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\(E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) for the reaction, \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\) \(\operatorname{sn}^{4+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) (a) Can a solution be prepared that is \(0.500 \mathrm{M}\) in each of the four ions at \(298 \mathrm{K} ?\) (b) If not, in what direction must a net reaction Occur?

Short Answer

Expert verified
No, a solution cannot be prepared that is 0.500M in each of the four ions at 298K because \(E_{\mathrm{cell}}\) is negative. Therefore, the net reaction must occur in the backward direction.

Step by step solution

01

Nernst Equation Formulation

First, express the Nernst equation, which provides a relation between the cell potential and the concentrations of the species involved in the cell reaction. It follows from the Second Law of thermodynamics. The formula is: \[E_{\mathrm{cell}}= E_{\mathrm{cell}}^{\circ} - \frac {0.0592}{\text{n}} \log\frac{[\text{Products}]^{\text{Stoichiometric coefficient}}}{[\text{Reactants}]^{\text{Stoichiometric coefficient}}}\] Here, \[E_{\mathrm{cell}}^{\circ}\] is the standard cell potential, \[n\] is the number of moles of electrons transferred in the reaction (for this equation, \(n = 2\)), and \[\text{Products}\] and \[\text{Reactants}\] denote the concentration of products and reactants, respectively.
02

Substitute Given Values

Substitute the values related to the exercise into the equation formulated in step 1. For this case, these values are: \(E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\), concentrations are all \(0.500 \mathrm{M}\), and since Cu and Sn transformation involve loss and gain of two electrons respectively, \(n = 2\). The modified Nernst equation for this problem is: \[E_{\mathrm{cell}}= -0.0050 - \frac {0.0592}{2} \log\left(\frac{(0.500)^2}{(0.500)^2}\right)\]
03

Simplify and Solve for E-cell

Simplify the expression and solve for \(E_{\mathrm{cell}}\). Since the logarithm of 1 is zero, the equation simplifies to: \[E_{\mathrm{cell}}= -0.0050 - 0\] Therefore, \(E_{\mathrm{cell}}= -0.0050 \mathrm{V}\]
04

Determine the Direction of the Reaction

Examine the sign of E-cell to determine the direction of the reaction. Here, since \(E_{\mathrm{cell}}\) is negative, it implies that the reaction cannot go in forward direction, but instead has to proceed in backward direction or to the left.

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Key Concepts

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

Electrochemistry
Electrochemistry is the branch of chemistry that explores the relationship between electricity and chemical reactions. It involves the study of reactions in which charged particles called ions participate, often generating or consuming electrical current. These chemical reactions can occur spontaneously, driving electrical work, or can be driven by an external electic source. A key area of focus in electrochemistry is how these reactions can convert chemical energy into electrical energy and vice versa.

Electrochemical cells, like voltaic cells and electrolytic cells, are the heart of electrochemistry. In a typical electrochemical cell, there are two electrodes: the anode (where oxidation occurs) and the cathode (where reduction occurs). These electrodes are immersed in electrolytes containing ions that facilitate the redox reactions. The flow of ions and electrons initiates the generation of current.

Understanding electrochemistry is not only vital for scientific purposes but also for practical applications. It provides insights into how batteries function, aids in metal plating and corrosion prevention, and even supports new advancements in energy storage technologies.
Standard Cell Potential
The standard cell potential, denoted as \(E_{\mathrm{cell}}^{\circ}\), is a critical concept in electrochemistry. It represents the electrical potential difference between two half-cells in an electrochemical cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C). This potential difference provides a measure of the cell's ability to drive an electrochemical reaction.Standard cell potential is determined using standard reduction potentials, which are established under standard conditions. These potentials allow chemists to predict the feasibility and direction of a redox reaction by comparing the electron affinities of different elements and ions. The equation to find the standard cell potential is:\[E_{\mathrm{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\]Where \(E_{\text{cathode}}^{\circ}\) and \(E_{\text{anode}}^{\circ}\) are the standard reduction potentials for the cathode and anode, respectively. If the value of \(E_{\mathrm{cell}}^{\circ}\) is positive, the reaction tends to occur spontaneously as written. When it is negative, like in our original exercise, it suggests that the reaction may not proceed forward spontaneously under standard conditions.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are essential processes in electrochemistry. They involve the transfer of electrons between two substances. One substance undergoes oxidation (loss of electrons), while the other undergoes reduction (gain of electrons). Together, these reactions release or require energy, often in the form of electricity. To balance redox reactions, it is important to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This balance of charge is fundamental to maintaining electric neutrality in the reactions, and it is crucial for the proper functioning of electrochemical cells. In the Nernst Equation problem provided, the redox reaction involves Cu and Sn ions where Sn^4+ is reduced to Sn^2+, and Cu^+ is oxidized to Cu^2+. By understanding the specific roles of oxidation and reduction, it becomes clearer how the electrons flow through the system driving the redox reaction. In practical applications, balanced redox reactions are foundational for technologies such as batteries, fuel cells, and various industrial processes where reduction and oxidation are leveraged to produce energy or synthesize materials.

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Most popular questions from this chapter

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

A test for completeness of electrodeposition of \(\mathrm{Cu}\) from a solution of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is to add \(\mathrm{NH}_{3}(\mathrm{aq}) .\) A blue color signifies the formation of the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\left(K_{\mathrm{f}}=1.1 \times 10^{13}\right) .\) Let \(250.0 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{CuSO}_{4}(\text { aq })\) be electrolyzed with a \(3.512 \mathrm{A}\) current for 1368 s. At this time, add a sufficient quantity of \(\mathrm{NH}_{3}(\text { aq })\) to complex any remaining \(\mathrm{Cu}^{2+}\) and to maintain a free \(\left[\mathrm{NH}_{3}\right]=0.10 \mathrm{M} .\) If \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is detectable at concentrations as low as \(1 \times 10^{-5} \mathrm{M}\) should the blue color appear?

Construct a concept map illustrating the principles of electrolysis and its industrial applications.

If a lead storage battery is charged at too high a voltage, gases are produced at each electrode. (It is possible to recharge a lead-storage battery only because of the high overpotential for gas formation on the electrodes.) (a) What are these gases? (b) Write a cell reaction to describe their formation.

The following voltaic cell registers an \(E_{\text {cell }}=0.108 \mathrm{V}\) What is the pH of the unknown solution? $$\operatorname{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(x \mathrm{M}) \| \mathrm{H}^{+}(1.00 \mathrm{M}) |$$ $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) | \mathrm{Pt}$$

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