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Chapter 20: Problem 51

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\) by \(\mathrm{N}_{2} \mathrm{O}(a q),\) and by \(\mathrm{VO}_{2}^{+}(a q)\) (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Short Answer

Expert verified
The balanced chemical equations for the oxidation of Fe虏鈦� by the given reactants are: (a) \(2\text{Fe}^{2+} + \text{S}_{2}\text{O}_{6}^{2-} + 4\text{H}^{+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_{2}\text{SO}_{3}\) (b) \(\text{Fe}^{2+} + \text{N}_{2}\text{O} + 2\text{H}^{+} \rightarrow \text{Fe}^{3+} + \text{N}_{2} + \text{H}_{2}\text{O}\) (c) \(\text{Fe}^{2+} + 2\text{VO}_{2}^{+} + 4\text{H}^{+} \rightarrow \text{Fe}^{3+} + 2\text{VO}^{2+} + 2\text{H}_{2}\text{O}\) The standard Gibbs free energy change (螖G掳) for each reaction at 298 K are: 螖G掳鈧� = -32,841 J/mol, 螖G掳鈧� = -245,099 J/mol, and 螖G掳鈧� = -22,193 J/mol. The equilibrium constants (K) for each reaction at 298 K are: K鈧� = 5.37 脳 10鹿虏, K鈧� = 8.68 脳 10鈦仿�, and K鈧� = 4.95 脳 10鈦�.

Step by step solution

01

Identify the redox couples

First, let's identify the redox couples involved in the given reactions. 1. Fe鲁鈦�/Fe虏鈦� 2. S鈧侽鈧喡测伝/H鈧係O鈧� 3. N鈧侽/N鈧� 4. VO鈧傗伜/VO虏鈦� We will write balanced chemical equations for the oxidation of Fe虏鈦�(aq) by S鈧侽鈧喡测伝(aq), N鈧侽(g), and VO鈧傗伜(aq).
02

Balance each redox reaction

For each redox reaction, the half-reactions need to be combined, making sure that the electrons cancel out, and the substances correctly balanced. (a) Oxidation of Fe虏鈦� by S鈧侽鈧喡测伝 Oxidation half-reaction: Fe虏鈦� -> Fe鲁鈦� + e鈦� (E掳 = +0.77 V) Reduction half-reaction: S鈧侽鈧喡测伝 + 4H鈦� + 2e鈦� -> 2H鈧係O鈧� (E掳 = +0.60 V) Multiply the first half-reaction by 2 to cancel the electrons, and then combine both half-reactions: 2(Fe虏鈦� -> Fe鲁鈦� + e鈦�) + S鈧侽鈧喡测伝 + 4H鈦� + 2e鈦� -> 2H鈧係O鈧� Resulting reaction: 2Fe虏鈦� + S鈧侽鈧喡测伝 + 4H鈦� -> 2Fe鲁鈦� + 2H鈧係O鈧� (b) Oxidation of Fe虏鈦� by N鈧侽 Oxidation half-reaction: Fe虏鈦� -> Fe鲁鈦� + e鈦� (E掳 = +0.77 V) Reduction half-reaction: N鈧侽 + 2H鈦� + 2e鈦� -> N鈧� + H鈧侽 (E掳 = -1.77 V) Combine both half-reactions (no need to multiply as the electrons are already equal): Fe虏鈦� + N鈧侽 + 2H鈦� -> Fe鲁鈦� + N鈧� + H鈧侽 (c) Oxidation of Fe虏鈦� by VO鈧傗伜 Oxidation half-reaction: Fe虏鈦� -> Fe鲁鈦� + e鈦� (E掳 = +0.77 V) Reduction half-reaction: VO鈧傗伜 + 2H鈦� + e鈦� -> VO虏鈦� + H鈧侽 (E掳 = +1.00 V) Multiply the second half-reaction by 2 to cancel the electrons, and then combine both half-reactions: Fe虏鈦� -> Fe鲁鈦� + e鈦� + 2(VO鈧傗伜 + 2H鈦� + e鈦�) -> 2VO虏鈦� + 2H鈧侽 Resulting reaction: Fe虏鈦� + 2VO鈧傗伜 + 4H鈦� -> Fe鲁鈦� + 2VO虏鈦� + 2H鈧侽
03

Calculate 螖G掳 for each reaction

To calculate 螖G掳 for each reaction at 298 K, we need to utilize the following equation: 螖G掳 = -nFE掳 where n is the number of electrons transferred in the reaction, F is Faraday's constant (96485 C/mol), and E掳 is the standard cell potential for the reaction. 螖G掳 values for each reaction: (a) 螖G掳鈧� = -2 (96485 C/mol) (0.77 V - 0.60 V) = -32,841 J/mol (b) 螖G掳鈧� = -1 (96485 C/mol) (0.77 V - (-1.77 V)) = -245,099 J/mol (c) 螖G掳鈧� = -1 (96485 C/mol) (0.77 V - 1.00 V) = -22,193 J/mol
04

Calculate the equilibrium constant K for each reaction

To determine the equilibrium constant K for each reaction at 298 K, we need to use the relationship between 螖G掳 and K: 螖G掳 = -RT ln K where R is the gas constant (8.314 J/(mol鈭橩)) and T is the temperature (298 K). We can solve for K by using the calculated 螖G掳 values. K values for each reaction: (a) K鈧� = exp(-(-32,841 J/mol) / (8.314 J/(mol鈭橩) 脳 298 K)) = 5.37 脳 10鹿虏 (b) K鈧� = exp(-(-245,099 J/mol) / (8.314 J/(mol鈭橩) 脳 298 K)) = 8.68 脳 10鈦仿� (c) K鈧� = exp(-(-22,193 J/mol) / (8.314 J/(mol鈭橩) 脳 298 K)) = 4.95 脳 10鈦� So, the equilibrium constants K at 298 K for each reaction are: K鈧� = 5.37 脳 10鹿虏, K鈧� = 8.68 脳 10鈦仿�, and K鈧� = 4.95 脳 10鈦�.

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Key Concepts

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

Electrochemistry
Electrochemistry is a fascinating branch of chemistry that deals with the interaction between electrical energy and chemical changes. It plays a crucial role in many everyday processes, including battery operation and corrosion control.

Electrochemical reactions are either spontaneous or non-spontaneous. Spontaneous reactions can generate electrical energy, whereas non-spontaneous requires an input of electrical energy to occur. A classic example of electrochemical processes includes the redox reaction, where oxidation and reduction processes occur simultaneously.
  • Oxidation: The loss of electrons from a chemical species.
  • Reduction: The gain of electrons by a chemical species.
By balancing these reactions and combining them effectively, we get an overall cell reaction that produces electrical energy, which can be seen in the exercise provided. Understanding electrochemical principles is essential for calculating the energy changes associated with these reactions, establishing the foundation for more complex analyses, such as determining the Gibbs free energy and equilibrium constants.
Standard Reduction Potential
Standard reduction potential (\(E_{ ext{red}}^{ ext{掳}}\)) is a measure of the tendency of a chemical species to gain electrons, i.e., to be reduced. It is expressed in volts and measured under standard conditions: 25掳C, 1 atm pressure, and 1 M concentration.

Higher standard reduction potential values indicate a greater likelihood for reduction to occur. In the context of redox reactions, these values are used to predict the direction of electron flow between different species.
  • A more positive potential implies a strong tendency to be reduced.
  • A more negative potential indicates a tendency to lose electrons and therefore be oxidized.
In the given exercise, understanding and using these potentials is vital for calculating the cell potential and subsequently determining the Gibbs free energy (螖G) of the reactions. When combining half-reactions, the standard reduction potentials are subtracted (\(E^{ ext{掳}}_{ ext{cell}} = E^{ ext{掳}}_{ ext{red, cathode}} - E^{ ext{掳}}_{ ext{red, anode}}\)), offering insight into the reaction's feasibility.
Equilibrium Constant
The equilibrium constant (\(K\)) gives us insight into the extent of a chemical reaction under equilibrium conditions. It is a ratio of the concentrations of the products to the reactants, raised to their respective stoichiometric coefficients, at equilibrium.

In electrochemical cells, the equilibrium constant is linked to the Gibbs free energy change (袄(螖骋^掳\)) for the reaction. The relationship is expressed via the equation:\[螖G^掳 = -RT \, \ln(K)\]where:
  • \(R\)is the universal gas constant (8.314 J/(mol路K)).
  • \(T\)is the temperature in Kelvin.
This means that knowing \(K\) allows us to deduce how far a reaction will proceed before reaching equilibrium. Large \(K\) values indicate the reaction lies far towards products, while small \(K\) values suggest minimal product formation at equilibrium. The calculation of \(K\)is crucial for understanding the feasibility and directionality of chemical processes, as seen in the reactions given in the exercise.
Gibbs Free Energy
Gibbs free energy (袄(螖骋\)) is a thermodynamic quantity that reflects the amount of energy available to do work during a process at constant temperature and pressure. It serves as a powerful predictor of reaction spontaneity:
  • 袄(螖骋 < 0\): the process is spontaneous and can occur without energy input.
  • 袄(螖骋 = 0\): the process is at equilibrium.
  • 袄(螖骋 > 0\): the process is non-spontaneous and requires energy input.
In electrochemistry, 袄(螖骋^掳\) relates to the cell potential (\(E^{ ext{掳}}_{ ext{cell}}\)) as follows:\[螖G^掳 = -nF \, E^{ ext{掳}}_{ ext{cell}}\]where:
  • \(n\)is the number of moles of electrons transferred.
  • \(F\)is Faraday's constant (96485 C/mol).
Calculating 袄(螖骋^掳\) gives a quantitative measure of the driving force behind a chemical reaction. In the exercise provided, understanding 袄(螖骋^掳\)is essential to predict and evaluate the thermodynamic efficiency of the given redox reactions.

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Most popular questions from this chapter

At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.63 \mathrm{~V}\). The equilibrium constant for the reaction is \(3.8 \times 10^{10}\). What is the value of \(n\) for the reaction?

Li-ion batteries used in automobiles typically use a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode in place of the \(\mathrm{LiCoO}_{2}\) cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a \(\mathrm{LiCoO}_{2}\) cathode, approximately \(50 \%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=&+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

You may have heard that "antioxidants" are good for your health. Is an "antioxidant" an oxidizing agent or a reducing agent?

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?
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