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. Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+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).

Tutorial Answer A17.4.3

1. Hg(l)+S₂−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)

To find the Standard Cell Potential:

a) E^o=E^oCathode-E^oAnode

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

b)You use The formula listed above. For that, you will need to look at the Standard Cell Potentials chart. Remember that within electrochemical reactions, oxidation occurs in the Anode and reduction occurs in the Cathode. In which for this case indicates that the oxidation half-reaction is:

Hg(l)+H₂(g)⟶HgS(s)+2H⁺+2e^- E^o=-0.72V

\[ \text{Hg}(\mathit{l})+\text{H}_2(\mathit{g}) \rightarrow \text{HgS}(\mathit{s})+\text{2H}^{+}+\text{2e}^- \]

\[ \text{E}^o(\mathit{Anode})=-0.72V \]

The reduction half reaction is:

c)Ag⁺(aq)+e^- ⟶Ag(s) (This is a reduction because of gain of electrons)

\[ \text{Ag}^{+}(\mathit{aq})+\text{e}^- \rightarrow \text{Ag}(\mathit{s}) \]

E^oCathode=0.7994

\[ \text{E}^o(\mathit{Cathode})=0.7994V \]

Plugging the two cell potentials into the equation:

E^o=0.7994-(-0.72)

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

\[ \text{E}^o= (0.7994) - (-0.72) \]

d)E^o=1.5V

\[ \text{E}^o= 1.5V \]

Because the Standard Cell Potential is positive, the reaction is spontaneous!

To Find the Cell potential

Because the given chemical reaction is not under standard conditions, to find the cell potential of the reaction we must use the Nernst equation!

\({E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}\)

n=number of electrons, in this case 2
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.10)(0.25)² =0.00625

Then we plug it all into the formula:

\({E_{cell}}={1.51}-{\frac{0.0592}{2}}{(log(0.00625))}\)

\({E_{cell}}={1.43V}\)

It is Spontaneous since it the E_cell is positive.

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.

First, write the corresponding half-reactions:

Al³⁺(aq) + 3 e^– Al(s) E^o=-1.662

\[ \text{Al}^{3+}(\mathit{aq})+\text{3e}^- \rightarrow \text{Al}(\mathit{s}) \; \text{E}^o = -1.662V \]

Ni²⁺(aq) + 2 e^– Ni(s) E^o=-0.257

\[ \text{Ni}^{2+}(\mathit{aq}) + \text{2e}^- \rightarrow \text{Ni}(\mathit{s}) \; \text{E}^o = -0.257V \]

Since Nickel is the most positive in terms of Standard Potential Values is the reduction, which will be in the cathode and Aluminum in the Anode, oxidation.

Looking at the given cell potential of the half-reactions we can identify which reaction would take place at the Cathode and Anode. Because the Nickel has a more positive Standard Cell Potential value we can infer that it would most likely be reduced, while the more negative Standard Cell Potential from Aluminum would indicate it would be oxidized. Thus, reduction of Nickel occurs at the cathode and the oxidation of Aluminum occurs at the Anode.

To find the Standard Cell Potential:

a) E^o=E^oCathode-E^oAnode

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

E^o=-0.25-(-1.66)=1.41V

\[ \text{E}^o= (-0.257) - (-1.662) \]

\[ \text{E}^o = 1.405V \]

Because the Standard Cell Potential is positive, the reaction is spontaneous!

To Find the Cell potential

a) \({E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}\)

n=number of electrons, in this case 6
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.25)³(0.015)² =3.5156E-6

\[ Q= \frac{(0.015)^2}{(0.25)^3} \]

\[ Q= 0.0144 \]

All this was from the balanced formula of the reaction, which indicates 6 electrons and the nickel with a coefficient of 3 and Al with a coefficient of 2.

Then we plug it all into the formula:

\({E_{cell}}={1.41}-{\frac{0.0592}{6}}{(log(3.516E-6)}\)

\({E_{cell}}={1.405}-{\frac{0.0592}{6}}{(log(.0144)}\)

E_cell=1.46V

\[ \text{E}(\mathit{Cell}) = 1.423V \]

Spontaneous since the Standard Cell is positive.

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)

a) First write the formula:

Al³⁺(aq) + 3 e^– Al(s) E^o=-1.662V

\[ \text{Al}^{3+}(\mathit{aq})+\text{3e}^- \rightarrow \text{Al}(\mathit{s}) \; \text{E}^o = -1.662V \]

Br₂(l) + 2 e^– 2 Br^–(aq) E^o=+1.066 1.0873

\[ \text{Br}_2(\mathit{l})+\text{2e}^- \rightarrow \text{2Br}^-(\mathit{aq}) \; \text{E}^o = 1.0873V \]

To find the Standard Cell Potential:

a) E^o=E^oCathode-E^oAnode

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

E^o=-1.66-1.066=-2.726V

\[ \text{E}^o= (-1.662) - (1.0873) \]

\[ \text{E}^o= -2.749V \]

Because the Cell Potential is negative the reaction is non-spontaneous!

To Find the Cell potential

\({E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}\)

n=number of electrons, in this case 6
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.11)³(0.023)² =7.04E-7

\[ Q= \frac{(0.11)^3}{(0.023)^2} \]

\[ Q= 2.52 \]

All this was from the balanced formula of the reaction, which indicates 6 electrons and the Br with a coefficient of 3 and Al with a coefficient of 2.

Then we plug it all into the formula:

E_cell=-2.726-\(frac{0.0592}{6})(log(7.04E-7\)

\({E_{cell}}={2.726}-{\frac{0.0592}{6}}{(log(7.04E-7)}\)

\({E_{cell}}={-2.749}-{\frac{0.0592}{6}}{(log(2.52)}\)

E_cell=-2.665V

\[ \text{E}(\mathit{Cell}) = -2.753V \]

Non Spontaneous since Standard Cell potential was negative

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

_{Tutorial Answer A12.1.3}

To find out the rate, first for the reactant it will be a rate of dissolving over time so it will be negative and for the products it will be positive. In the rate, the coefficients matter since they are part of the ratio over Δt so you write them in the numerator times Δt.

In order for us to write the rate expression for the question above, we have to understand what a reaction rate tell us. Recall from section 4.2, that reactions rates detail how much reactant is consumed per unit of time and the amount of product formed per a unit of time. Also remember, that a reaction rate is equal throughout; the change of reactants and product per a unit of time should be equivalent to the overall rate.

Looking at the given equation, we can see that Cl₂ has the smallest coefficient in the chemical reaction. Thus, the reaction rate should be defined as the rate of change of [Cl₂]. Since Cl₂ and F₂ are reactants, their reaction rates should be negative due them being consumed, while ClF₃ should have a positive rate due to it being a product of the reaction.

So you write the rate equation:

\({rate}=+\frac{1}{2}\frac{ΔCIF3}{Δt}=-\frac{ΔCl2}{Δt}=\frac{1}{3}\frac{ΔF2}{Δt}\)

\({rate}=+\frac{1}{2}\frac{Δ[CIF_3]}{Δt}=-\frac{Δ[Cl_2]}{Δt}=\frac{1}{3}\frac{Δ[F_2]}{Δt}\)

Q12.5.1

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

Tutorial A12.5.1

For reactions to collide they have to have an orientation and enough energy for them to collide. To prevent collision, two factors might be not have an orientation that will make them collide and not have enough or too much energy.

Recall from section 4.7, that in order for a chemical reaction to occur, reactant particles must collide with enough energy required to overcome the activation energy and with the proper orientation. Two factors that can prevent a collision from producing chemical reactions are:

1. The reactant particles that collided, did not contain enough energy to overcome the activation energy to break chemical bonds.
2. The reactant particles collided with an orientation that did not allow the proper atoms to align with one another and break/form bonds for the correct product.

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 {_1^2{H}} bombardment

Tutorial A21.3.5

1. α is the same as \(_4^2He\), so when finding a balanced equation both sides have to equal the numbers. Then for the atomic number for N is 7 and for Oxygen is 8, that is a 1 difference, so that is why you add \(_1^1H\) so the atomic numbers match up and then you put 17 for O since you have to account for the 1 in Hydrogen

\(^{14}N\) + \(_4^2He\) ⟶ \(_8^{17}O\)+\(_1^1H\)

2. neutron has 0 for atomic number and 1 for mass \(_0^1n\). The atomic number for N is 7 and for Carbon is 6, since both sides have to equal the same you add \(_1^1H\) so the atomic numbers match up and then you put 14 for C since you have to account for the 1 in Hydrogen.

\(^{14}N\) + \(_0^1He\) ⟶ \(_6^{14}O\)+\(_1^1H\)

3. neutron has 0 for atomic number and 1 for mass \(_0^1n\). The atomic number for Th is 90 since the neutron has 0 for atomic number, then it is still Th but you increase the atomic mass from 233 to 232.

\(_90^{232}Th\)+\(_0^1n\)⟶\(_90^{233}Th\)

\[ {^{90}_{232}Th}+_0^1n \rightarrow _90^{233}Th \]

4. \(_1^2H\) had atomic number 1 and 2 for atomic mass, so by adding it but keep Uranium as an element, then you add \(_1^1H\) so the atomic number continues the same and only add 1 for the atomic mass.

\(_92^{238}U\)+ \(_1^{2}H\)⟶\(_92^{239}U\)+\(_1^1H\)

\[ {^{238}_{292}U}+ {^{2}_{1}H} \rightarrow {^{239}_{92}U}+{^{1}_{1}H} \]

Q20.2.1

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

Tutorial A20.2.1

Halogens are good oxidants because they only have one vacant space in outer orbit, they are pull strongly electrons from other elements. Alkali metals have only one electron in outer orbit, and they can loose it very easily, indicating they oxidize very easily making them good reducing agents/reductants.

Halogens are good oxidants due to their large electronegativity, caused by their almost filled electron shell. This allows Halogens to easily remove electrons from other elements to obtain a filled electron shell, making them effective oxidants. Alkali metals are considered good reductants due to them only having one valence electron. Because Alkali metals have only one valence electron, their ionization energy value is low and can easily lose their electron to other elements, making them good reducing agents!

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)

Tutorial A20.4.20

For both questions, first, find the half equations and its standard potentials, then find the Standard Potential Cell with E^o=E^oCathode-E^oAnode. Then depending if the standard potential is positive, it is spontaneous and if negative, is nonspontaneous.

1. a)Find Half Equations

2 H⁺(aq) + 2 e^– H₂(g) (reference electrode) E^o=0

\[ \text{2H}^+(\mathit{aq})+\text{2e}^- \rightarrow \text{H}_2(\mathit{g}) \text{(Reference Standard Reduction Potential)} \; \text{E}^o=0V \]

Cu⁺(aq) + e^– Cu(s) E^o=+0.521

\[ \text{Cu}^{2+}(\mathit{aq}) + \text{e}^- \rightarrow \text{Cu}(\mathit{s}) \; \text{E}^o = 0.337V \]

From the overall equation, you can see that H is being reduced while Cu is being oxidized. Then H will be in the cathode and Cu in anode.

Then find the E^o.

b)E^o=E^oCathode-E^oAnode

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

E^o=0-0.521=-0.521

\[ \text{E}^o= (0) - (0.337) \]

\[ \text{E}^o= -0.337V \]

Which indicates it is nonspontaneous

2.a) Find Half equations

Zn²⁺(aq) + 2 e^– Zn(s) E^o=-0.763

\[ \text{Zn}^{2+}(\mathit{aq})+\text{2e}^- \rightarrow \text{Zn}(\mathit{s}) \; \text{E}^o=-0.763 \]

Pb²⁺(aq) + 2 e^– Pb(s) E^o=-0.126

\[ \text{Pb}^{2+}(\mathit{aq})+\text{2e}^- \rightarrow \text{Pb}(\mathit{s}) \; \text{E}^o=-0.126 \]

From the overall reaction, you can tell that Zn is being reduced since it is gaining electrons then it ill be cathode and Pb then is being oxidized so it will be the anode.

b)E^o=E^oCathode-E^oAnode

\[ \text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode}) \]

E^o=-0.763-(-0.126)=-0.637

\[ \text{E}^o= (-0.763) -(-0.126) \]

\[ \text{E}^o= -0.637V \]

It is nonspontaneous.

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

_{Tutorial A20.9.5}

1. For this salt, you find half equation of Al and Cl

Al³⁺(l) + 3e- Al(s)

3Cl^- (aq) Cl₃ (s) + 3e^-

From the half equations, you can see that there are three electrons, so you set a ratio to convert the moles of electron to grams of metal.

\(\frac{3\,moles\,e^-}{1}\frac{1 mole\,of\,AlCl_3}{1\,mole\,of\,e^-}\frac{133.34g}{1\,mole\,of\,AlCl_3}\)\(={400.02g\,of\,AlCl_3}\)

\(\frac{1\,moles\,e^-}{1}\frac{1 mole\,of\,Al}{3\,mole\,of\,e^-}\frac{26.98g}{1\,mole\,of\,Al}\)\(={8.99 g\,of\,Al}\)

2. For this salt, you find half equation of Mg and Cl

2Cl^- (aq) Cl₂ (s) + 2e^-

Mg²⁺(aq)+2e^-Mg(s)

From the half equations, you can see that there are two electrons, so you set a ratio to convert the moles of electron to grams of metal.

\(\frac{2\,moles\,e^-}{1}\frac{1\,mole\,of\,MgCl_2}{1\,mole\,of\,e^-}\frac{95.211g}{1\,mole\,of\,MgCl_2}={190.422g\,of\,MgCl_2}\)

\(\frac{1\,moles\,e^-}{1}\frac{1\,mole\,of\,Mg}{2\,mole\,of\,e^-}\frac{24.31g}{1\,mole\,of\,Mg}={12.155g\,of\,Mg}\)

3. For this salt, you find half equation of Fe and Cl

3Cl^- (aq) Cl₃ (s) + 3e^-

Fe³⁺ (aq)+ 3e^-Fe(s)

From the half equations, you can see that there are three electrons, so you set a ratio to convert the moles of electron to grams of metal.

\(\frac{3\,moles\,e^-}{1}\frac{1\,mole\,of\,FeCl_3}{1\,mole\,of\,e^-}\frac{162.204g}{1\,mole\,of\,FeCl_3)}={486.612g\,of\,FeCl_3}\)

\(\frac{1\,moles\,e^-}{1}\frac{1\,mole\,of\,Fe}{3\,mole\,of\,e^-}\frac{55.845g}{1\,mole\,of\,Fe)}={18.615g\,of\,Fe}\)

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?

Tutorial A14.5.1

Activation energy is required for a collision between molecules to result in a chemical reaction, as well it is related to the rate due to the Arrhenius equation

Activation energy is the threshold of energy needed in order for a reaction to occur. Reactant particles must collide with enough energy to be able to break chemical bonds, that will then allow the creation of new bonds. If the particles do not have enough energy, when they collide they will simply bounce off one another.

k=Ae^(-Ea/RT). The increase in temperature increases the rate of reaction despite the fact the average kinetic is less than the activation energy since it increase the rate of collision between molecules.

With an increase in temperature, there is a greater distribution of kinetic energy among reactant particles. This increase of temperature allows the rate in which particles collide with one another to increase. Although the average kinetic energy is still lower than the activation energy, the increase of collisions among particles increases the chance of particles that contain enough energy to overcome the energy barrier to collide. Thus, the reaction rate increases due to this increased rate of collisions.