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Use the data in Appendix D to calculate the standard cell potential for each of the following reactions. Which reactions will occur spontaneously? (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq})\) (b) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Ba}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ba}(\mathrm{s})\) (c) \(3 \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Fe}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq})\) (d) \(\mathrm{Hg}(1)+\mathrm{HgCl}_{2}(\mathrm{aq}) \longrightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{s})\)

Short Answer

Expert verified
The standard cell potentials for each reaction were calculated and the spontaneous reactions were determined to be those reactions with positive standard cell potentials.

Step by step solution

01

Identify Half-Reactions

Using Appendix D, the standard reduction potentials documented, the half-reactions for each chemical reaction are identified. For example, for reaction (a), the half-reactions involved are: \[2 H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\] and \[-2 F^{-}(aq) \rightarrow F_{2}(g) + 2e^{-}\] Corresponding potentials are taken from Appendix D.
02

Calculate Cell Potentials

Using the potentials obtained, the standard cell potential (\(E^{0}\)) of each reaction is calculated by subtracting the potential of the oxidation half-reaction from the reduction half-reaction. Each reaction will give: (a) \(E_{cell}^{a}\), (b) \(E_{cell}^{b}\), (c) \(E_{cell}^{c}\), and (d) \(E_{cell}^{d}\).
03

Determine Spontaneity

The reactions that occur spontaneously are those with positive \(E^{0}\). Determine which of \(E_{cell}^{a}\), \(E_{cell}^{b}\), \(E_{cell}^{c}\), and \(E_{cell}^{d}\) are positive.

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Key Concepts

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

Electrochemical Cell
An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical changes. These cells consist of two half-cells connected by a salt bridge, with each half-cell containing an electrode and an electrolyte.

In our exercise, several electrochemical reactions are presented, and in order to understand how to calculate the standard cell potential, a student must first comprehend the basic components of these cells. Each reaction can be separated into two half-reactions: one representing oxidation (loss of electrons) and another representing reduction (gain of electrons).

Electrochemical cells are broadly classified into two types:
  • Galvanic (voltaic) cells - These are cells in which spontaneous chemical reactions produce electrical energy.
  • Electrolytic cells - Cells that consume electrical energy to drive non-spontaneous chemical reactions.
Understanding this distinction is pivotal for determining the spontaneity of reactions in electrochemical cells.
Reduction Potentials
The concept of reduction potentials is central to understanding electrochemistry. The reduction potential, often represented by the symbol \( E^{\circ} \), measures the tendency of a chemical species to gain electrons and thereby be reduced. Each half-reaction in an electrochemical cell has an associated standard reduction potential, which is typically measured under standard conditions (1 M concentration of all aqueous species, 1 atm pressure for gases, and 25°C temperature).

To calculate the cell's overall potential, we use the half-cell potentials listed in reference materials such as Appendix D from our exercise. The

Standard Cell Potential

\( E^{\circ}_{cell} \) of the entire electrochemical reaction can be found by taking the difference between the reduction potential of the cathode (where reduction occurs) and the anode (where oxidation occurs).

It's important to remember that the more positive the reduction potential, the greater the species' tendency to be reduced. During calculations, reversing the half-reaction for oxidation requires also changing the sign of the associated reduction potential. Students often forget this which can lead to incorrect results.
Spontaneous Reactions
A spontaneous reaction is one that occurs naturally without external intervention. In the context of electrochemical cells, the spontaneity of a reaction can be determined by the sign of the standard cell potential (\(E^{\circ}_{cell}\)). When the standard cell potential is positive, the reaction is spontaneous and can produce electrical energy, as is the case with galvanic cells.

The exercise we are looking at asks students to calculate the standard cell potential for several reactions to identify which ones are spontaneous. The calculation of standard cell potentials involves using values obtained from a standard reduction potential table and applying the Nernst equation if conditions are not standard.

Furthermore, it is worth noting that not only does a positive cell potential indicate spontaneity, but a negative cell potential suggests that energy must be supplied for the reaction to proceed, characteristic of electrolytic cells. Hence, understanding the standard cell potential and its relation to spontaneity is crucial for anyone studying electrochemistry. It is also essential for students to practice the calculation and conceptual understanding thoroughly to confidently apply these principles to solve problems.

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Most popular questions from this chapter

An aqueous solution of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is electrolyzed by means of Pt electrodes. (a) Which of the following gases should form at the anode: \(\mathrm{O}_{2}, \mathrm{H}_{2}, \mathrm{SO}_{2}, \mathrm{SO}_{3} ?\) Explain. (b) What product should form at the cathode? Explain. (c) What is the minimum voltage required? Why is the actual voltage needed likely to be higher than this value?

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.

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.

From the observations listed, estimate the value of \(E^{\circ}\) for the half- reaction \(\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}(\mathrm{s})\) (a) The metal M reacts with HNO \(_{3}(\text { aq })\), but not with \(\mathrm{HCl}(\mathrm{aq}) ; \mathrm{M}\) displaces \(\mathrm{Ag}^{+}(\mathrm{aq}),\) but not \(\mathrm{Cu}^{2+}(\mathrm{aq})\) (b) The metal \(M\) reacts with \(\mathrm{HCl}(\mathrm{aq}),\) producing \(\mathrm{H}_{2}(\mathrm{g}),\) but displaces neither \(\mathrm{Zn}^{2+}(\text { aq })\) nor \(\mathrm{Fe}^{2+}(\mathrm{aq})\).

The value of \(E_{\text {cell }}^{\circ}\) for the reaction \(\mathrm{Zn}(\mathrm{s})+\) \(\mathrm{Pb}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s})\) is \(0.66 \mathrm{V} .\) This means that for the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Pb}^{2+}(0.01 \mathrm{M})\) \(\rightarrow \mathrm{Zn}^{2+}(0.10 \mathrm{M})+\mathrm{Pb}(\mathrm{s}), E_{\text {cell }}\) equals \((\mathrm{a}) 0.72 \mathrm{V}\) (b) \(0.69 \mathrm{V} ;\) (c) \(0.66 \mathrm{V} ;\) (d) \(0.63 \mathrm{V}\)

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