Extra Credit 11

Q19.3B

Write cell reactions for the electrochemical cells diagrammed here, and use data from Table P2 to calculate E°cell for each reaction.

1. \(\mathrm{Al(s)|Al^{3+}||Zn^{2+}(aq)|Zn(s)}\)

2. \(\mathrm{Pt(s)|Fe^{2+}(aq),Fe^{3+}||Cu^{2+}(aq)|Cu(s)}\)

3. \(\mathrm{Pt(s)|Cr_2O_7,Cr^{3+}(aq)||Ag^+(aq)|Ag(s)}\)

4. \(\mathrm{O^{-2}(aq)|O_2(g)||H^+(aq)|H_2(g)|C(s)}\)

S19.3B

Cell notations (also known as cell diagrams) describe voltaic or galvanic (spontaneous) cells. In these diagrams, the anode half-cell is described first on the left, followed by the cathode half-cell on the right. Single vertical lines in cell diagrams are used to denote phase changes, whereas double vertical lines indicate a salt bridge or porous membrane that separates the individual half-cells. Keep in mind that in voltaic/galvanic cells, reduction occurs at the cathode and oxidation occurs at the anode.

For a more detailed explanation of cell diagrams, follow this link: Cell Diagrams

Cell diagrams describe the individual half-reactions that make up a reduction-oxidation reaction, and for this reason can be used to find the cell potential under standard and non-standard conditions of a reduction-oxidation reaction. In this example, since the concentrations of reactants and products in each cell diagram are not explicitly stated, the cell is assumed to be under standard conditions of 1 M for solutions. Therefore, we can use the following equation to find the standard cell potential, E∘cell, for each reaction:

E°_cell=E°_cathode−E°_anode or standard cell potential = reduction potential (cathode) - reduction potential (anode)

This link can be followed for more information on Cell Potential Under Standard Conditions. Furthermore, parts (1) through (4) utilize the inert electrodes Pt(s) and C(s) in their cell diagrams. More information on inert electrodes and their application is described here.

Using this information, we can write the cell reactions for each of the electrochemical cells and calculate E∘cell for each reaction.

1. \(\mathrm{Al(s)|Al^{3+}||Zn^{2+}(aq)|Zn(s)}\)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{Al(s)→Al^{3+}(aq)+3e^-}\)

three electrons must be added to the product side in order for the charges on the product side and reactant side to be equal. This step is taken in order to balance the oxidation half reaction. This link provides more instruction on balanced oxidation-reduction reactions.

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = -1.676 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{Zn^{2+}(aq)+2e^-→Zn(s)}\)

again, two electrons must be added to the reactant side in order for the charges on the product side and reactant side to be equal.

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = -0.7618 V

The overall cell reaction, when the electron numbers of each reaction are balanced, is:

\(\mathrm{3Zn^{2+}(aq)+2Al(s)→2Al^{3+}(aq)+3Zn(s)}\)

The stoichiometry of the reaction, however, does not affect the calculations of the standard cell potential. Thus, the standard cell potential for this reaction is:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=(-0.7618V)-(1.676V)=0.9142V}\)

2. \(\mathrm{Pt(s)||Fe^{2+}(aq),Fe^{3+}(aq)||Cu^{2+}(aq)|Cu(s)}\)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{Fe^{2+}(aq)→Fe^{3+}(aq)+e^-}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = 0.771 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{Cu^{2+}(aq)+2e^-→Cu(s)}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = 0.3419 V

The overall cell reaction, when the electron numbers of each reaction are balanced, is:

\(\mathrm{Cu^{2+}(aq)+2Fe^{2+}(aq)→Cu(s)+2Fe^{3+}(aq)}\)

The stoichiometry of the reaction, however, does not affect the calculations of the standard cell potential. Thus, the standard cell potential for this reaction is:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=(0.3419V)-(0.771V)=(-0.4291V)}\)

3. \(\mathrm{Pt(s)|Cr_2O_7,Cr^{3+}(aq)||Ag^+(aq)|Ag(s)}\)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{Cr_2O^{2-}_7(aq)+14H^++6e^-→2Cr^{3+}(aq)+7H_2O}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = 1.36 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{Ag(s)→Ag^{3+}(aq)+e^-}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = 0.7996 V

The overall cell reaction, when the electron numbers of each reaction are balanced, is:

\(\mathrm{Cr_2O_7^{2-}+14H^++6Ag(s)→2Cr^{3+}(aq)+7H_2O+6Ag^{3+}(aq)}\)

The stoichiometry of the reaction, however, does not affect the calculations of the standard cell potential. Thus, the standard cell potential for this reaction is:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=(0.7996V)-(1.36V)=-0.5604V}\)

4. \(\mathrm{O^{2-}|O_2(g)||H^+(aq)|H_2(g)|C(s)}\)

O−2(aq)|O2(g)||H+(aq)|H2(g)|C(s)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{O^-_2(aq)→O_2(aq)+e^-}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = -1.0 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{2^+(aq)+2e-→H_2(g)}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = 0.00 V

The overall cell reaction, when the electron numbers of each reaction are balanced, is:

\(\mathrm{2O^-_2(aq)+2H^+(aq)→2O_2(aq)+H_2(g)}\)

The stoichiometry of the reaction, however, does not affect the calculations of the standard cell potential. Thus, the standard cell potential for this reaction is:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=(0.00V)-(1.00V)=1.00V}\)

Q19.37B

Use Nernst's Equation to calculate E cell for:

1. \(\mathrm{Al(s)|Al^{3+}(aq)(0.20M)||Sn^{2+}(aq)(0.90M)|Sn(s)}\)

2. \(\mathrm{Zn(s)|Zn^{2+}(aq)(0.42M)||C;^-(0.02M),C;_2(g,0.5atm)||Pt(s)}\)

S19.37B

Nerst's equation is used to calculate the cell potential under non-standard conditions (E cell) by utilizing the values of the cell potential in standard conditions (E∘cell), the number of electrons transferred from the oxidation half-cell to the reduction half-cell in the balanced overall reaction (n), Faraday's constant (F), the temperature (T), the reaction quotient (Q), the temperature of the and the gas constant (R). The generalized equation is:

\(\mathrm{E=E_0-(2.303RTlogQ)/(nF)}\)

Under standard temperature of 298 K, however, the equation is simplified to:

\(\mathrm{E=E_0-(2.303VlogQ)(n)}\)

1. \(\mathrm{Al(s)|Al^{3+}(aq)(0.20M)||Sn^{2+}(aq)(0.90M)|Sn(s)}\)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{Al(s)→Al^{3+}(Aq)+3e^-}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = -1.676 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{Sn^{2+}(aq)+2e^-→Sn(s)}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = -0.14 V

The standard cell potential does not depend on the stoichiometry of the reaction, so the the standard cell potential for this reaction can be calculated directly from the standard reduction potentials of the reduction and oxidation half-cells using the equation:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=(-0.14V)-(-1.676V)=1.536V}\)

The Nerst equation, however, does take into account the total number of electrons transferred in the reaction and the reaction quotient, both which depend on the stoichiometry of the reaction. So the reaction must be balanced in order for the calculated cell potentials under non-standard conditions to be accurate. By balancing the number of electrons transferred, the overall reaction equates to:

\(\mathrm{2Al(s)+3Sn^{2+}(aq)→2Al^{3+}(aq)+3Sn(s)}\) with 6 electrons transferred from the anode to the cathode

The reaction quotient is thus:

\(\mathrm{Q=[Al^{3+}]^2/[Sn^{2+}]^3=[0.20M]^2/[0.90M]^3=0.0549}\)

and the number of electrons transferred is:

\(\mathrm{n=6}\)

Since the temperature of the reaction is not specified, standard temperature of 298 K can be assumed and the Nerst Equation simplifies to:

\(\mathrm{E=E_0-(0.0592V*logQ)(n)}\)

By plugging in the values of E₀, Q, and n, the calculated E cell for this reaction under non-standard conditions is:

\(\mathrm{E=(1.536V)-(0.0592V*log(0.0549))/6}\)

\(\mathrm{=1.548V}\)

2. \(\mathrm{Zn(s)|Zn^{2+}(aq)(0.42M)||Cl^-(0.02M),Cl_2(g,0.5atm)|Pt(s)}\)

The first half cell described is the anode, where oxidation occurs. The oxidation half-reaction can be written as:

\(\mathrm{Zn(s)→Zn^{2+}(aq)+2e^-}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_anode = -0.7618 V

The second half cell described is the cathode, where reduction occurs. The reduction half-reaction can be written as:

\(\mathrm{Cl_2(g)+2e^-→2Cl^-(aq)}\)

From Table P2 , the standard reduction potential for this half-reaction, in volts, is: E°_cathode = 1.396 V

Again, the standard cell potential does not depend on the stoichiometry of the reaction, so the the standard cell potential for this reaction can be calculated directly from the standard reduction potentials of the reduction and oxidation half-cells using the equation:

\(\mathrm{E°_{cell}=E°_{cathode}-E°_{anode}=1.396V-(-0.7618V)=2.1578V}\)

The Nerst equation, however, does take into account the total number of electrons transferred in the reaction and the reaction quotient, both which depend on the stoichiometry of the reaction. So the reaction must be balanced in order for the calculated cell potentials under non-standard conditions to be accurate. Since the number of electrons transferred from the anode to the cathode is already equal, which can be seen by the half-cell reactions, the overall equation is:

\(\mathrm{Zn(s)+Cl_2→(g)Zn^{2+}(aq)+2Cl^-(aq)}\) with 2 electrons transferred from the anode to the cathode

The reaction quotient is thus:

\(\mathrm{Q=[Cl^-]^{2}[Zn^{2+}]/[Cl_2]=[0.02M]^2[0.42M]/[0.05atm]=0.00336}\)

and the number of electrons transferred is:

\(\mathrm{n=2}\)

Since the temperature of the reaction is not specified, standard temperature of 298 K can be assumed and the Nerst Equation simplifies to:

\(\mathrm{E=E_0-(0.0592V*logQ)(n)}\)

By plugging in the values of E₀, Q, and n, the calculated E cell for this reaction under non-standard conditions is:

\(\mathrm{E=(2.1578V)-(0.0592V*log(0.00336))/2}\)

\(\mathrm{=2.231V}\)

Q20.27B

Create a standard electrode potential diagram for the reduction of Fe³⁺ to Fe²⁺ in acidic solution.

S20.27B

The balanced reaction for this reduction in acidic solution (which occurs at the cathode) is:

\(\mathrm{Fe^{3+}(aq)+e^-→Fe^{2+}(aq)}\)

Since neither Fe³⁺(aq) nor Fe²⁺(aq) are metals, an inert electrode such as graphite(carbon), platinum, gold, or rhodium can be used that does not participate in the reaction. Thus a standard electrode potential diagram for the reduction of Fe³⁺ to Fe²⁺ in acidic solution is:

\(\mathrm{Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)}\)

Q24.1C

For the reaction \(\mathrm{3A+2B→C+2D}\) reactant A is found to disappear at the rate of 4.6 X 10^-5 Ms^-1

1. What is the rate of reaction?
2. What is the rate of disappearance of the reactant B?
3. What is the rate of appearance for product D?

S24.1C

Reaction rates for chemical reactions measure the change in concentration of the reactants or the change in concentration of the products per unit time. The reaction rate for this chemical reaction is equal to:

the change in concentration of reactant A per unit time = the change in concentration of reactant B per unit time = the change in concentration of product C per unit time = the change in concentration of product D per unit time

The concentration of the reactants always decreases with time, and the concentration of products increases with time. Since rates are always positive, however, rates expressed in terms of a reactant concentration are always preceded by a minus sign to make the rate come out positive.

Furthermore, a reaction in which the coefficients of reactions and products are different must account for this in the rate reaction expression. It is clear that in this reaction, [B] decreases twice as rapidly as [C] appears, so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient.

1. The reaction rate is thus:

\(\mathrm{Rate=-(Δ[A])/(3Δt)=-(Δ[B])/(2Δt)=(Δ[C]/(Δt)=(Δ[D])/(2Δt)}\)

using the rate of disappearance of A, which can also be denoted as (Δ[A])/(Δt) and is a negative value since A is decreasing in concentration, the rate of reaction is equal to:

\(\mathrm{Rate=-((1/3)Δ[A])/(Δt)=-(1/3)/(-4.6*10^{-5}Ms^{-1}}\)

\(\mathrm{1.53*10^{-5}Ms^{-1}}\)

2. The rate of disappearance of B is denoted as (Δ[B])/(Δt), and can be found from its relationship with A via the reaction rate formula:

\(\mathrm{Rate=-(Δ[A])/(3Δt)=-(Δ[B])/(2Δt)}\)

\(\mathrm{(Δ[B])/(Δt)=-(Δ[A])/((3Δt)(1/2))}\)

\(\mathrm{=((2/3)(Δ[A]))/(Δt)}\)

\(\mathrm{=(2/3)(-4.6*10^{-5})}\)

\(\mathrm{=-3.07*10^{-5}Ms^{-1}}\)

3. The rate of appearance of D is denoted as \(\mathrm{(Δ[D])/(Δt)}\), and can be found from its relationship with A via the reaction rate formula:

\(\mathrm{Rate=-(Δ[A])/(3Δt)=(Δ[D])/(2Δt)}\)

\(\mathrm{Rate=(Δ[D])/(Δt)=-(2/3)((Δ[A]/(Δt)}\)

\(\mathrm{=-(2/3)(-4.46*10^{-5}Ms^{-1}}\)

\(\mathrm{=3.07*10^{-5}Ms^{-1}}\)

Q24.45B

Answer the following:

1. What two factors does the rate of a reaction depend on other than the frequency of collisions?
2. Why does the rate of reaction increase dramatically with temperature?
3. What is the net effect of the addition of a catalyst?

S24.45B

1 Other than the frequency of collisions, the collision theory predicts that the rate of reaction depends on the ration of collision with sufficient “Activation Energy” (function of temperature) and the the orientation effects (not a function of temperature).

2 The rate of reaction increases dramatically with temperature because increasing the temperature of the system makes molecules move faster and collide more vigorously, which increases the likelihood of bond breakage upon collision. Furthermore, the collisions must be energetic enough (due to kinetic energy) to break chemical bonds; this energy is known as the activation energy. Since a temperature increase will lead to more molecules possessing a greater kinetic energy and since the molecules are colliding more often, the rate of reaction will increase. More collisions will occur with sufficient "activation energy" and the frequency of collisions will also increase.

3 A catalyst lower the activation energy required for a reaction to occur, which means that the number of collisions that successfully occur with sufficient activation energy is increased. This in turn leads to an increase in the rate of reaction, as the frequency of collisions with sufficient activation energy is increased.

Q25.21C

The rate of disintegration for a sample with Astatine-210 is 9871 dis day^-1. The half-life of Astatine-210 is 4.3 years. Estimate the mass of Astatine-210 in the sample.

S25.21C

The rate of decay of a radioactive sample, also known as its activity, is equal to the decrease in the number of the radioisotope’s nuclei per unit time:

\(\mathrm{A=ΔN/Δt}\)

The activity of a sample is also directly proportional to the number of atoms of the radioactive isotope in the sample:

\(\mathrm{A=kN}\)

Thus, the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample is:

\(\mathrm{-ΔN/Δt=kN}\)

Radioactive decay is a first-order process, it can be described in terms of the integrated rate law and the half-life of a first-order reaction:

\(\mathrm{N=N_0e^{-kt}}\)

\(\mathrm{t_{1/2}=0.693/k}\)

Astatine-210 has a half-life of 4.3 years, which can be used to solve for the reaction constant, k, by the half-life equation:

\(\mathrm{4.3years=0.693/k}\)

\(\mathrm{k=0.693/4.3 years=0.161year^{-1}}\)

In order to solve for the mass of Astatine-210 present in the sample, k must be converted to units of day^-1 so that it can be used alongside the decrease in the number of the radioisotope’s nuclei per unit time (9871 dis day^-1) to solve for the mass quantity.

\(\mathrm{k=0.161year^{-1}*(1year/365 days)=0.000441day^{-1}}\)

to solve for the mass of Astatine-210 present in the sample, multiply the activity by the reaction constant to obtain the number of atoms present in the sample. Then, convert this number to grams using Avogadro's number and the molar mass of Astatine-210:

\(\mathrm{mass=9871atoms/day)/(4.41*10^{-4}d^{-1})(1mol/6.022*10^{23})(210g/1mol)=7.81*10^{-15}g present

Q19.5

Write balanced nuclear equations for:

1. positron emission by Sr-83
2. the fusion of two C-12 nuclei to give another nucleus and a neutron.
3. the fission of U-235 to give Ba-140, another nucleus and an excess of two neutrons.

S19.5

1. A positron is a beta particle with a positive 1 (+1) charge and no mass. When an atom undergoes positron emission, the nucleus produced has a mass number (the sum of the number of protons and the number of neutrons in the atom) that is the same as Sr-83 but a proton number (atomic number) of one greater than Sr-83. Sr-83 has an atomic number of 38, and thus the nucleus produced will have an atomic number of 39. The balanced nuclear equation is thus:

\(\mathrm{^{83}Sr→^{83}Y+positron}\)

2. When two C-12 nuclei are fused, they are joined together to form a heavier nucleus (or pairs of nuclei). Thus, the fusion of two nuclei will result in a nucleus and neutron whose sum of atomic mass is equal to twice the atomic mass of a single C-12 nuclei. This mass sum is thus 2*12 amu= 24 amu. Furthermore, the nucleus and neutron must have a summed atomic number that is equal to twice the atomic number of a single C-12 nuclei. This atomic number sum is 2*(6 protons)= 12 protons. Since a neutron is a chargeless nucleus with a mass of 1, the nucleus produced by the fusion of the two C-12 nuclei must have a mass number of 24-1= 23 amu and will have am atomic number equal to the summed atomic number of the two C-12 nuclei, which is equal to 12 protons. The element on the periodic table with an atomic number of 12 is Magnesium, and thus the balanced nuclear equation is :

\(\mathrm{^{12}C+^{12}C →neutron+^{23}Mg}\)

3. Fission involves the bombardment of a nucleus with neutrons to cause a nuclear to absorb a neutron and then split into two large fragments. A neutron is a nucleus with no charge and a mass number of 1, so bombarding a U-235 nucleus with a neutron produces a nucleus of the same element (because the atomic number remains unchanged) with a different mass number (a mass number that is one less than the original mass number, since a neutron has a mass of one amu). In this example, the bombardment of a U-235 nucleus with a neutron produces U-234:

\(\mathrm{^{235}U →neutron+^{234}U}\)

The U-234 nucleus will then spontaneously decay into Ba-140, another nucleus and an excess of two neutrons by the balanced nuclear equation:

\(\mathrm{^{234}U →^{140}Ba+2 neutrons+^{92}Kr}\)

U has an atomic number of 92 protons, thus the sum atomic number of the two nuclei produced must equal 92 due to nuclear accounting. The neutrons do not affect this calculation because they have no charge, thus no protons, thus no atomic number. Ba has an atomic number of 56, so the other nucleus produced has an atomic number of 92-56=36, which corresponds to the element Krypton (Kr). In order to find out the atomic mass of the Krypton produced, the fact that the sum of the product masses must equal the sum of the reactant mass is taken into account. U-234 has a mass of 234 amu, so the sum of Ba-140 (mass of 140amu), the two neutrons (total mass of 2 amu), and the other nucleus (unknown mass, denote y) must equal 234. Thus, 234 = 2 + 140 + y. Solving for y, we find that y is equal to 92, so the Krypton nucleus has a mass of 92 amu.

Q21.3.4

During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.

S21.3.4

Nuclear fusion is the process by which multiple nuclei join together to form a heavier nucleus. It is accompanied by the release or absorption of energy depending on the masses of the nuclei involved. In a star, the net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos, and energy. The fusion of helium and hydrogen nuclei in stars is provides that energy and synthesizes new nuclei as a byproduct of that fusion process. Stars are formed by the aggreagation of interstellar dust. In the beginning of its life, it is powered by a series of nuclear fusion reactions in which hydrogen is converted to helium. This overall reaction is paired with the release of two positrons, two gamma rays, and a large amount of energy. These reactions are responsible for its release of solar heat.