Extra Credit 30

Q17.4.3

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

1. \(\ce{Hg}(l)+\ce{S^2-}(aq,\: 0.10\:M)+\ce{2Ag+}(aq,\: 0.25\:M)⟶\ce{2Ag}(s)+\ce{HgS}(s)\)
2. The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br₂(l) is the same as that of Br₂(aq).

Solution:

a) The overall reaction is:

\[anode: Hg(l)+S^{2-}(aq, 0.25M)\rightarrow HgS(s)+2e^- \space \space \space \space \space E^{o}_{anode}=-0.7V\]

\[cathode: 2\times(Ag^+(aq, 0.25M)+e^-\rightarrow Ag(s))\space \space \space \space \space E^{o}_{cathode}=0.7996V\]

\[overall: Hg(l)+S^{2-}(aq)+2Ag^+(aq) \rightarrow 2Ag(s)+HgS(s)\]

To calculate standard cell potential, we use the equation:

\[\begin{align}
E^{o}_{cell}&=E^{o}_{cathode}-E^{o}_{anode}\\
&= 0.7996V-(-0.7V)\\
&= 1.50V \\
\end{align}\]

Since the value of E ^o_cell is positive, the reaction is spontaneous. Therefore, the reaction will move towards the products because of a negative delta G.

b) The overall reaction is:

\[anode: 2\times(Al(s) \rightarrow Al^{3+}(aq, 0.015M)+3e^-) \space \space \space \space \space E^{o}_{anode}=-1.662V\]

\[cathode: 3\times(Ni^{2+}(aq, 0.25M)+2e^-\rightarrow Ni(s))\space \space \space \space \space E^{o}_{cathode}=-0.257V\]

\[overall: 2Al(s)+3Ni^{2+}(aq, 0.25M) \rightarrow 2Al^{3+}(aq, 0.015M)+3Ni(s)\]

To calculate standard cell potential, we use the equation:

\[\begin{align}
E^{o}_{cell}&=E^{o}_{cathode}-E^{o}_{anode}\\
&= -0.257-(-1.662V)\\
&= 1.405V \\
\end{align}\]

Since the value of E ^o_cell is positive, the reaction is spontaneous.

c) The overall reaction is:

\[anode: 3\times(Br_{2}(aq, 1.0M) \rightarrow 2Br^{-}(aq, 0.11M)+2e^-) \space \space \space \space \space E^{o}_{anode}=1.0873V\]

\[cathode: 2\times(Al^{3+}(aq, 0.023M)+3e^-\rightarrow Al(s))\space \space \space \space \space E^{o}_{cathode}=-1.662V\]

\[overall: 2Al^{3+}(aq, 0.023M)+3Br_{2}(aq, 1.0M) \rightarrow 6Br^{-}(aq, 0.11M)+2Al(s)\]

To calculate standard cell potential, we use the equation:

\[\begin{align}
E^{o}_{cell}&=E^{o}_{cathode}-E^{o}_{anode}\\
&= -1.662V-1.0873V\\
&= -2.749V \\
\end{align}\]

Since the value of E ^o_cell is negative, the reaction is non-spontaneous, meaning it will not favor the products.

Q12.1.3

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Cl₂(g)+3F₂(g)⟶2ClF₃(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl₂ and F₂ and the formation of ClF₃.

Solution:

From the equation

\[Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)\]

We can see that both Cl₂ and F₂ are decomposing to form ClF₃, so the rate equation can be written as the disappearance (negative sign) of Cl₂ and F₂ and formation (positive sign) of ClF_3.

_{When writing the rate expressions, we must take coefficients into consideration which tell us the rate at which the reactant is disappearing or forming.}

That is, considering to the stoichiometry:

\[rate = -\dfrac{\Delta [Cl_2]}{\Delta t}=-\dfrac{1}{3} \dfrac{\Delta [F_2]}{\Delta t}=\dfrac{1}{2}\dfrac{\Delta [ClF_3]}{\Delta t}\]

Q12.5.1

Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

Solution:

The two factors that may prevent a collision from producing a chemical reaction are:

1. Kinetic energy of the molecule

In order for chemical reactions to occur, molecules require enough velocity to overcome the minimum activation energy needed to break the old bonds and form new bonds with other molecules. At higher temperatures, the molecules possess the minimum amount of kinetic energy needed which ensures the collisions will be energetic enough to lead to a reaction.

2. The orientation of molecules during the collision

Two molecules have to collide in the right orientation in order for the reaction to occur. Molecules have to orient properly for another molecule to collide at the right activation state.

Q21.3.5

Write a balanced equation for each of the following nuclear reactions:

1. the production of ¹⁷O from ¹⁴N by α particle bombardment
2. the production of ¹⁴C from ¹⁴N by neutron bombardment
3. the production of ²³³Th from ²³²Th by neutron bombardment
4. the production of ²³⁹U from ²³⁸U by \(^{2}_{1}H\) bombardment

Solution:

\[ \begin{align}&1. \ce{^{14}_{7}N} + \ce{^{4}_{2}He} \rightarrow \ce{^{17}_{8}O} +\ce{^{1}_{1}H}\\&2. \ce{^{14}_{7}N} + \ce{^{1}_{0}n} \rightarrow \ce{^{14}_{6}C} +\ce{^{1}_{1}H}\\&3. \ce{^{232}_{90}Th} + \ce{^{1}_{0}n} \rightarrow \ce{^{233}_{90}Th}\\&4. \ce{^{238}_{90}Th} + \ce{^{2}_{1}H} \rightarrow \ce{^{239}_{90}U} + \ce{^{1}_{1}H}\end{align}\]

A hydrogen atom must be added as a product in order to properly balance the following reactions according to the mass number as well as the atomic number while taking the type of radiation into consideration.

Q20.2.1

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

Solution:

Elements with high electronegativity tend to GAIN electrons, and elements with low electronegativity tend to LOSE electrons. Therefore elements closer to the upper right corner of the periodic table are good oxidants because they tend to go reduction reactions. Elements closer to the bottom left corner are good reductants because they tend to go oxidation reactions. Group 16 has many good oxidizing agents, getting stronger as you move from bottom to top of the group.

Q20.4.20

Will each reaction occur spontaneously under standard conditions?

1. Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
2. Zn²⁺(aq) + Pb(s) → Zn(s) + Pb²⁺(aq)

Solution:

To determine whether a reaction occurs spontaneously under standard conditions, we first have to calculate the standard cell potential \(E^{o}_{cell}\) using the standard reduction potential of the reactants. If the E^° value is positive, then it will occur spontaneously.

a. \[anode: Cu(s) \rightarrow Cu^{2+}(aq)+2e^- \space \space \space \space \space E^{o}_{anode}=0.337V\]

\[cathode: 2H^{+}(aq)+2e^-\rightarrow H_2(g)\space \space \space \space \space E^{o}_{cathode}=0V\]

\[overall: Cu(s)+2H^{+}(aq) \rightarrow Cu^{2+} + H_2(g)\]

To calculate standard cell potential, we use the equation:

\[\begin{align}
E^{o}_{cell}&=E^{o}_{cathode}-E^{o}_{anode}\\
&= 0-(0.337V)\\
&= -0.337V \\
\end{align}\]

Since the value of E ^o_cell is negative, the reaction is non-spontaneous.

b.

\[anode: Pb(s) \rightarrow Pb^{2+}(aq)+2e^- \space \space \space \space \space E^{o}_{anode}=-0.126V\]

\[cathode: Zn^{2+}(aq)+2e^-\rightarrow Zn(s)\space \space \space \space \space E^{o}_{cathode}=-0.7618V\]

\[overall: Pb(s)+Zn^{2+}(aq) \rightarrow Pb^{2+} + Zn(s)\]

To calculate standard cell potential, we use the equation:

\[\begin{align}
E^{o}_{cell}&=E^{o}_{cathode}-E^{o}_{anode}\\
&= -0.7618-(-0.126V)\\
&= -0.6358V \\
\end{align}\]

Since the value of E ^o_cell is negative, the reaction is non-spontaneous.

Q20.9.5

The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?

1. AlCl₃
2. MgCl₂
3. FeCl₃

Solution:

The number of electrons must be written as a fraction over 1 and multiplied by the molar mass of the metal present during the electrolysis.

1. \[AlCl_3(s) \rightarrow Al^{3+}(aq)+3Cl^{-}(aq)\]

\[Al^{3+}(aq)+3e^{-} \rightarrow Al(s)\]

Since we need 3 moles of electrons to form one mole of Al, one mole of electron can only make \(\dfrac{1}{3}\) of Al. Hence,

\[ \dfrac{1}{3}mole \times 26.982 grams/mole = 8.994 grams\]

2. \[MgCl_2(s) \rightarrow Mg^{2+}(aq)+2Cl^{-}(aq)\]

\[Mg^{2+}(aq)+2e^{-} \rightarrow Mg(s)\]

Since we need 2 moles of electrons to form one mole of Mg, one mole of electron can only make \(\dfrac{1}{2}\) of Al. Hence,

\[ \dfrac{1}{2}mole \times 24.305 grams/mole = 12.1525 grams\]

3. \[FeCl_3(s) \rightarrow Fe^{3+}(aq)+3Cl^{-}(aq)\]

\[Fe^{3+}(aq)+3e^{-} \rightarrow Fe(s)\]

Since we need 3 moles of electrons to form one mole of Mg, one mole of electron can only make \(\dfrac{1}{3}\) of Mg. Hence,

\[ \dfrac{1}{3}mole \times 55.933 grams/mole = 18.644 grams\]

Q14.5.1

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

Solution:

An increase of temperature results in an increase of kinetic energy. With enough kinetic energy to climb the activation energy barrier, molecules are able to react. Although the average kinetic energy is less than the activation energy, not all the molecules have the same kinetic energy. Since the average kinetic energy is increased, the energy needed to overcome the activation energy is much smaller, hence more molecules can overcome the activation energy, despite the average kinetic energy is less than the activation energy. Reactants have enough energy because they have overcome the activation energy threshold needed to collide and form the products.