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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, \(1973,\) Vol. 50,61 ). (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
The number of coulombs of electrical charge required to generate the hydrogen gas is approximately \(1.35 \times 10^{13} \mathrm{~C}\). The minimum voltage required for electrolysis is 1.01 V. The minimum electrical energy required to raise the Titanic by electrolysis is approximately \(1.36 \times 10^{13} \mathrm{~J}\), and the minimum cost of the electrical energy required to generate the necessary hydrogen gas is approximately 3.21 million dollars.

Step by step solution

01

(a) Find the number of coulombs required to generate the hydrogen gas

To find the number of coulombs required, we must first convert the number of moles of hydrogen gas into individual hydrogen atoms. Then, we'll use Faraday's Law to relate the number of moles to the required electrical charge. To generate hydrogen gas, we need \(2 \times 7 \times 10^8 \mathrm{~mol}\) of hydrogen ions. Using Faraday's constant (\(1 \mathrm{F} = 96,485 \mathrm{~C/mol}\)), we calculate the required coulombs: Charge (Q) = \(\mathrm{Moles~of~Hydrogen~Ions \times F}\) Q = \((2 \times 7 \times 10^8 \mathrm{~mol}) \times (96485 \mathrm{~C/mol})\) Q = \(1.35 \times 10^{13} \mathrm{~C}\) Thus, the number of coulombs of electrical charge required is approximately \(1.35 \times 10^{13} \mathrm{~C}\).
02

(b) Calculate the minimum voltage required for electrolysis

The minimum voltage is given by the Nernst Equation: \(E = E^o - \frac{RT}{nF} \ln \frac{[P(\mathrm{H_2})][P(\mathrm{O_2})^{1/2}]}{[P(\mathrm{H_2O})]}\) Assuming standard conditions (\(E^o = 1.23\) V), E = \(1.23 - \frac{8.314 \times 298}{2 \times 96485} \ln \frac{(300)^{1/2}}{1}\) E = \(1.23 - 0.040 \ln (300)\) E = \(1.01 \mathrm{~V}\) The minimum voltage to generate the hydrogen and oxygen gas at 300 atm pressure is approximately 1.01 V.
03

(c) Calculate the minimum electrical energy required

The energy required (in joules) can be calculated using the formula: Energy (E) = Q * V E = \(1.35 \times 10^{13} \mathrm{~C} \times 1.01 \mathrm{~V}\) E = \(1.36 \times 10^{13} \mathrm{~J}\) The minimum electrical energy required to raise the Titanic by electrolysis is approximately \(1.36 \times 10^{13} \mathrm{~J}\).
04

(d) Compute the minimum cost of electrical energy

Convert the energy in joules to kilowatt-hours: Energy (in kilowatt-hours) = \(\frac{1.36 \times 10^{13} \mathrm{~J}}{3.6 \times 10^6 \mathrm{~J/kWh}}\) Energy = \(3.78 \times 10^6 \mathrm{~kWh}\) Now, calculate the cost of the energy required: Cost ($) = Energy (in kilowatt-hours) * Cost per kilowatt-hour Cost = \(3.78 \times 10^6 \mathrm{~kWh} \times 0.85 \mathrm{~\$ /kWh}\) Cost = \(3.21 \times 10^6 \mathrm{\$}\) The minimum cost of the electrical energy required to generate the necessary hydrogen gas is approximately 3.21 million dollars.

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Key Concepts

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

Faraday's Law
Faraday's Law provides a key insight into the process of electrolysis, particularly how electrical charge relates to a chemical change during the reaction. When it comes to electrolysis, like in raising the Titanic scenario, Faraday's Law tells us how much charge is needed to produce a specific amount of substance at an electrode. This is calculated using the formula:

\[ Q = n \cdot F \]
where \( Q \) is the charge in coulombs, \( n \) is the number of moles of electrons needed, and \( F \) is Faraday's constant, which is approximately 96,485 C/mol. In our exercise, to produce hydrogen gas, this law helps us determine that we need about \(1.35 \times 10^{13} \mathrm{~C}\). This insight is crucial for knowing how much current and for how long it needs to be applied to generate the desired amount of hydrogen.
Nernst Equation
The Nernst Equation allows us to calculate the minimum voltage needed for an electrochemical reaction under non-standard conditions, such as high pressures experienced at deep underwater sites. It is represented as:

\[ E = E^o - \frac{RT}{nF} \ln \frac{[P(\mathrm{H_2})][P(\mathrm{O_2})^{1/2}]}{[P(\mathrm{H_2O})]} \]
where:
  • \(E\) is the electrode potential under non-standard conditions.
  • \(E^o\) is the standard electrode potential.
  • \(R\) is the universal gas constant (8.314 J/mol K).
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons transferred.
  • \(F\) is Faraday's constant.
In the Titanic exercise, for hydrogen generation at 300 atm, using standard conditions, the Nernst Equation gives us a minimum potential of approximately 1.01 V, indicating the energy efficiency of the process under high pressure.
Electrical Energy Calculation
Calculating the necessary electrical energy for electrolysis involves finding the total energy, measured in joules, required to achieve the desired chemical transformation. The formula used is:

\[ E = Q \cdot V \]
where \( E \) is energy in joules, \( Q \) is the charge in coulombs, and \( V \) is the potential in volts. From the exercise, with a charge of \(1.35 \times 10^{13} \mathrm{~C}\) and a voltage of 1.01 V, the energy comes out to be \(1.36 \times 10^{13} \mathrm{~J}\). This calculation is vital because it gives us insight into how much electrical energy we need to input into the system, which also ties into cost calculations.
Hydrogen Generation
The process of hydrogen generation during electrolysis is an essential element in many nuclear, industrial, and even historical applications like the Titanic example. It involves splitting water molecules into hydrogen and oxygen gas using electric current. The electrolysis reaction occurring is:

\[ 2 \mathrm{H_2O(l)} \rightarrow 2 \mathrm{H_2(g)} + \mathrm{O_2}(g) \]
Hydrogen is generated at the cathode by gaining electrons, and it's crucial for various uses, including buoyancy. The amount of hydrogen required and produced can be determined through stoichiometry and electrochemical principles, ensuring sufficient gas generation to achieve objectives like raising the Titanic by displacing its weight with the buoyant hydrogen gas. This underlines the intersection of chemistry and engineering in practical applications.

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Most popular questions from this chapter

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

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

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 \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)\) (acidic solution) (b) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q)\) (acidic solution) (c) \(\mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q)\) (acidic solution) (e) \(\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g)\) (basic solution) (f \(\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (basic solution)

(a) Suggest an explanation for why liquid water is needed in an alkaline battery. (b) What is the advantage of using highly concentrated or solid reactants in a voltaic cell?

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