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 \((3 \mathrm{~km})\) is \(30 \mathrm{MPa} ?(\mathbf{c})\) 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 required electrical charge is approximately 1.35 × 10^14 coulombs. The minimum voltage required to generate H2 and O2 is approximately 1.28 V. The minimum electrical energy required is approximately 1.73 × 10^14 joules. The minimum cost of the electrical energy required to generate the necessary H2 is approximately 40.9 million dollars.

Step by step solution

01

Calculate the required electrical charge (coulombs)

We need to find the number of coulombs required to produce 7 × 10^8 mol of H2. The number of coulombs can be calculated using Faraday's laws of electrolysis: Q = n × F Where: - Q is the electrical charge in coulombs - n is the number of moles of electrons involved in the electrolysis reaction - F is Faraday's constant, approximately 96,485 C/mol To calculate n, we need to know the stoichiometry of the electrolysis reaction. The balanced equation for the electrolysis of water is as follows: 2H2O → 2H2 + O2 From this equation, we can see that 2 moles of H2 are produced for every 1 mole of O2. Therefore, the required number of moles of electrons for producing 7 × 10^8 mol of H2 is: n = 2 × (7 × 10^8) = 1.4 × 10^9 mol Now, we can calculate Q: Q = n × F = (1.4 × 10^9 mol) × (96,485 C/mol) ≈ 1.35 × 10^14 C So, the required electrical charge is approximately 1.35 × 10^14 coulombs.
02

Calculate the minimum voltage required

We need to find the minimum voltage required to generate H2 and O2 gas to raise the Titanic. We will use the Nernst equation to find the minimum voltage: E = E° - (RT/nF)lnQ where - E is the cell potential (voltage) at non-standard conditions - E° is the standard cell potential (voltage) - R is the gas constant (8.314 J/(mol·K)) - T is the temperature in Kelvin (assume 298 K) - n is the number of moles of electrons transferred in the reaction - F is Faraday's constant (96,485 C/mol) - Q is the reaction quotient, which can be simplified to the ratio of the product pressures (since the electrolysis reaction involves gases only) E° of H2O electrolysis under standard conditions is 1.23 V. Since we're given the pressure (30 MPa) at the depth of the wreckage, we can calculate Q using the ideal gas equation: 2H2O → 2H2 + O2 Q = P(H2)^2 / P(O2) At standard conditions, pressure is 100 kPa. Therefore, Q = (0.3)^2 / (100) = 9 × 10^(-4) Now, we can plug in the values for the Nernst equation. E = 1.23 - ((8.314 × 298) / (4 × 96,485)) × ln(9 × 10^(-4)) E ≈ 1.28 V So, the minimum voltage required to generate H2 and O2 is approximately 1.28 V.
03

Calculate the minimum electrical energy required

Now, we need to find the minimum electrical energy required to raise the Titanic by electrolysis. The electrical energy, E (in joules), can be calculated as: E = Q × V Where: - Q is the electrical charge in coulombs (from step 1, 1.35 × 10^14 C) - V is the cell potential (from step 2, 1.28 V) So, the minimum electrical energy required is: E = (1.35 × 10^14 C) × (1.28 V) ≈ 1.73 × 10^14 J Therefore, the minimum electrical energy required is approximately 1.73 × 10^14 joules.
04

Calculate the minimum cost of the electrical energy

Finally, we need to calculate the minimum cost of the electrical energy required to generate the necessary H2. We know that the cost of the electricity is 85 cents per kilowatt-hour, and the minimum electrical energy required is 1.73 × 10^14 joules (from step 3). First, we need to convert joules to kilowatt-hours: kilowatt-hours = (1.73 × 10^14 J) × (1 kWh/ 3.6 × 10^6 J) ≈ 4.81 × 10^7 kWh Now, we can calculate the minimum cost of the electrical energy: cost = (4.81 × 10^7 kWh) × \(0.85/kWh ≈ \)4.09 × 10^7 So, the minimum cost of the electrical energy required to generate the necessary H2 is approximately 40.9 million dollars.

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Most popular questions from this chapter

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to $0.15 \mathrm{M}\(, at what concentration of \)\mathrm{I}^{-}$ would the cell have zero potential?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) $2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{HNO}_{3}(a q)$ (b) $\mathrm{FeS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{Fe}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+ 2 \mathrm{NO}_{2}(g)+\mathrm{FeO}(s)$

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) $2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ (c) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). \((\mathbf{b})\) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)$.

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