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).

A17.4.3

Given: Components of an electrochemical cell at 295.15 K

Asked For: Standard cell potential and cell potential under given conditions; spontaneity

Strategy:

1. Identify the oxidation half-reaction and the reduction half-reaction. Assign their standard reduction potential with this table or the table below. Then identify the anode and cathode from the half-reaction that occurs at each electrode.

2. Use the equation (ε^o_Cell= ε^o_Cathode- ε^o_Anode) to determine ε^o, standard cell potential. If ε^o is positive, the reaction is spontaneous. If it is negative, than it is not spontaneous.

3. Use the Nernst equation ε_Cell = ε^o_Cell - (0.0592/n)*logQ to find cell potential under the given conditions.

SOLUTION

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

Split this into the half reactions:

Hg(l)+S²⁻(aq,0.10M) ⟶ HgS(s) + 2e^- ε^o=-0.7 V (anode because electrons are being lost in given reaction)

2Ag⁺(aq,0.25M) + e^- ⟶ 2Ag(s) ε^o=0.8 V (cathode because electrons are being gained in given reaction)

Multiply the Ag reaction by 2 to balance the electrons transferred. Combining these two reactions gives you

\(\ce{Hg}(l)+\ce{S^{2-}}(aq)+\ce{2Ag+}(aq)+2e^{-}⟶\ce{2Ag}(s)+\ce{HgS}(s)+2e^{-}\)

Determine the standard cell potential:

\(ε^o_{Cell}= 0.8 - (-.0.7)=1.5V\) Because ε^o is positive, the reaction is spontaneous.

Find the cell potential using the Nernst equation:

\(ε_{Cell}= 1.5 - \frac{0.0592}{2}log(\frac{1}{(0.1)(0.25)^2})=1.43V\) Because ε is positive, the reaction is spontaneous under given conditions.

2. By definition, a galvanic cell is one driven by chemical reactions and is spontaneous.

Get the aluminum and nickel half reactions:

\(\ce{Al^{3+}}(aq)+\ce{3e^{-}}⟶\ce{Al}(s)\) \(ε^o=-1.66V\)

\(\ce{Ni2+}(aq)+2e^{-}⟶\ce{Ni}(s)\) \(ε^o=-0.26V\)

To make this cell spontaneous, we need to assign the anode and cathode so that \(ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}\) results in a positive value. Therefore, the aluminum electrode is the anode and the nickel electrode is the cathode. and \(ε^o_{Cell}= -0.26 - (-1.66)=1.4 \)

Next, multiply the aluminum anode by 2(to balance electrons) and flip it because the anode is the electrode in which electrons are being lost. Multiply the nickel electrode by 3(to balance electrons) and combine the two half reactions to get the complete cell.

\(\ce{3Ni^{2+}}(aq)+\ce{2Al}(s)+6e^{-}⟶\ce{3Ni}(s)+\ce{2Al^{3+}}(aq)+6e^{-}\)

Plug in given concentrations to get: \(\ce{3Ni^{2+}}(aq, 0.25M)+\ce{2Al}(s)+6e^{-}⟶\ce{3Ni}(s)+\ce{2Al^{3+}}(aq, 0.015M)+6e^{-}\)

Find the cell potential using the Nernst equation:

\(ε_{Cell}= 1.4 - \frac{0.0592}{6}log(\frac{0.015^{2}}{(0.25)^3})=1.42V\) ε is still positive and spontaneous for the given concentrations.

3. Set up the half cell reactions with the information given.

\(\ce{2Br^{-}}(aq, 1M)⟶\ce{Br_2}(aq, 0.11M)+\ce{2e^{-}}\) \(ε^o=+1.08V\) (loss of electrons happens at the anode)

\(\ce{Al^{3+}}(aq, 0.023M)+3e^{-}⟶\ce{Al}(s)\) \(ε^o=-1.66V\) (gain of electrons happens at the cathode)

Determine the standard cell potential:

\(ε^o_{Cell}= -1.66 - (1.08)=-2.41V\) Because ε^o is negative, the standard reaction without given concentrations is nonspontaneous.

Combine the two to get the complete cell:

\(\ce{2Al^{3+}}(aq, 0.023M)+\ce{6Br^{-}}(aq, 1.0M)+6e^{-}⟶\ce{2Al}(s)+\ce{3Br_2}(aq, 0.11M)+6e^{-}\)

Find the cell potential using the Nernst equation:

\(ε_{Cell}= -2.41 - \frac{0.0592}{6}log(\frac{0.11^{3}}{(0.023)^{2}})=-2.41V\) ε is still negative and nonspontaneous.

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 \(\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)\). 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.

A12.1.3

This problem only asks for the rate expressions of the given balanced chemical reaction \(\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)\), all other information is extraneous.

Given: Balanced chemical reaction: \(\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)\)

Asked For: reaction rate expressions

Strategy:

1. Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
2. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.

SOLUTION

1. Because Cl₂ has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of Cl₂ and write that expression as \(\ce{rate}=−\dfrac{Δ[\ce{Cl2}]}{Δt}\)

2. The balanced chemical equation shows that 2 mol of ClF₃ are produced for each mol of Cl₂ decomposed and that 3 mol of F₂ is decomposed for each mol of Cl₂ decomposed. The molar ratios of Cl₂ to ClF₃ and F₂ is 1:2 and 1:3, respectively. Therefore, [ClF₃] and [F₂] must be divided by these stoichiometric coefficients to obtain equivalent expressions for reaction rate. The reaction rate expressions are as follows:

\(\ce{rate}=+\dfrac{1}{2}\dfrac{Δ[\ce{CIF3}]}{Δt}\)

\(\ce{rate}=−\dfrac{1}{3}\dfrac{Δ[\ce{F2}]}{Δt}\)

Combine these to achieve:

\(\ce{rate}=+\dfrac{1}{2}\dfrac{Δ[\ce{CIF3}]}{Δt}=−\dfrac{Δ[\ce{Cl2}]}{Δt}=−\dfrac{1}{3}\dfrac{Δ[\ce{F2}]}{Δt}\)

More information for reaction rates here.

Q12.5.1

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

A12.5.1

Chemical reaction rates are proportional to the rate of reactant collisions. When these reactants collide, they must do so in a specific orientation that allows corresponding atoms to bond together to form the product and with enough energy so that reactants' valence shells can be penetrated. There are various factors that affect chemical reaction rates when collisions occur. These are them below:

Factors from this table that can prevent a collision from producing a chemical reaction are: decreasing concentration(dilution), decreasing pressure, and decreasing temperature in the container in which a reaction occurs. (Choose two.)

Q21.3.5

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

1. the production of ¹⁷O from ¹⁴N by alpha 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 \(\ce{^2_1H}\) bombardment

A21.3.5

Understand that bombardment is the directing of a species against(type of bombardment is in the reactant side of an equation) and know the types of particles.

alpha (α) particle: \(\ce{^4_2He}\)

beta (β) particle: \(\ce{^0_{-1}e}\)

positron (β⁺) particle: \(\ce{^0_{+1}e}\)

gamma (γ) particle: \(\ce{^0_0γ}\)

neutron particle: \(\ce{^1_0n}\)

Given: Produced species, original species, types of bombardment

Asked For: Balanced equations for nuclear reactions

Strategy:

1. Set up an equation with the original species plus the type of bombardment and an arrow directing towards the produced species. Leave a blank space for any other species that may be produced.
2. Fill in the blanks, such as the atomic numbers of elements(using a periodic table)
3. Check if the numbers add up. Are the number of protons equal on both sides? Neutrons? If they are not, add in a secondary product that will balance the nuclear reaction.

SOLUTION

1. ¹⁴N undergoes alpha particle bombardment to produce ¹⁷O

\(\ce{^{14}_N + {^4_2He} ⟶ ^{17}_O}\) ; Nitrogen has an atomic number of 7 and oxygen, 8. Plug these in.

\(\ce{^{14}_7N + {^4_2He} ⟶ ^{17}_8O}\) ; 7+2=9 but oxygen only has 8 protons. 14+4=18 but this oxygen only has 17 nucleons. Add a second product with 1 proton and 1 nucleon.

\(\ce{^{14}_7N + {^4_2He} ⟶ ^{17}_8O + ^1_1H}\)

2. ¹⁴N undergoes neutron bombardment to produce ¹⁴C

\(\ce{^{14}_N + ^1_0n⟶ ^{14}_C }\) ; Nitrogen has an atomic number of 7 and carbon, 6.

\(\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C}\) ; 7+0=7 but carbon only has 6 protons. 14+1=15 but this carbon only has 14 nucleons. Add a second product with 1 proton and 1 nucleon.

\(\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\)

3. ²³²Th undergoes neutron bombardment to produce ²³³Th

\(\ce{^{232}_Th + ^1_0n⟶ ^{233}_Th}\) ; Thorium has an atomic number of 90.

\(\ce{^{232}_{90}Th + ^1_0n⟶ ^{233}_{90}Th}\) ; 232+1=233, the number of nucleons in the thorium product. 90+0=90, the atomic number of thorium. Equation does not need a second product.

4. ²³⁸U undergoes \(\ce{^2_1H}\) bombardment to produce ²³⁹U

\(\ce{^{238}_U + ^2_1H⟶ ^{239}_U }\) ; Uranium has an atomic number of 92.

\(\ce{^{238}_{92}U + ^2_1H⟶ ^{239}_{92}U + ^1_1H}\) ; 238+2=240 but the uranium product has 239 nucleons. 92+1=93 but the product thorium has an atomic number of 92. Add a second product with 1 nucleon and 1 proton.

\(\ce{^{238}_{92}U + ^2_1H⟶ ^{239}_{92}U + ^1_1H}\)

More information on nuclear chemistry here.

Q20.2.1

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

A20.2.1

A good oxidant is an element that easily oxidizes another species. This means this species takes an electron from the other species or is easily reduced. A good reductant is the opposite. A good reductant is an element that easily reduces another species. This means this species gives an electron to the other species or is easily oxidized.

Elements that are good oxidants or easily reduced are highly electronegative, meaning they want electrons. These elements tend to be the elements on the far right side of the periodic table(excluding the halogens, period 18) because their outermost electron shells are nearly full and being full is very "desirable." Therefore, they readily accept electrons to become reduced whenever they get the chance to.

Elements that are good reducers or are easily oxidized have very low electronegativities, meaning they would much rather lose electrons than gain any. These elements tend to be the elements on the far left side of the periodic table because their outermost electron shells have very few electrons in them. It is easier to obtain a full outermost electron shell by losing their few outer electrons and dropping down a shell than gaining all electrons needed to fill their current shell.

Q20.4.20

Will each reaction occur spontaneously under standard conditions?

1. \(\mathrm{Cu(s)+2H^{+}(aq)⟶Cu^{2+}(aq)+H_2(g)}\)
2. \(\mathrm{Zn^{2+}(aq)+Pb(s)⟶Zn(s)+Pb^{2+}(aq)}\)

A20.4.20

Spontaneity is a trait of a reaction in which ε is greater than 0. Treat these reactions as electrochemical cells and solve for ε.

Given: balanced redox chemical reaction for electrochemical cell

Asked For: reaction rate expressions

Strategy:

1. Identify the oxidation half-reaction and the reduction half-reaction. Assign their standard reduction potential with this table. Then identify the anode and cathode from the half-reaction that occurs at each electrode.

2. Use the equation \(ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}\) to determine if ε is positive or negative. If ε is positive, the reaction is spontaneous. If it is negative, than it is not.

SOLUTION

1. \(\mathrm{Cu(s)+2H^{+}(aq)⟶Cu^{2+}(aq)+H_2(g)}\)

Split into half reactions and identify each reactions standard reduction potential:

\(\mathrm{Cu(s)⟶Cu^{2+}(aq)}\) (+0.34) and \(\mathrm{2H^{+}(aq)⟶H_2(g)}\) (+0.00)

Because copper is losing electrons, it is the oxidation half reaction and is the anode. Hydrogen is gaining electrons in the reduction half reaction, making it the cathode.

Plug in \(ε^o_{Cathode,2H^{+}}={+0.34}\) and \(ε^o_{Anode,Cu}={+0.00}\) to \(ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}\) to get \(ε^o_{Cell}= {+0.34} - {+0.00}= {+0.34}\)

Because the solution is positive, the reaction is spontaneous.

2. \(\mathrm{Zn^{2+}(aq)+Pb(s)⟶Zn(s)+Pb^{2+}(aq)}\)

Split into half reactions and identify each reactions standard reduction potential:

\(\mathrm{Zn^{2+}(aq)⟶Zn(s)}\) and \(\mathrm{Pb(s)⟶+Pb^{2+}(aq)}\)

Because lead is losing electrons, it is the oxidation half reaction and is the anode. Zinc is gaining electrons in the reduction half reaction, making it the cathode.

Plug in \(ε^o_{Cathode,Zn}={-0.76}\) and \(ε^o_{Anode,Pb}={-0.13}\) to \(ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}\) to get \(ε^o_{Cell}= {-0.76} - {-0.13}= {-0.63}\)

Because the solution is negative, the reaction is nonspontaneous.

More information on electrochemical cells can be found here.

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

A20.9.5

Each of these salts partakes in an electrochemical cell. These salts are molten, meaning we can assume they are in a liquid state, but for the purpose of the looking them up in a standard reduction table, we can say they are aqueous.

Given: molten salt; one mole of electron transfer

Asked For: grams of metal deposited from salt

Strategy:

1. Identify the oxidation and reduction half-reactions to compose the balanced redox reaction the molten salt undergoes in electrolysis.

2. Use stoichiometry to convert one mole of electrons to grams of deposited metal using the redox reaction just composed.

SOLUTION

1. AlCl₃ is made from the reactions \(\mathrm{Al^{3+}(aq)+{3e^{-}}⟶Al(s)}\) and \(\mathrm{2Cl^{-}(aq)⟶Cl_2(g)+2e^{-}}\)

To balance electrons transferred, multiply the aluminum reaction by 2 and the chlorine reaction by 3 to get \(\mathrm{2Al^{3+}(aq)+{6e^{-}}⟶2Al(s)}\) and \(\mathrm{6Cl^{-}(aq)⟶3Cl_2(g)+6e^{-}}\)

Combine these two reactions to get the balanced redox reaction: \(\mathrm{2Al^{3+}(aq)+{6Cl^{-}(aq)} +{6e^{-}}⟶2Al(s)+3Cl_2(g)+{6e^{-}}}\) or \(\mathrm{2AgCl_3(l)⟶2Al(s)+3Cl_2(g)}\)

Using stoichiometry, convert one mole of electrons to grams of metal:

\((1\text{ mol}\text{ of}\text{ e}^-)\times\frac{2\text{ mol Al}}{6\text{ mol}\text{ of}\text{ e}^-} \times\frac{26.98\text{ grams Al}}{1\text{ mol}\text{ of}\text{ Al}} = 8.99\text{ grams Al}\)

2. MgCl₂ is made from the reactions \(\mathrm{Mg^{2+}(aq)+{2e^{-}}⟶Mg(s)}\) and \(\mathrm{2Cl^{-}(aq)⟶Cl_2(g)+2e^{-}}\)

There is no balancing required between these two equations, so simply combine them to get \(\mathrm{Mg^{2+}(aq)+{2Cl^{-}(aq)} +{2e^{-}}⟶Mg(s)+Cl_2(g)+{2e^{-}}}\) or \(\mathrm{MgCl_2(l)⟶Mg(s)+Cl_2(g)}\)

Using stoichiometry, convert one mole of electrons to grams of metal:

\((1\text{ mol}\text{ of}\text{ e}^-)\times\frac{1\text{ mol Al}}{2\text{ mol}\text{ of}\text{ e}^-} \times\frac{24.31\text{ grams Al}}{1\text{ mol}\text{ of}\text{ Al}} = 12.16\text{ grams Al}\)

3. FeCl₃ is made from the reactions \(\mathrm{Fe^{3+}(aq)+{e^{-}}⟶Fe^2+(aq)}\), \(\mathrm{Fe^{2+}(aq)+{2e^{-}}⟶Fe(s)}\) and \(\mathrm{2Cl^{-}(aq)⟶Cl_2(g)+2e^{-}}\)

Combine the iron reactions to get \(\mathrm{Fe^{3+}(aq)+{3e^{-}}⟶Fe(s)}\)

Multiply the iron reaction by 2 and the chlorine reaction by 3 to get the balanced redox reaction: \(\mathrm{2Fe^{3+}(aq)+{6Cl^{-}(aq)} +{6e^{-}}⟶2Fe(s)+3Cl_2(g)+{6e^{-}}}\) or \(\mathrm{2FeCl_3(l)⟶2Fe(s)+3Cl_2(g)}\)

Using stoichiometry, convert one mole of electrons to grams of metal:

\((1\text{ mol}\text{ of}\text{ e}^-)\times\frac{2\text{ mol Fe}}{6\text{ mol}\text{ of}\text{ e}^-} \times\frac{55.85\text{ grams Fe}}{1\text{ mol}\text{ of}\text{ Fe}} = 18.62\text{ grams Fe}\)

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?

A14.5.1

Activation energy is the minimum amount of energy necessary for a reaction to occur. This energy requirement comes from the need to overcome electrostatic repulsion and breaking chemical bonds to replace them with new ones. When molecules collide, they need to do so with enough energy to overcome the activation energy. This energy comes from kinetic energy and energy between the bonds of their current molecule. Molecules with less that the threshold energy needed will simply bounce off each other instead of reacting. An increase in temperature increases the reaction rate despite the fact that the average kinetic energy is still less than the activation energy because kinetic energy is not completely uniform throughout a medium. There are molecules that are transferred with more kinetic energy, and those that have less. By increasing temperature, one is making it so that the likelihood of a single molecule having enough kinetic energy to overcome the activation energy is higher, resulting in an inreased reaction rate.

More information on temperature and reaction rates can be found here.