Which of the following reactions is (are) spontaneous at standard conditions? (a) \(\mathrm{Zn}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Fe}^{2+}(a q)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{H}_{2}(g)\) (c) \(2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Br}_{2}(l)+2 \mathrm{I}^{-}(a q)\)

Short Answer

Expert verified
(a) Zn + Fe3+ → Zn2+ + Fe2+, (b) Cu + 2H+ → Cu2+ + H2, (c) 2Br- + I2 → Br2 + 2I-. Answer: All three reactions a, b, and c are spontaneous at standard conditions.

Step by step solution

01

Look up the standard reduction potentials

First, look up the standard reduction potentials for each half-reaction involved in the three reactions. These values can be found in a table of standard reduction potentials. Note down the half-reactions and their corresponding potentials. For reaction (a): Zn \(\to\) Zn\(^{2+}\) + 2e\(^{-}\), \(E^\circ = -0.76\,\text{V}\) Fe\(^{3+}\) + e\(^{-}\) \(\to\) Fe\(^{2+}\), \(E^\circ = 0.77\,\text{V}\) For reaction (b): Cu \(\to\) Cu\(^{2+}\) + 2e\(^{-}\), \(E^\circ = -0.34\,\text{V}\) 2H\(^{+}\) + 2e\(^{-}\) \(\to\) H\(_2\), \(E^\circ = 0\,\text{V}\) For reaction (c): 2Br\(^{-}\) \(\to\) Br\(_2\) + 2e\(^{-}\), \(E^\circ = -1.09\,\text{V}\) I\(_2\) + 2e\(^{-}\) \(\to\) 2I\(^{-}\), \(E^\circ = 0.54\,\text{V}\)
02

Calculate the cell potentials

To calculate the cell potential for each reaction, subtract the standard reduction potential of the oxidized species from the standard reduction potential of the reduced species: For reaction (a): \(E^\circ_{cell} = E^\circ_{\text{reduced}} - E^\circ_{\text{oxidized}} = 0.77\,\text{V} - (-0.76\,\text{V}) = 1.53\,\text{V}\) For reaction (b): \(E^\circ_{cell} = E^\circ_{\text{reduced}} - E^\circ_{\text{oxidized}} = 0\,\text{V} - (-0.34\,\text{V}) = 0.34\,\text{V}\) For reaction (c): \(E^\circ_{cell} = E^\circ_{\text{reduced}} - E^\circ_{\text{oxidized}} = 0.54\,\text{V} - (-1.09\,\text{V}) = 1.63\,\text{V}\)
03

Calculate \(\Delta G^\circ\) for each reaction

Plug the values obtained in Steps 1 and 2 into the formula \(\Delta G^\circ = -nFE^\circ_{cell}\) to find \(\Delta G^\circ\) for each reaction. A negative \(\Delta G^\circ\) indicates a spontaneous reaction. For reaction (a): \(\Delta G^\circ = -2(96,485\,\text{C/mol})(1.53\,\text{V}) \approx -295,502\,\text{J/mol}\), which is negative. For reaction (b): \(\Delta G^\circ = -2(96,485\,\text{C/mol})(0.34\,\text{V}) \approx -65,609\,\text{J/mol}\), which is negative. For reaction (c): \(\Delta G^\circ = -2(96,485\,\text{C/mol})(1.63\,\text{V}) \approx -314,840\,\text{J/mol}\), which is negative.
04

Determine spontaneity

Since all three reactions have negative \(\Delta G^\circ\) values, it can be concluded that all three reactions a, b, and c are spontaneous at standard conditions.

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Key Concepts

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

Understanding Standard Reduction Potentials
When we talk about standard reduction potentials, we're referring to the inherent tendency of a substance to gain electrons or be reduced. These values are essential in determining the spontaneity of reactions in electrochemistry.

Standard reduction potentials, usually cited as values, are measured in volts (V) and based on a standard set of conditions: a 1 M concentration for each aqueous species, a 1 atm pressure for any gases that are involved, and a temperature of 298 K. The standard hydrogen electrode (SHE) is chosen as the reference point, with an assigned potential of 0 V.

For a spontaneous reaction, the cell potential must be positive, which means the reduction potential of the cathode (where reduction occurs) must be greater than the reduction potential of the anode (where oxidation occurs). If you see a reaction with a higher reduction potential driving another with a lower potential, you're witnessing spontaneity in action!
Calculating Cell Potential
The step of cell potential calculation is crucial in predicting whether an electrochemical reaction will occur spontaneously. The formula we use is cell = E°cathode - E°anode. By subtracting the standard reduction potential of the anode from that of the cathode, we obtain the overall cell potential.

This difference tells us the electromotive force driving the reaction. A positive result indicates a tendency towards spontaneous reaction. For example, in reaction (a) from the exercise, the positive cell potential of 1.53 V suggests that zinc is more likely to be oxidized while iron is reduced, making the overall reaction spontaneous under standard conditions.
Deciphering Gibbs Free Energy
The concept of Gibbs free energy, symbolized as ΔG, is paramount in determining reaction spontaneity. It’s a thermodynamic quantity that combines enthalpy, entropy, and temperature to give us valuable information about whether a process will proceed without external energy input.

In electrochemistry, we use the formula ΔG° = -nFE°cell to relate Gibbs free energy to the cell potential. Here n represents the number of moles of electrons exchanged in the reaction, and F is Faraday’s constant (approximately 96,485 C/mol).

A negative ΔG° means the reaction is spontaneous. It is important to remember that ΔG° only predicts spontaneity under standard conditions; real-world conditions may vary and could affect the actual spontaneity of a reaction.
Spontaneity in Electrochemistry
The topic of spontaneity in electrochemistry is integral in understanding which reactions will occur without external input of energy. This is closely linked to the sign of the ΔG° value.

When we find all of our ΔG° values to be negative, as in the example exercise, we confirm that the reactions are spontaneous at standard conditions. We also incorporate the second law of thermodynamics here: spontaneous processes are characterized by an overall increase in the entropy of the universe.

In the context of an electrochemical cell, a positive cell potential leads to a negative Gibbs free energy change and thus a spontaneous reaction. Contemplate this deeply in learning electrochemistry: every spontaneous reaction has an electrical work potential, usually harnessed in batteries or other electrochemical cells. The beauty of spontaneity lies in nature's predisposition to favor processes that release energy and increase disorder – and our understanding of this principle helps us to harness that energy in practical applications.

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Most popular questions from this chapter

The electrolysis of an aqueous solution of \(\mathrm{NaCl}\) has the overall equation $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{OH}^{-}(a q) $$ During the electrolysis, \(0.228\) mol of electrons passes through the cell. (a) How many electrons does this represent? (b) How many coulombs does this represent? (c) Assuming \(100 \%\) yield, what masses of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) are produced?

Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+}\) ? (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?

Which species in each pair is the stronger oxidizing agent? (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\) (c) \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\) (d) \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution

Consider the following species. $$ \begin{array}{llll} \mathrm{Cr}^{3+} & \mathrm{Hg}(l) & \mathrm{H}_{2} \text { (acidic) } & \mathrm{Sn}^{2+} \end{array} $$ \(\mathrm{Br}_{2}\) (acidic) Classify each species as oxidizing agent, reducing agent, or both. Arrange the oxidizing agents in order of increasing strength. Do the same for the reducing agents.

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