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Which of the following reactions occur spontaneously, and which can be brought about only through electrolysis, assuming that all reactants and products are in their standard states? For those requiring electrolysis, what is the minimum voltage required? (a) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\left[\text { in } 1 \mathrm{M} \mathrm{H}^{+}(\mathrm{aq})\right]\) (b) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})\) (c) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{s}) \longrightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})\) (d) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\)

Short Answer

Expert verified
Please note this is a generic procedure, the actual results of the calculations depend on the provided standard reduction potentials (E°) for the reactants and products involved in each reaction.

Step by step solution

01

Setup the Standard Reduction Potentials (E°)

For this problem, the standard reduction potentials (E°) are provided for reactants and products involved in each of the four reactions: (a) \(H_2O(l) \to 2 H_2(g) + O_2(g)\), (b) \(Zn(s) + Fe^{2+}(aq) \to Zn^{2+}(aq) + Fe(s)\), (c) \(2Fe^{2+}(aq) + I_2(s) \to 2 Fe^{3+}(aq) +2 I^-(aq)\), and (d) \(Cu(s) + Sn^{4+}(aq) \to Cu^{2+}(aq) + Sn^{2+}(aq)\).
02

Calculation of Cell Potential (E°)

Calculate ΔE° for each reaction using the formula: ΔE° = E°(cathode) - E°(anode). Here cathode refers to the reduction half-reaction and anode refers to the oxidation half-reaction.
03

Decide on Spontaneity or Non-Spontaneity

If the calculated ΔE° is positive, the reaction is spontaneous. If ΔE° is negative, the reaction is non-spontaneous and requires electrolysis for completion, and the absolute value is the minimum voltage required for electrolysis.
04

Repeat for each Chemical Reaction

Repeat steps 2 and 3 for reactions (a), (b), (c), and (d) to determine the spontaneity and for non-spontaneous reactions the voltage required for electrolysis.

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Key Concepts

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

Electrolysis
Electrolysis is a fascinating process where electrical energy is used to drive a chemical reaction that wouldn't occur naturally. This is especially useful for reactions that are non-spontaneous. To understand whether a reaction requires electrolysis, we look at its cell potential.

- If the cell potential is negative, this means the reaction isn't spontaneous under normal conditions. - Therefore, it would require an external source of electricity to proceed, hence, electrolysis is needed. - Think of electrolysis as using electricity to "push" a reaction forward—similar to how charging a battery works.

It's often used in various applications such as metal plating, purification, or decomposition of compounds. A classic example is splitting water into hydrogen and oxygen gases, which requires an external voltage. Understanding electrolysis helps in recognizing how we can control and direct chemical reactions using electric energy.
Standard Reduction Potential
Standard reduction potential, denoted as E°, is a way to measure the tendency of a chemical species to gain electrons and be reduced. These potentials are measured under standard conditions, which means concentrations of 1 M, pressures of 1 atm, and a standard temperature of 25°C (298 K).

- Each half-reaction has its own E°, and they are typically given in tables for quick reference. - These values help predict the direction of electron flow in an electrochemical cell. - The more positive the E° value, the greater the species' ability to gain electrons and be reduced.

For example, in a spontaneous reaction, the species with a higher standard reduction potential serves as the cathode (where reduction occurs), while the one with a lower potential is the anode (where oxidation occurs).

Understanding standard reduction potential is crucial for assessing the likelihood of a reaction occurring spontaneously and plays a vital role in electrochemistry.
Cell Potential
Cell potential, or electromotive force (EMF), is symbolized as E° and indicates the voltage difference between two half-cells in an electrochemical cell.

- It's calculated using the formula: \[ \Delta E^\circ = E^\circ(\text{cathode}) - E^\circ(\text{anode}) \]- A positive cell potential means a reaction is spontaneous, whereas a negative value signifies a non-spontaneous reaction.

In simple terms, cell potential tells us if a reaction can "just happen" or if it needs a bit of a push (like electrolysis) to get going. It's a helpful tool for predicting the behavior of chemical reactions and understanding the energy requirements needed to achieve a desired chemical change.

Evaluating cell potential helps not only in academic exercises but also in designing batteries and other devices involving redox reactions. Through it, we see the intersection of chemistry and energy, illustrating how they work together in both natural and industrial processes.

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Most popular questions from this chapter

The quantity of electric charge that will deposit \(4.5 \mathrm{g}\) Al at a cathode will also produce the following volume at STP of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}^{+}(\) aq) at a cathode: (a) \(44.8 \mathrm{L} ;\) (b) \(22.4 \mathrm{L} ;\) (c) \(11.2 \mathrm{L} ;\) (d) \(5.6 \mathrm{L}\).

Describe how you might construct batteries with each of the following voltages: (a) \(0.10 \mathrm{V} ;\) (b) \(2.5 \mathrm{V} ;\) (c) \(10.0 \mathrm{V}\). Be as specific as you can about the electrodes and solution concentrations you would use, and indicate whether the battery would consist of a single cell or two or more cells connected in series.

Natural gas transmission pipes are sometimes protected against corrosion by the maintenance of a small potential difference between the pipe and an inert electrode buried in the ground. Describe how the method works.

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.

Your task is to determine \(E^{\circ}\) for the reduction of \(\mathrm{CO}_{2}(\mathrm{g})\) to \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) in two different ways and to explain why each gives the same result. (a) Consider a fuel cell in which the cell reaction corresponds to the complete combustion of propane gas. Write the half-cell reactions and the overall reaction. Determine \(\Delta G^{\circ}\) and \(E_{\text {cell }}^{\circ}\) for the reaction, then obtain \(E_{\mathrm{CO}_{2} / \mathrm{C}_{3} \mathrm{H}_{8}^{*}}^{\circ}\) (b) Without considering the oxidation that occurs simultaneously, obtain \(E_{\mathrm{CO}_{2} / \mathrm{C}_{3} \mathrm{H}_{8}}^{\circ}\) directly from tabulated thermodynamic data for the reduction half-reaction.

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