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Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., Vol. \(50,1973,61\) ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

Short Answer

Expert verified
(a) The number of coulombs (Q) required can be calculated using Faraday's constant (F) and the moles of hydrogen gas (n) required: \[ Q = n_{H_2} \times F = (7 \times 10^8 \, \text{mol}) \times (96485 \, \frac{\text{C}}{\text{mol}}) = 6.75 \times 10^{13} \, \text{C} \] (b) Using the Nernst equation with the standard cell potential (\(E_{cell}^0\)) and partial pressures of gases, we can calculate the minimum voltage (E_cell) required for electrolysis. (c) The minimum electrical energy (\(E_{min}\)) required can be calculated by multiplying the charge and the minimum voltage: \[ E_{min} = Q \times E_{cell} \] (d) To find the minimum cost of the electrical energy, convert the energy into kilowatt-hours and multiply by the cost per kilowatt-hour: \[ \text{Cost} = E_{min} (\text{kWh}) \times 0.85 \frac{\$}{\text{kWh}}\]

Step by step solution

01

Calculate the number of coulombs of electrical charge required

Faraday's law of electrolysis states that the amount of substance produced at the electrode is directly proportional to the amount of electrical charge passed. We are given that we need \( 7 \times 10^8 \) moles of hydrogen gas, the number of coulombs needed can be calculated using Faraday's constant (\(F = 96485 \, C \, mol^{-1}\)). Since the electrolysis of water generates hydrogen gas and each hydrogen ion has one elementary charge, we can calculate the charge needed as follows: \[ Q = n_{H_2} \times F \] Where: \(Q\) = charge required (in coulombs) \(n_{H_2}\) = moles of hydrogen gas \(F\) = Faraday's constant (96485 C/mol)
02

Calculate the minimum voltage required to generate hydrogen and oxygen gas

To find the minimum voltage required, we will need to calculate the standard cell potential of the electrolysis and apply the Nernst equation. Standard cell potential (\(E_{cell}^0\)) for the electrolysis of water can be calculated as follows: \[ E_{cell}^0 = E_{reduction}^0 - E_{oxidation}^0 \] Here, \(E_{reduction}^0\) is the standard reduction potential for the hydrogen half-cell (0 V) and \(E_{oxidation}^0\) is the standard reduction potential for the oxygen half-cell (-1.229 V). We will then use the Nernst equation to find the minimum voltage: \[E_{cell} = E_{cell}^0 + \frac{RT}{nF} \ln\frac{P_{O_2} P_{H_2}^{1/2}}{P_{H_2O}}\] Here, \(R\) is the gas constant (8.314 J/mol K), \(n\) is the number of transferred electrons (4 for this reaction), \(T\) is the temperature in Kelvin (assume 298K), \(P_{O_2}\) and \(P_{H_2}\) are partial pressures of the gases and \(P_{H_2O}\) is the vapor pressure of liquid water at the same temperature. We can find the partial pressures using the ideal gas law.
03

Calculate the minimum electrical energy

The minimum electrical energy required is the product of the charge and the minimum voltage: \[ E_{min} = Q \times E_{cell} \]
04

Calculate the minimum cost of electrical energy

To calculate the minimum cost of electrical energy, first convert the energy into kilowatt-hours: \[ E_{min} (\text{kWh}) = \frac{E_{min} (\text{J})}{3.6 \times 10^6 \frac{\text{J}}{\text{kWh}}} \] And then multiply it by the cost per kilowatt-hour: \[ \text{Cost} = E_{min} (\text{kWh}) \times 0.85 \frac{\$}{\text{kWh}}\] Follow these steps for a detailed solution to each part of the exercise.

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Key Concepts

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

Faraday's Law of Electrolysis
Faraday's Law of Electrolysis is a fundamental principle that enables us to understand and quantify the process of electrolysis, such as the production of hydrogen and oxygen gas from water. In essence, it establishes the link between electrical charge and chemical change. According to this law, the amount of a substance that undergoes oxidation or reduction at each electrode is proportional to the total electric charge passed through the substance.

For water electrolysis, the formation of hydrogen gas can be represented by the reaction \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). Faraday's first law quantifies this reaction; it tells us that the mass of hydrogen (\(m_{H_2}\)) produced is proportional to the amount of charge (\(Q\)) passed through the water, as shown in the equation \(m_{H_2} = (Q/F) \times (M_{H_2}/z)\) where \((F\)) is Faraday's constant, representing the charge of one mole of electrons (\(96,485 C \cdot mol^{-1}\)), \((M_{H_2}\)) is the molar mass of hydrogen, and \((z\)) is the valence number of the reaction which, for hydrogen production, is 2 because two electrons are transferred per hydrogen molecule formed.

Electrolysis applications range from industrial processes to innovative concepts like the proposed raising of the Titanic, where hydrogen gas production through electrolysis could provide the necessary buoyancy.
Nernst Equation
The Nernst Equation is critical for predicting the behavior of electrochemical cells under non-standard conditions. While the standard reduction potentials provide info for reactions at standard conditions (\(1M\) concentration, \(298 K\) temperature, and \(1 atm\) pressure), real-world applications often operate differently. The Nernst Equation allows us to calculate the actual cell potential (\(E_{cell}\)) when the concentrations, temperature, or pressure differ from the standard state.

The formula \[E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q\] incorporates \((E_{cell}^0\)), the standard cell potential, \((R\)), the universal gas constant, \((T\)), the temperature in Kelvin, \((n\)), the number of moles of electrons transferred, \((F\)), Faraday's constant, and \((Q\)), the reaction quotient indicating the ratio of product and reactant activities or concentrations when not at standard conditions.

When applied to electrolysis under pressures vastly different from \(1 atm\), such as the depth at which the Titanic lies, the Nernst equation can determine the minimum voltage necessary to carry out the electrolysis while considering the increased gases' pressure. The resulting voltage accounts for real pressures and confirms that electrolysis can occur under the specified conditions.
Electrochemical Cell Potential
Electrochemical cell potential, often simply called cell potential (\(E_{cell}\)), is the driving force behind the movement of electrons in an electrochemical cell. It determines if a redox reaction will occur spontaneously. The cell potential is measured in volts (V) and is essentially the difference in potential energy between the anode and the cathode within the cell.

For the electrolysis of water, the cell potential must be positive, and sufficient to cause the decomposition of water into hydrogen and oxygen gases. The standard cell potential is calculated under standard conditions (all reactants and products at \(1M\) concentrations, a temperature of \(298 K\), and a pressure of \(1 atm\)), and reflects the inherent tendency of the reaction to proceed. In situations where conditions are non-standard (like at the ocean's depths), the actual cell potential may differ and must be determined using adjustments such as the Nernst Equation.

Understanding the electrochemical cell potential is critical for designing systems to harness chemical reactions to produce electricity or, conversely, to use electrical energy to drive chemical changes, as in the case of electrolysis for hydrogen gas production.
Standard Reduction Potential
The standard reduction potential (\(E^0_{red}\)) is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. Each half-reaction in an electrochemical cell has a standard reduction potential, and these values are key in predicting the direction of electron flow and the feasibility of a reaction. Tabulated values for standard reduction potentials are determined at standard conditions, which include a temperature of \(298 K\), a pressure of \(1 atm\), and \(1M\) concentration for all the aqueous species involved.

Within the context of water electrolysis, two half-reactions occur: one where water is reduced to produce hydrogen gas (with an \(E^0_{red}\) of \(0 V\) for \(2H^+ + 2e^- \rightarrow H_2\)) and another where water is oxidized to produce oxygen gas (with an \(E^0_{red}\) for the converse of \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\), which is typically \(1.229 V\)). By understanding and using standard reduction potentials, scientists and engineers can calculate the standard cell potential and make predictions about the energy requirements for chemical processes.

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Most popular questions from this chapter

The capacity of batteries such as thetypical AA alkaline battery is expressed in units of milliamp-hours (mAh). An "AA" alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

The \(K_{\infty}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-28} .\) By using this value together with an electrode potential from Appendix \(\mathrm{E}\), determine the value of the standard reduction potential for the reaction $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-}-\cdots \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\) and \(\mathrm{Fe}^{3+}(a q) / \mathrm{Fe}^{2+}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode, and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\).

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 \(\mathrm{NaCl}\) 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?

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{~K}\) : $$ \mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 M\), and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

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