Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) 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 metal. c. Calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accident at Three Mile Island, Pennsylvania, in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{~kg} \mathrm{Zr}\) reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at \(1.0 \mathrm{~atm}\) and \(1000 .{ }^{\circ} \mathrm{C}\) is produced? e. At Chernobyl, USSR, in 1986 , hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ A chemical explosion involving the hydrogen gas did occur at Chernobyl. In light of this fact, 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

a. Capability of zirconium to reduce water

To determine if zirconium metal can reduce water to form hydrogen gas at standard conditions, we need to consider the standard reduction potential of water. For the reduction of water to hydrogen gas, the half-reaction is: \(2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}\) (Eº = -0.829V) Comparing the standard reduction potentials, we can observe that the given half-reaction of zirconium (Eº = -2.36V) is much more negative than that of water (Eº = -0.829V). This indicates that zirconium has a stronger tendency to be oxidized; thus, zirconium metal can indeed reduce water to form hydrogen gas at standard conditions.

02

b. Balanced equation for the reduction of water

To write the balanced equation for the reduction of water by zirconium metal, we need to combine the half-reaction of zirconium and the half-reaction of water: Zirconium half-reaction: \(ZrO_{2} \cdot H_{2}O + H_{2}O + 4e^{-} \rightarrow Zr + 4OH^{-}\) Water half-reaction (multiplied by 2 to balance electrons): \(4H_{2}O + 4e^{-} \rightarrow 2H_{2} + 4OH^{-}\) Upon adding both half-reactions, we obtain the following balanced equation: \(Zr + ZrO_{2} \cdot H_{2}O + 2H_{2}O \rightarrow Zr + 2H_{2} + 4OH^{-}\)

03

c. Standard cell potential, Gibbs free energy change, and equilibrium constant

To calculate the standard cell potential (Eº) for the reaction, we subtract the standard reduction potential of water (Eº_water) from the standard reduction potential of zirconium (Eº_zirconium): Eº = Eº_zirconium - Eº_water Eº = -2.36V - (-0.829V) = -1.531V Now we can determine the Gibbs free energy change (ΔGº) using the equation ΔGº = -nFEº, where n is the number of electrons transferred in the reaction (in this case, n=4) and F is Faraday's constant (96,485 C/mol): ΔGº = -(4)(96,485 C/mol)(-1.531V) = 592,860 J/mol To find the equilibrium constant (K), we can use the relationship ΔGº = -RTln(K), where R is the gas constant (8.314 J/mol*K) and T is the temperature in Kelvin (T=298K, as we assume standard conditions): 592,860 J/mol = -(8.314 J/mol*K)(298K) * ln(K) ln(K) = -24.55 K = \(e^{-24.55}\) ≈ 2.68 × 10^{-11}

04

d. Mass and volume of hydrogen produced

We have the balanced equation for the reaction: \(Zr + ZrO_{2} \cdot H_{2}O + 2H_{2}O \rightarrow Zr + 2H_{2} + 4OH^{-}\) Given 1.00 × 10^3 kg of Zr, we can determine the mass of hydrogen produced. First, convert the mass of Zr to moles using the molar mass of Zr (91.22 g/mol): \[(1.00 * 10^{3}\,kg)\,(\frac{1000\,g}{1\,kg})\,(\frac{1\,mol\,Zr}{91.22\,g\,Zr}) = 10,960\,mol\,Zr\] From the balanced equation, we see that 1 mole of Zr produces 2 moles of H2. Therefore, we can determine the number of moles of H2 produced from 10,960 moles of Zr: \[10,960\,mol\,Zr\,(\frac{2\,mol\,H_2}{1\,mol\,Zr}) = 21,920\,mol\,H_2\] To convert this amount of H2 to mass, use the molar mass of H2 (2.02 g/mol): \[21,920\,mol\,H_2\,(\frac{2.02\,g\,H_2}{1\,mol\,H_2}) = 44,300\,g\,H_2\] We are asked to determine the volume of hydrogen gas produced at 1.0 atm and 1000 ºC. Using the ideal gas law (PV=nRT), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin: \[V\,=\,\frac{nRT}{P}\,=\,\frac{(21,920\,mol)\,(0.0821\,\frac{L\,atm}{mol\,K})\,(1,273\,K)}{(1.0\,atm)}\,=\,2.28\,×\,10^6\,L\] The mass of H2 produced is 44,300 g, and the volume produced is 2.28 × 10^6 L.

05

e. Decision-making in light of the Chernobyl accident

At Chernobyl, a chemical explosion involving hydrogen gas occurred, but no such explosion occurred at Three Mile Island, where hydrogen was vented along with other radioactive gases. While it can be argued that releasing radioactive gases into the atmosphere is harmful to the environment, it may have been the correct decision at the time to vent the hydrogen and radioactive gases in order to prevent a potentially catastrophic chemical explosion. Ultimately, due to the different circumstances and reactor designs at Chernobyl and Three Mile Island, the decision to vent the gases likely helped prevent a more severe accident and reduce the potential for harm to the surrounding population.

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