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Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Sn}\) to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}\) (to form \(\mathrm{I}_{2}\) ), (c) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2},\) (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\) ).

Short Answer

Expert verified
In acidic solution under standard conditions, the spontaneity of the given reactions is as follows: (a) The oxidation of Sn to Sn虏鈦 by I鈧 (to form I鈦) is spontaneous with E掳 = 0.68 V. (b) The reduction of Ni虏鈦 to Ni by I鈦 (to form I鈧) is non-spontaneous with E掳 = -0.79 V. (c) The reduction of Ce鈦粹伜 to Ce鲁鈦 by H鈧侽鈧 is non-spontaneous with E掳 = -0.15 V. (d) The reduction of Cu虏鈦 to Cu by Sn虏鈦 (to form Sn鈦粹伜) is spontaneous with E掳 = 0.19 V.

Step by step solution

01

Determine the Standard Reduction Potentials for Each Half-Reaction

Look up the standard reduction potentials for each half-reaction in a table. These values are generally provided in textbooks or can be found online.
02

Calculate the E掳 value for the cell

For each reaction, plug in the values of standard reduction potentials (E掳) into the formula: \[ E掳_{cell} = E掳_{cathode} - E掳_{anode} \]
03

Determine the spontaneity of the reaction

Compare the calculated E掳 values with 0 to determine the spontaneity of each reaction: - If E掳 > 0, the reaction is spontaneous - If E掳 < 0, the reaction is non-spontaneous Using the standard reduction potentials and the above steps, let's predict the spontaneity of the given reactions in acidic solution under standard conditions: (a) Oxidation of Sn to Sn虏鈦 by I鈧 (to form I鈦): E掳(Sn虏鈦/Sn) = -0.14 V (This is the anode) E掳(I鈧/I鈦) = 0.54 V (This is the cathode) E掳(cell) = 0.54 - (-0.14) = 0.68 V Since E掳 > 0, the reaction is spontaneous. (b) Reduction of Ni虏鈦 to Ni by I鈦 (to form I鈧): E掳(Ni虏鈦/Ni) = -0.25 V (This is the cathode) E掳(I鈧/I鈦) = 0.54 V (This is the anode) E掳(cell) = -0.25 - 0.54 = -0.79 V Since E掳 < 0, the reaction is non-spontaneous. (c) Reduction of Ce鈦粹伜 to Ce鲁鈦 by H鈧侽鈧: E掳(Ce鈦粹伜/Ce鲁鈦) = 1.61 V (This is the cathode) E掳(H鈧侽鈧/H鈧侽) = 1.76 V (This is the anode) E掳(cell) = 1.61 - 1.76 = -0.15 V Since E掳 < 0, the reaction is non-spontaneous. (d) Reduction of Cu虏鈦 to Cu by Sn虏鈦 (to form Sn鈦粹伜): E掳(Cu虏鈦/Cu) = 0.34 V (This is the cathode) E掳(Sn鈦粹伜/Sn虏鈦) = 0.15 V (This is the anode) E掳(cell) = 0.34 - 0.15 = 0.19 V Since E掳 > 0, the reaction is spontaneous.

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Key Concepts

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

Standard Reduction Potentials
Understanding standard reduction potentials is crucial when predicting the spontaneity of redox reactions. A reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-reaction has a standard reduction potential (\( E^\text{o} \) value), which are typically found in electrochemistry tables and indicate the ease with which a species is reduced.

The standard reduction potentials are measured under standard conditions, which is typically 25掳C, 1 M concentration for each ion participating in the reaction, and a pressure of 1 atm for any gases that are involved.

In a redox reaction, the substance that has the higher (more positive) standard reduction potential will tend to undergo reduction (gain electrons), while the other substance with the lower standard reduction potential will tend to get oxidized (lose electrons). The difference in potential (\( E^\text{o}_\text{cell} \) value) can be calculated using the formula: \[ E^\text{o}_\text{cell} = E^\text{o}_\text{cathode} - E^\text{o}_\text{anode} \]
If this value is positive (\( E^\text{o}_\text{cell} > 0 \) ), the redox reaction is considered to be spontaneous under standard conditions. Conversely, if the value is negative (\( E^\text{o}_\text{cell} < 0 \) ), the reaction is non-spontaneous.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes, which is at the core of redox reactions. It's primarily concerned with how chemical energy is converted to electrical energy and vice versa.

In the context of spontaneous reactions, electrochemistry provides us with the tools to quantify and predict the flow of electrons, which are the source of the chemical changes. The standard reduction potentials are a key concept within electrochemistry that enables these predictions. By understanding the flow of electrons, we can harness these reactions to do useful work, for example in batteries or corrosion prevention.

The principles of electrochemistry are used to develop sensors, electroplating techniques, and electrochemical cells, where chemical energy is transformed into electrical energy as a means of power generation. Understanding how electrochemical gradients work allows scientists and engineers to innovate in power storage and generation technologies.
Galvanic Cells
Galvanic cells, also known as voltaic cells, are the physical structures that allow for spontaneous redox reactions to be harnessed for electrical energy. These cells are designed to separate the oxidation and reduction reactions into different compartments, connected by a salt bridge and an external circuit.

The anode is the electrode where oxidation occurs, while the cathode is the electrode where reduction takes place. Electrons flow from the anode to the cathode through an external circuit, thereby generating an electric current that can be used to do work. The salt bridge allows for the flow of ions to maintain a charge balance as the redox reaction proceeds.

In a galvanic cell, the spontaneous reaction's electrical potential can be measured using a voltmeter and is a direct result of the differences in standard reduction potentials of the electrodes. When dealing with exercises involving spontaneous reactions, it's often helpful to visualize the reaction occurring within a galvanic cell to understand better where oxidation and reduction are taking place and the direction of electron flow.

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Most popular questions from this chapter

Consider the half-reaction \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\). Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of \(\mathrm{Ag}^{+} ?(\mathrm{~b})\) What is the value of \(E_{\mathrm{red}}\) when \(\log \left[\mathrm{Ag}^{+}\right]=0 ?\) [Section \(\left.20.6\right]\)

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.

In the Br酶nsted-Lowry concept of acids and bases, acid-base reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. In what ways are redox reactions analogous? [Sections 20.1 and 20.2\(]\)

Using the standard reduction potentials listed in Appendix \(\mathrm{E}_{2}\) calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}^{\circ}\).

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