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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 \(7 \times 10^{4} \mathrm{~mol}\) 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 \- - the electricity costs 85 cents per kilowatt-

Short Answer

Expert verified
(a) The required electrical charge (Q) to produce the hydrogen gas is \(Q = 1.35 \times 10^9 \mathrm{~C}\). (b) The minimum voltage (V) required to generate hydrogen and oxygen gases is \(E \approx 0.019 \mathrm{~V}\). (c) The minimum electrical energy (E) required to raise the Titanic by electrolysis is \(E \approx 2.57 \times 10^7 \mathrm{~J}\). (d) The total cost of electricity to raise the Titanic is approximately $578.45.

Step by step solution

01

(a) Finding the Required Electrical Charge

To find the required electrical charge to produce the necessary hydrogen gas, we need to multiply the moles of hydrogen gas required by the charge of one mole of electron (Faraday's constant). Here, we are given the necessary moles of hydrogen gas, \(7 \times 10^{4} \mathrm{~mol}\), and the Faraday constant, \(F = 96485 \mathrm{~C/mol}\). Let's denote the required electrical charge as \(Q\). The reaction to produce hydrogen gas is: \[2 H^{+} + 2 e^{-} \rightarrow H_{2}\] From the stoichiometry of the reaction, we can see that 2 moles of electrons are required to produce 1 mole of hydrogen gas. Therefore, we can determine the charge needed using the following equation: \[Q = \frac{(7 \times 10^{4} \mathrm{~mol})\times ( 2 \mathrm{~mol~of~electrons}) \times ( 96485 \mathrm{~C/mol})}{(1 \mathrm{~mol~of~H_{2}})}\]
02

(b) Finding the Minimum Voltage

To find the minimum voltage required to generate both hydrogen and oxygen gas, we need to first calculate the cell potential using the Nernst equation: \[E = E^{0} - \frac{RT}{nF} \ln Q\] At the depth of the wreckage, we are given the pressure, \(P = 300 \mathrm{~atm}\). We are interested in the generation of hydrogen gas, so we will analyze the reaction half-cell: \[2 H^{+} + 2 e^{-} \rightarrow H_{2}\] For this reaction, the standard cell potential, \(E^{0} = 0\). The number of electrons being transferred, \(n = 2\), and we assume the temperature to be \(T = 298 \mathrm{~K}\). The gas constant, \(R = 8.314 \mathrm{~J/(mol \cdot K)}\). Finally, the reaction quotient, \(Q = \frac{[H_{2}]}{[H^{+}]^{2}}\). We will assume [\(H^{+}\)] = 1 M and [\(H_{2}\)] = 300 atmospheres. (since the pressure on the gases is 300 atm) Then we can plug these values into the Nernst equation to find the minimum cell potential: \[E = 0 - \frac{(8.314 \mathrm{~J/(mol \cdot K)})\times (298 \mathrm{~K})}{(2 \times 96485 \mathrm{~C/mol})} \ln \frac{300}{1^{2}}\]
03

(c) Finding the Minimum Electrical Energy

To calculate the minimum electrical energy needed to raise the Titanic using electrolysis, we will use the following equation (where energy is denoted by \(E\), charge by \(Q\), and voltage by \(V\)): \[E = Q \times V\] We already found \(Q\) and \(V\), so we can plug those values into this equation.
04

(d) Calculating the Electricity Cost

Finally, we need to calculate the cost of the electricity needed to raise the Titanic. We are given that the cost is 85 cents per kilowatt-hour. First, we need to convert the energy we found earlier in (c), which is in Joules, to kilowatt-hours by using the following conversion factors: \[1 \mathrm{~kW} = 1000 \mathrm{~W}\] \[1 \mathrm{~Wh} = 3600 \mathrm{~J}\] Considering these conversion factors, we can convert the energy as follows: \[\mathrm{Energy~(kWh)} = \frac{ \mathrm{Energy~(J)} }{(1000 \mathrm{~W/kW}\ ) \times (3600 \mathrm{~J/Wh})}\] Now that we have the energy in kilowatt-hours, we can multiply it by the given cost of electricity to find the total cost: \[ \mathrm{Total~Cost} = \mathrm{Energy~(kWh)} \times 0.85 \mathrm{~$/kWh}\]

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Key Concepts

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

Faraday's Constant
Understanding the electrolysis of water is underpinned by a key concept known as Faraday's constant. This is a fundamental value in electrochemistry, representing the amount of electrical charge carried by one mole of electrons. Faraday's constant, denoted as \(F\), has a value of approximately \(96,485 \mathrm{~C/mol}\).

During electrolysis, electrical current causes chemical reactions at the electrodes. For instance, the electrolysis of water produces hydrogen and oxygen gases. When quantifying the amount of charge required to produce a certain amount of gas, Faraday's constant is used to translate moles of electrons into coulombs of charge.

For example, when considering the proposal to raise the Titanic, the calculated moles of \(\mathrm{H}_{2}\) needed to be converted into the total charge required using Faraday's constant in this manner: \[\begin{equation}Q = \text{moles of } \mathrm{H}_{2} \times \text{moles of electrons per mole of } \mathrm{H}_{2} \times F\end{equation}\]Here, the stoichiometry reveals that it takes two moles of electrons to produce one mole of hydrogen gas, which means Faraday's constant is fundamental in linking the chemical reaction happening at the molecular level to the electrical charge needed for the reaction to occur.
Nernst Equation
The Nernst equation is a key principle in electrochemistry that allows us to calculate the electric potential of an electrochemical cell under non-standard conditions. It takes into consideration the effect of ion concentration on the voltage of the cell.

The equation is expressed as:\[\begin{equation}E = E^{0} - \frac{RT}{nF} \ln Q\end{equation}\]where \(E\) is the cell potential, \(E^{0}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient, which reflects the ratio of the concentrations of the products to reactants.

Applying the Nernst equation in our Titanic example is vital for determining the minimum voltage needed to conduct electrolysis deep underwater. We need to adjust the equation for the increased pressure, which affects the concentration of gases, and ultimately, the required voltage to initiate the electrolysis process to produce hydrogen and oxygen gases.
Electrical Energy
Electrical energy is the energy derived from electric potential energy or kinetic energy. It is essentially harnessed and used to produce various forms of work, in this case, driving the electrolysis of water. When calculating the minimum electrical energy required to raise the Titanic via electrolysis, we consider the voltage and the charge necessary for the reaction.

The formula used is:\[\begin{equation}E = Q \times V\end{equation}\]where \(E\) is the energy in Joules, \(Q\) is the total charge in coulombs, and \(V\) is the voltage in volts. Calculating the electrical energy is crucial in this context because it translates the electrochemical reaction into tangible work - lifting the Titanic using hydrogen gas generated from electrolysis. By understanding the electrical energy involved, we also pave the way for calculating the costs associated with such an endeavor.
Stoichiometry
Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. In the context of electrolysis, stoichiometry allows us to predict the amount of reactants required and the amount of products formed.

The electrolysis of water involves a specific stoichiometric ratio, where two moles of water yield two moles of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{O}_{2}\). Thus, for every mole of \(\mathrm{H}_{2}\) produced, two moles of electrons are consumed. Understanding this ratio is vital when calculating the total charge needed for the reaction.

The exercise to calculate the charge needed to produce the hydrogen gas for the Titanic project relies on a clear grasp of stoichiometry to ensure accurate estimation of the required materials and the consequent cost. It's the foundation that bridges our understanding of the principles of chemistry with practical applications and outcomes.

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Most popular questions from this chapter

For a spontaneous reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(\mathrm{at})\), answer the following questions: (a) If you made a voltaic cell out of this reaction, what halfreaction would be occurring at the cathode, and what halfreaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text {cell? }}^{\text {? }}\) [Section 20.3]

A voltaic cell that uses the reaction $$ \mathrm{T1}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tr}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(\mathrm{E}\), determine \(E_{\mathrm{ed}}^{0}\) for the reduction of \(\mathrm{Ti}^{3+}(a q)\) to \(\mathrm{Ti}^{+}(a q)\). (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Gold exists in two common positive oxidation states, \(+1\) and \(+3\). The standard reduction potentials for these oxidation states are $$ \begin{aligned} A u^{*}(a q)+\mathrm{e}^{-} & \longrightarrow A u(s) \quad E_{\text {red }}^{o}=+1.69 \mathrm{~V} \\ A u^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow A u(s) \quad E_{\text {red }}^{o}=+1.50 \mathrm{~V} \end{aligned} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqucous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{gathered} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{gathered} $$ What is being oxidized and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being exidized, and what is being reduced?

Indicate whether each of the following statements is true or false: (a) If something is axidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\) \(\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

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