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 \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) 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. 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 \(1.0 \mathrm{~atm}\) 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: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$

Step by step solution

01

Compare standard reduction potentials

To determine if Zirconium can reduce water to hydrogen gas, we must compare the standard reduction potential of Zirconium to the standard reduction potential of water. For water, the standard reduction potential is \(E^{\circ}_{H_2O} = 0V\). Since the given standard potential of Zirconium is \(E^{\circ}_{Zr} = -2.36V\), it is less than that of water. Therefore, Zirconium can reduce water to hydrogen gas under standard conditions, since a negative reduction potential implies that the reaction is spontaneous and feasible. #b. Balanced equation for reduction of water#

02

Write balanced equation

To write the balanced equation for the reduction of water by Zirconium, we first express the half-reactions for Zirconium and water: Zirconium: \[ZrO_2 \cdot H_2O + H_2O + 4e^- \to Zr + 4OH^-\] Water: \[2H_2O + 2e- \to H_2 + 2OH^-\] Now we must balance the half-reactions by equating the number of electrons in each: Zirconium: \(4e^-\) Water: \(2 \times 2e^-=4e^-\) Finally, we add the balanced half-reactions: \[2(ZrO_2 \cdot H_2O + H_2O) \to 2Zr + 8OH^-\] \[4H_2O + 4e^- \to 2H_2 + 4OH^-\] \square \[\boxed{2ZrO_2 \cdot H_2O + 2H_2O \longrightarrow 2Zr + 2H_2 + 4OH^- }\] #c. Calculation of E, ΔG, and K for reduction of water by Zirconium#

03

Nernst equation and standard equation

To calculate these values, we utilize the Nernst equation for E: \[E = E^{\circ} - \frac{RT}{nF} \ln{Q}\] For the standard potential, since the only electron transfer involved is of Zirconium reducing water which has \(E^{\circ}_{Zr} = -2.36V\), we have \(E^{\circ} = -2.36V\). For the Gibbs free energy, we use the relation: \[\Delta G^{\circ} = -nFE^{\circ}\] Where: n = 4 moles of electrons F = 96485 C/mol (Faraday's constant) Plugging the values, we get: \[\Delta G^{\circ} = -4(96485)(-2.36) J/mol = 909,717 J/mol\] For the equilibrium constant K, we use the relationship: \[K = e^{-\Delta G^{\circ}/(RT)}\] Where: R = 8.314 J/mol ⋅ K (gas constant) T = 298 K (standard conditions) \[K = e^{-909717/(8.314 × 298)} = 3.53 × 10^{50}\] Hence, for the reduction of water by Zirconium, we have: \(\mathscr{E}^{\circ} = -2.36 V\) \(\Delta G^{\circ} = 909,717 J/mol\) \(K = 3.53 × 10^{50}\) #d. Calculation of mass and volume of hydrogen produced#

04

Moles and volume of hydrogen

Given the reaction involves \(1.00 \times 10^3 kg\) of Zirconium, we can determine the amount of hydrogen produced using stoichiometry: Molecular weight of Zr: \[91g/mol\] Moles of Zirconium: \[\frac{1.00 \times 10^3 kg}{91g} \times \frac{1mol}{1000g} = 11mol\] Since the balanced equation is \(2ZrO_2 \cdot H_2O + 2H_2O \longrightarrow 2Zr + 2H_2 + 4OH^-\), every 2 moles of Zirconium will produce 2 moles of hydrogen gas Moles of hydrogen produced: \[\frac{11mol}{2} \times 2 = 11mol\] Mass of hydrogen produced: \[11mol \times \frac{2g}{1mol} = 22g\] To find the volume of hydrogen, we use the ideal gas law equation \(PV=nRT\): Volume of hydrogen produced: \[V=\frac{nRT}{P}\] Where P = 1 atm R = 0.0821 L ⋅ atm/mol ⋅ K T = 1273 K (given as 1000°C, but we must convert to Kelvin) \[V = \frac{11(0.0821)(1273)}{1} L = 920 L\] Hence, the produced hydrogen has: Mass \(= 22g\) Volume \(= 920 L\) #e. Hydrogen production at Chernobyl#

05

Mention reaction at Chernobyl

The hydrogen production at Chernobyl occurred due to the following reaction: \[C(s) + H_2O(g) \longrightarrow CO(g) + H_2(g)\] At Chernobyl, the reaction occurred between superheated steam and the graphite reactor core, producing hydrogen gas.

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!