Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{C}^{\circ}=-2.36 \mathrm{V}$$a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{kg}\) Zr reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 atm and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core:$$\mathbf{C}(s)+\mathbf{H}_{2} \mathbf{O}(g) \longrightarrow \mathbf{C O}(g)+\mathbf{H}_{2}(g)$$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

Step by step solution

01

Determine if zirconium can reduce water to hydrogen gas

The half-reaction for the reduction of water to hydrogen gas is: \(2 H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq)\) and its standard reduction potential, \(E^0_H\), is -0.83 V. To determine if zirconium can reduce water to hydrogen gas, we need to compare its standard reduction potential, \(E^0_Z\) = -2.36 V (given), with the standard reduction potential of the reduction of water to hydrogen gas, \(E^0_H\) = -0.83 V. Since \(E^0_Z < E^0_H\), the reduction of zirconium is more favorable than the reduction of water. This means that zirconium metal can indeed reduce water to form hydrogen gas.

02

Balanced equation for the reduction of zirconium

We can find the balanced equation for the reduction of water by zirconium by combining the given half-reaction and the reaction for the formation of hydrogen gas. First, we must balance both half-reactions: Reduction of zirconium: \(ZrO_2 \cdot H_2O + H_2O + 4 e^{-} \rightarrow Zr + 4 OH^{-}\) Reduction of water to hydrogen gas: \(2H_2O + 2 e^{-} \rightarrow H_2 + 2OH^{-}\) Next, multiply the second equation by 2 to make the number of electrons in both equations equal: Reduction of water to hydrogen gas (multiplied by 2): \(4H_2O + 4 e^{-} \rightarrow 2H_2 + 4OH^{-}\) Now, add both equations: Reduction of zirconium + Reduction of water to hydrogen gas: \(ZrO_2 \cdot H_2O + H_2O + 4 e^{-} + 4H_2O + 4 e^{-} \rightarrow Zr + 4 OH^{-} + 2H_2 + 4OH^{-}\) Simplifying this yields the following balanced equation for the reduction of water by zirconium: \(ZrO_2 \cdot H_2O + 5 H_2O + 4 e^{-} \rightarrow Zr + 2H_2 + 8 OH^{-}\)

03

Calculate standard cell potential, Gibbs free energy change, and equilibrium constant

First, let's calculate the standard cell potential (\(E^0\)) for the overall cell reaction. Since we found that zirconium can reduce water, we need to reverse the half-reaction for the reduction of water to hydrogen gas: Oxidation of hydrogen gas: \(H_2(g) \rightarrow 2H_2O(l) + 2e^{-}\) with \(E^0 = 0.83 V\). Now, we can calculate the standard cell potential (\(E^0\)) for the overall cell reaction: \(E^0_{cell} = E^0_{Z} - E^0_{H} = -2.36 V - 0.83 V = -3.19 V\) Next, let's calculate the standard Gibbs free energy change (\(\Delta G^0\)): \(\Delta G^0 = -nFE^0_{cell}\) \(n\) = number of moles of electrons transferred (\(n = 4\)) \(F\) = Faraday's constant \(= 96500 C/mol\) \(\Delta G^0 = -4 \times 96500 C/mol \times -3.19 V = 1.23 \times 10^6 J/mol\) Now we can calculate the equilibrium constant, \(K\), using the following equation: \(\Delta G^0 = -RT \ln K\) \(R\) = gas constant (\(8.314 J/(K \cdot mol)\)) \(T\) = temperature in Kelvin (assume standard temperature, 298 K) \(K = e^{\frac{-\Delta G^0}{RT}} = e^{\frac{1.23 \times 10^6 J/mol}{8.314 J/(K \cdot mol) \times 298 K}} = 6.68 \times 10^{63}\)

04

Calculate the mass and volume of hydrogen produced

Given that 1.00x10^3 kg of Zirconium reacts, we need to find the mass of hydrogen produced. To do this, we must first convert mass of Zirconium into moles: Moles of Zirconium = \(\frac{mass}{molar mass} = \frac{1.00 \times 10^3 kg}{91.22 g/mol} = 1.10 \times 10^4 mol\) Using the stoichiometry from the balanced equation in step 2, the number of moles of hydrogen produced is half the moles of zirconium: Moles of Hydrogen = \(\frac{1}{2} \times 1.10 \times 10^4 mol = 5.48 \times 10^3 mol\) Now convert moles of hydrogen into mass: Mass of hydrogen = moles × molar mass = \(5.48 \times 10^3 mol \times 2.02 g/mol = 1.11 \times 10^4 g\) Next, we need to calculate the volume of hydrogen produced at 1.0 atm and 1000°C. First, convert the temperature to Kelvin: Temperature = \(1000°C + 273.15K = 1273.15 K\) Using the ideal gas law, PV=nRT, find the volume of hydrogen: \(V = \frac{nRT}{P} = \frac{5.48 \times 10^3 mol \times 8.314 J/(K \cdot mol) \times 1273.15 K}{1.0 atm \times 101.3 J/(L \cdot atm)} = 6.97 \times 10^4 L\) The volume of hydrogen produced is \(6.97 \times 10^4 L\).

05

Discuss implications of venting the hydrogen

At Three Mile Island, hydrogen produced by the reaction of zirconium with water was successfully vented, preventing a chemical explosion. In the case of Chernobyl, hydrogen was produced by the reaction of superheated steam with the graphite reactor core, and it was not possible to vent it, leading to an explosion. Venting hydrogen and other radioactive gases into the atmosphere at Three Mile Island was the correct decision. It prevented a massive chemical explosion within the reactor itself, minimizing the release of radioactive substances and damage to the reactor. Although releasing hydrogen and radioactive gases into the atmosphere presents an environmental and health risk, the risk of a catastrophic explosion would have been significantly higher if the gases were not vented.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!