Chapter 18, Problem 82CP

(a)

Interpretation Introduction

Interpretation: The given questions based upon the redox properties of zirconium are to be answered.

Concept introduction: Zirconium is one of the few metals that retain its structural integrity upon exposure to radiation. The fuel rods in most of the nuclear reactors are therefore often made up of zirconium.

To determine: whether zirconium metal is capable of reducing water to form hydrogen gas at standard conditions.

Answer to Problem 82CP

Answer

The zirconium metal is capable of reducing water to form hydrogen gas at standard conditions.

Explanation of Solution

The zirconium metal is capable of reducing water to form hydrogen gas at standard conditions.

The given half reaction is,

${\text{ZrO}}_{\text{2}}{\text{.H}}_{\text{2}}\text{O}+{\text{H}}_{\text{2}}\text{O}+{\text{4e}}^{-}\to \text{Zr}+{\text{4OH}}^{-}{E}_1{}^{\text{ο}}=-2.3\text{6 V}$ (1)

The half reaction for reduction of water is,

${\text{2H}}_{\text{2}}\text{O}+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}+2{\text{OH}}^{-}{E}_2{}^{\text{ο}}=-0.83\text{ V}$ (2)

The equation (1) is reversed, equation (2) is multiplied by 2 and both the equations are then added to get the resultant equation given below.

$\text{Zr}+{\text{3H}}_{\text{2}}\text{O}\to {\text{ZrO}}_2.{\text{H}}_{\text{2}}\text{O}+{\text{2H}}_2$ (3)

This equation shown that zinc metal is capable of reducing water to form hydrogen gas at standard conditions.

(b)

Interpretation Introduction

Interpretation: The given questions based upon the redox properties of zirconium are to be answered.

Concept introduction: Zirconium is one of the few metals that retain its structural integrity upon exposure to radiation. The fuel rods in most of the nuclear reactors are therefore often made up of zirconium.

Answer to Problem 82CP

Answer

The equation has been stated below.

Explanation of Solution

To determine: The balanced chemical equation for the reduction of water by zirconium.

The equation has been stated below.

The balance chemical equation for the reduction of water by zirconium is,

$\text{Zr}+{\text{3H}}_{\text{2}}\text{O}\to {\text{ZrO}}_2.{\text{H}}_{\text{2}}\text{O}+{\text{2H}}_2$

Water is reduced by zirconium to form hydrogen gas.

(c)

Interpretation Introduction

Interpretation: The given questions based upon the redox properties of zirconium are to be answered.

Concept introduction: Zirconium is one of the few metals that retain its structural integrity upon exposure to radiation. The fuel rods in most of the nuclear reactors are therefore often made up of zirconium.

To determine: The values of $E^{\text{ο}}$, $\Delta G^{\text{ο}}$ and $K$ for the reduction of water by zirconium metal.

Answer to Problem 82CP

Answer

The values of $E^{\text{ο}}$ is $\underset{\_}{1.53V}$, values of $\Delta G^{\text{ο}}$ is $\underset{\_}{5.90\times 10^{-5}J}$ and value of $K$ $\underset{\_}{3.2\times 10^{103}}$. The mass of hydrogen produced is $\underset{\_}{2.21\times 10^{4}g}$.

Explanation of Solution

The value of $E^{\text{ο}}$ is $\underset{\_}{1.53V}$.

The values of electrode potential, $E^{\text{ο}}$, for the reduction of water by zirconium metal is calculated by the formula,

$\begin{array}{l}{E}^{\text{ο}}=E_{\text{cathode}}{}^{\text{ο}}-E_{\text{anode}}{}^{\text{ο}}\\ E^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}\end{array}$

Where,

• $E_1{}^{\text{ο}}$ is the standard electrode potential of equation (1).
• $E_2{}^{\text{ο}}$ is the standard electrode potential of equation (2).

Substitute the value of $E_1{}^{\text{ο}}$ and $E_2{}^{\text{ο}}$ in the above formula.

$\begin{array}{c}{E}^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}\\ =-0.\text{83 V -}\left(\text{-2}\text{.36 V}\right)\\ =\underset{\_}{1.53V}\end{array}$

Therefore, the value of $E^{\text{ο}}$ is $\underset{\_}{1.53V}$.

The value of $\left(\Delta G^{\text{ο}}\right)$ is $\underset{\_}{5.90\times 10^{-5}J}$.

The standard free energy change $\left(\Delta G^{\text{ο}}\right)$ for the reaction is calculated by the formula,

$\Delta G^{\text{ο}}=-nF{E}^{\text{ο}}$

Where,

• $F$ is faraday constant $\left(9648{\text{5 C/mol e}}^{-}\right)$.
• $n$ is number of electrons.

The value of $n$ is $4$.

Substitute the value of $n$, $F$ and $E^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-4\times 96485\times 1.53\\ =\underset{\_}{5.90\times 10^{-5}J}\end{array}$

Therefore, the value of $\left(\Delta G^{\text{ο}}\right)$ is $\underset{\_}{5.90\times 10^{-5}J}$.

The mass of hydrogen produced is $\underset{\_}{2.21\times 10^{4}g}$.

The equilibrium constant $\left(K\right)$ is calculated by the formula,

$\text{log }K=\frac{n{E}^0}{0.0591}$

Substitute the value of $E^{\text{ο}}$, $n$ in the above equation.

$\begin{array}{c}\text{log }K=\frac{4\times 1.53}{0.0591}\\ K=\underset{\_}{3.2\times 10^{103}}\end{array}$

The value of equilibrium constant $\left(K\right)$ is $\underset{\_}{3.2\times 10^{103}}$.

Molar mass of $\text{Zr}$ is $\text{91}\text{.22 g/mol}$.

The conversion of $\text{Kg}$ to $\text{g}$ is done as,

$\text{1 Kg}={10}^{\text{3}}\text{ g}$

Therefore, the conversion of ${10}^{\text{3}}\text{ Kg Zr}$ to $\text{g}$ is done as,

$\begin{array}{c}{10}^{\text{3}}\text{ Kg Zr}={10}^3\times {10}^3\text{ g Zr}\\ {\text{=10}}^6\text{ g Zr}\end{array}$

The amount of hydrogen formed by $\text{91}\text{.22 g Zr}=\text{2 mol}$.

Therefore, the amount of hydrogen formed by $\begin{array}{c}{\text{10}}^6\text{ g Zr}=\frac{2}{91.22}\times {10}^{\text{6}}{\text{ mol H}}_{\text{2}}\\ =2.19\times {10}^4{\text{ mol H}}_{\text{2}}\end{array}$

Mass of hydrogen produced from ${\text{1 mol H}}_{\text{2}}=1.00\text{8 g}$.

Therefore, the mass of hydrogen produced from $\begin{array}{c}2.19\times {10}^4{\text{ mol H}}_{\text{2}}=2.19\times {10}^4\times 1.0\text{08 g}\\ \text{=}\underset{\_}{2.21\times 10^{4}g}{\text{ H}}_{\text{2}}\end{array}$

Therefore, the mass of hydrogen produced is $\underset{\_}{2.21\times 10^{4}g}$.

The volume of hydrogen gas is $\underset{\_}{228.8L}$.

The volume of ${\text{H}}_{\text{2}}$ is calculated by the ideal gas equation given below.

$PV=nRT$

Where,

• $P$ is the pressure of hydrogen gas.
• $V$ is the volume. of hydrogen gas
• $R$ is universal gas constant $\left(\text{0}{\text{.0821 L atm mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}\right)$.
• $T$ is the temperature.
• $n$ is number of moles.

The value of $P$ is $\text{1 atm}$.

The value of $T$ is ${\text{100 }}^{\text{ο}}\text{C}=\text{1273 K}$

The value of $n$ is $\text{2}{\text{.21×10}}^{\text{4}}$.

Substitute the value of $n$, $T$, $R$ and $P$ in the above equation.

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{\text{2}{\text{.21×10}}^{\text{4}}\times 0.0821\times 1273}{1}\\ =\underset{\_}{228.8L}\end{array}$

Therefore, the volume of hydrogen gas is $\underset{\_}{228.8L}$.

(d)

Interpretation Introduction

Interpretation: The given questions based upon the redox properties of zirconium are to be answered.

Concept introduction: Zirconium is one of the few metals that retain its structural integrity upon exposure to radiation. The fuel rods in most of the nuclear reactors are therefore often made up of zirconium.

To determine: whether it was correct decision to vent the hydrogen and other radioactive gases into the atmosphere or not.

Answer to Problem 82CP

Answer

The decision was not correct.

Explanation of Solution

The decision was not correct.

At Chernobyl in $1986$, hydrogen was produced by the reaction of superheated steam with the graphite reactor core. The reaction that took place was,

$\text{C}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\to \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$

It was not possible to prevent a chemical explosion at Chernobyl. It was not a correct decision to vent the hydrogen and other radioactive gases into the atmosphere because of safety and waste disposal issues.