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Calculate \(\Delta E^{\circ}\) for the following cells: (1) Cadmium and Hydrogen (2) Silver and Hydrogen (3) Cadmium and silver, using the following data: \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|} { Reaction } & \multicolumn{1}{|c|} {\(\mathrm{E}^{\circ}\) volts } \\ \hline \(\mathrm{Cd} \rightarrow \mathrm{Cd}^{+2}+2 \mathrm{e}^{-}\) & \(+.403\) \\\ \hline \(\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) & \(0.00\) \\\ \hline \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}\) & \(-.799\) \\ \hline \end{tabular}

Short Answer

Expert verified
The calculated standard cell potentials for the three cells are: (1) Cadmium and Hydrogen: \(\Delta E_{\text{Cd/H}}^{\circ} = -0.403 \, \text{volts}\) (2) Silver and Hydrogen: \(\Delta E_{\text{Ag/H}}^{\circ} = 0.799 \, \text{volts}\) (3) Cadmium and Silver: \(\Delta E_{\text{Cd/Ag}}^{\circ} = -1.202 \, \text{volts}\)

Step by step solution

01

Identify the Half-Reactions for each Cell

The first step is to identify the half-reactions for each of the following cells: 1. Cadmium and Hydrogen 2. Silver and Hydrogen 3. Cadmium and Silver
02

Determine the Cathode and Anode for each Cell

Determine the cathode and anode for each of the given cells. Remember, the reaction with the larger standard reduction potential occurs at the cathode, while the one with the lower standard reduction potential happens at the anode.
03

Calculate \(\Delta E^{\circ}\) for each Cell

Calculate the standard cell potential, \(\Delta E^{\circ}\), using the equation: \(\Delta E^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\) Now, calculate \(\Delta E^{\circ}\) for each cell: 1. For the cell with cadmium and hydrogen: \(\Delta E_{\text{Cd/H}}^{\circ} = E_{\text{H}}^{\circ} - E_{\text{Cd}}^{\circ} = 0.00 - 0.403 = -0.403 \, \text{volts}\) 2. For the cell with silver and hydrogen: \(\Delta E_{\text{Ag/H}}^{\circ} = E_{\text{H}}^{\circ} - E_{\text{Ag}}^{\circ} = 0.00 - (-0.799) = 0.799 \, \text{volts}\) 3. For the cell with cadmium and silver: \(\Delta E_{\text{Cd/Ag}}^{\circ} = E_{\text{Ag}}^{\circ} - E_{\text{Cd}}^{\circ} = -0.799 - 0.403 = -1.202 \, \text{volts}\) The signs of the \(\Delta E^{\circ}\) values (negative or positive) indicate which reaction is spontaneous and which is not, based on the principle that a spontaneous reaction has a positive \(\Delta E^{\circ}\).

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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 relationship between electricity and chemical change. This fascinating field is at the crossroads of physics and chemistry, investigating how electric currents can facilitate chemical reactions and, conversely, how chemical reactions can produce electrical energy. One of the most common applications of electrochemistry is in batteries, where chemical reactions produce electric currents that can power all sorts of devices.

At the heart of electrochemistry is the transfer of electrons from one substance to another, known as redox reactions. These reactions can be split into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained. By harnessing these reactions, electrochemistry provides the basis for various technologies from generating electricity to electroplating metals.
Galvanic Cells
Galvanic cells, also known as voltaic cells, are the powerhouses of chemistry that convert chemical energy into electrical energy through spontaneous redox reactions. These cells consist of two electrodes made of different materials (typically metals or metal compounds) submerged in solutions called electrolytes.

The two electrodes, the anode, and the cathode, have different propensities to lose or gain electrons. The anode undergoes oxidation (loss of electrons), while the cathode sees a reduction (gain of electrons). This difference in reduction potential between the two electrodes leads to the flow of electrons through an external circuit, creating electric current, while ions in the electrolytes move between the cells to balance the charge. The ability to harness this flow of electrons is pivotal for powering electronic devices and has extensive implications in manufacturing, metal recovery, and energy storage technologies.
Reduction Potential
Reduction potential, denoted as \( E^{\text{\textdegree}} \), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. In electrochemistry, it's a critical concept because it determines how different substances will behave at an electrode during a redox reaction.

Each half-reaction in a redox process has its own standard reduction potential, and these values are referenced to a standard hydrogen electrode (SHE), which is arbitrarily given a potential of 0.00 volts. The sign of the reduction potential is meaningful; a positive reduction potential indicates a greater tendency to gain electrons, while a negative value indicates a lesser tendency. By comparing the potential of various half-reactions, we can predict which way the electrons will flow and which substances will be oxidized or reduced in a galvanic cell.
Spontaneous Reactions
Spontaneous reactions are processes that occur naturally without any input of external energy. In electrochemistry, these reactions can be predicted by evaluating the standard cell potential, \( \Delta E^{\circ} \), calculated as the difference between the potentials of the cathode and anode.

A positive \( \Delta E^{\circ} \) indicates a spontaneous reaction, meaning the reaction can proceed on its own, generating an electric current as it goes. This is the principle behind batteries and other galvanic cells — they produce a flow of electrons, hence electric current, as long as the chemical reactions within them are spontaneous. On the other hand, a negative \( \Delta E^{\circ} \) suggests a non-spontaneous reaction, which will not occur under standard conditions without external input, such as electricity applied through an external circuit (electrolysis). In the context of galvanic cells, understanding spontaneous reactions is crucial for the design of efficient and functional electrochemical cells. This underpins the very operation of power sources from small-scale electronics to large industrial power systems.

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Most popular questions from this chapter

Two electrochemical cells were fitted with inert platinum electrodes and connected in series. The first cell contained a solution of a metal nitrate; the second cell, metal sulfate. After operation of the cell for a prescribed period of time, it was observed that \(675 \mathrm{mg}\) of the metal had been deposited in the first cell and \(76.3 \mathrm{ml}\) of hydrogen gas \(\left(25^{\circ}\right.\) and \(1 \mathrm{~atm}\) ) evolved in the second. Determine the equivalent weight of the metal.

Knowing that the \(\mathrm{K}_{3 p}\) for AgCl is \(1.8 \times 10^{-10}\), calculate \(\mathrm{E}\), the electrode potential, for a silver-silver chloride electrode immersed in \(1 \mathrm{M} \mathrm{KCI}\). The standard oxidation potential for the \(\left(\mathrm{Ag}, \mathrm{Ag}^{+}\right)\) half reaction is \(-0.799\) volts.

A voltaic cell is made up of a silver electrode in a \(1 \mathrm{M}\) silver nitrate solution and an aluminum electrode in a \(1 \mathrm{M}\) aluminum nitrate solution. The half reactions are 1) \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\) (s) \(\mathrm{E}^{\circ}=.80\) volt, and (2) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s})\) \(E^{\circ}=-1.66\) volt, calculate the standard free energy change, \(\Delta \mathrm{G}^{\circ}\), in \(\mathrm{Kcal} /\) mole. \(1 \mathrm{cal}=4.184\) joule \(1 \mathrm{~F}=96,500\) coul, \(1 \mathrm{Kcal}=1000 \mathrm{cal}\)

You have the following cell process: \(\mathrm{Fe}(\mathrm{s})+\mathrm{Co}^{2+}(.5 \mathrm{M}) \rightarrow \mathrm{Fe}^{2+}(1.0 \mathrm{M})+\mathrm{Co}(\mathrm{s})\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \leftrightarrows \mathrm{Fe}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.44 \mathrm{e}\) and \(\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \leftrightharpoons \mathrm{Co}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.28\), find the standard cell potential \(\Delta \mathrm{E}\), the cell potential \(\Delta \mathrm{E}\) and the concentration ratio at which the potential generated by the cell is exactly zero. which the potential generated by the cell is exactly zero.

Using the tables, of standard electrode potentials, list the fol lowing ions in order of decreasing ability as oxidizing agents: \(\mathrm{Fe}^{3+}, \mathrm{F}_{2}, \mathrm{~Pb}^{2+}, \mathrm{I}_{2}, \mathrm{Sn}^{4+}, \mathrm{O}_{2}\)

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