electrode potential of the M虏鈦�/M half-cell, we can compare it to the known redox potentials of various metals to identify M. Comparing our calculated value of 1.125 V to the standard electrode potentials of metals, we find that it is closest to the value for the Cu虏鈦�/Cu half-cell, which has a standard electrode potential of 1.118 V. So, it is very likely that the unknown metal M used to construct the standard galvanic cell is Copper (Cu).

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Key Concepts

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

Standard Electrode Potential
The standard electrode potential, denoted as \(E^\circ\), is a key concept in electrochemistry. It refers to the measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Standard electrode potentials are measured in volts and are determined under standard conditions: a concentration of 1 M for any aqueous species, a pressure of 1 bar for gases, and a temperature around 25掳C.
These potentials are crucial because they help predict the direction of redox reactions. Specifically, a more positive \(E^\circ\) indicates a greater tendency to gain electrons. By calculating and comparing the \(E^\circ\) of different substances, we can determine which will act as the oxidizing or reducing agent in a galvanic cell, which is a device that converts chemical energy into electrical energy through redox reactions.
Faraday Constant
The Faraday constant, represented as \(F\), is the amount of electric charge per mole of electrons. Its value is approximately 96,485 C/mol of electrons.
This constant plays a pivotal role in electrochemical calculations. It serves as a bridge between the electrical and chemical realms by linking the amount of substance that undergoes a redox reaction to the electric charge involved in the reaction.
  • In the relation \( \Delta G^\circ = -nFE^\circ \), \(F\) is used alongside the number of moles of electrons \(n\) to calculate changes in Gibbs free energy and the standard electrode potential.
  • Moreover, the Faraday constant ensures that we consider the precise quantity of electric change that can drive specific redox reactions.
Redox Reaction
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two chemical species. They consist of two half-reactions:
  • A reduction process, where a chemical species gains electrons.
  • An oxidation process, where a species loses electrons.
Understanding the electron transfer in redox reactions is fundamental in galvanic cells, which derive electrical current from these reactions.
Each half-reaction has its own standard electrode potential \(E^\circ\), and the overall cell potential can be determined by combining the potentials of the two half-reactions. The galvanic cell then uses this electron flow to perform work outside the electrochemical cell, such as powering a device.
Gibbs Free Energy
Gibbs free energy, often symbolized as \(\Delta G\), provides insights into the spontaneity of a reaction. A negative \(\Delta G\) indicates that a reaction is spontaneous, meaning it can occur without the input of external energy. For galvanic cells, the relationship between Gibbs free energy and the cell potential is expressed as:
\( \Delta G^\circ = -nFE^\circ \).
Here,
  • \(\Delta G^\circ\) is the standard change in Gibbs energy.
  • \(n\) is the number of moles of electrons transferred in the reaction.
  • \(F\) is the Faraday constant.
A typical galvanic cell has a negative \(\Delta G\), demonstrating that the cell can generate electrical energy spontaneously. This link between Gibbs free energy and cell potential allows us to predict and control the energy available from electrochemical processes.

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Most popular questions from this chapter

Consider the standard galvanic cell based on the following half-reactions: $$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$$ $$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$$ The electrodes in this cell are \(A g(s)\) and \(C u(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\operatorname{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) . ]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: Ag' reacts with Cl- to form AgCl(s). ] d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V}$$

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is \(\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \qquad \mathscr{E}^{\circ}=1.10 \mathrm{V}\) For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow\) Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

The Ostwald process for the commercial production of nitric acid involves the following three steps: $$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$ a. Which reactions in the Ostwald process are oxidation鈥搑eduction reactions? b. Identify each oxidizing agent and reducing agent.
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