Extra Credit 10

Q17.1.9

a. Why is it not possible for hydroxide ion (\(OH^-\)) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?

b. Why is it not possible for hydrogen ion (\(H^+\)) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?

S17.1.9

a. Hydroxide ion does not appear anywhere in the reaction because in an acidic solution, hydroxide ions do not need to combine with hydrogen ions to produce water. Only in a basic solution is it necessary to react hydroxide ions with hydrogen ions.

For example, to balance \(I^-(aq)+NO_2^-(aq)\rightarrow I_2(s)+NO(g)\) in acidic conditions,

Step 1: Split reaction into half-reactions (reduction and oxidation)

\(I^-(aq)\rightarrow I_2(s)\)

\(NO_2^-(aq)\rightarrow NO(g)\)

Step 2: Balance the charge or oxidation number with electrons

\(2I^-\rightarrow I_2+2e^-\)

\(e^-+NO_2^-\rightarrow NO\)

Step 3: Add O by adding \(H_2O\) & balance H by adding \(H^+\)

\(2I^-\rightarrow I_2+2e^-\)

\(e^-+NO_2^-+2H^+\rightarrow NO+H_2O\)

Step 4: Multiply by some integer to make electrons (lost) = electrons (gained)

\(2I^-\rightarrow I_2+2e^-\)

\([e^-+NO_2^-+2H^+\rightarrow NO+H_2O]*2\Longrightarrow 2e^-+2NO_2^-+4H^+\rightarrow 2NO+2H_2O\)

Step 5: Add half reactions and cancel substances on both sides

\(2I^-\rightarrow I_2+2e^-\)

\(2e^-+2NO_2^-+4H^+\rightarrow 2NO+2H_2O\)

--------------------------------------------------------

\(4H^+ +2NO_2^-+2I^-\rightarrow 2NO+2H_2O+I_2\)

This is the balanced net reaction. As shown, hydroxide does not appear anywhere in the process.

b. Hydrogen ion cannot appear anywhere in a basic solution because in a basic solution, the hydrogen ions will react with hydroxide ion to produce water.

For example, to balance \(I^-(aq)+NO_2^-(aq)\rightarrow I_2(s)+NO(g)\) in basic conditions, first do the same steps as in acidic conditions.

Step 1: Split reaction into half-reactions (reduction and oxidation)

\(I^-(aq)\rightarrow I_2(s)\)

\(NO_2^-(aq)\rightarrow NO(g)\)

Step 2: Balance the charge or oxidation number with electrons

\(2I^-\rightarrow I_2+2e^-\)

\(e^-+NO_2^-\rightarrow NO\)

Step 3: Add O by adding \(H_2O\) & balance H by adding \(H^+\)

\(2I^-\rightarrow I_2+2e^-\)

\(e^-+NO_2^-+2H^+\rightarrow NO+H_2O\)

Step 4: Multiply by some integer to make electrons (lost) = electrons (gained)

\(2I^-\rightarrow I_2+2e^-\)

\([e^-+NO_2^-+2H^+\rightarrow NO+H_2O]*2\Longrightarrow 2e^-+2NO_2^-+4H^+\rightarrow 2NO+2H_2O\)

Step 5: Add half reactions and cancel substances on both sides

\(2I^-\rightarrow I_2+2e^-\)

\(2e^-+2NO_2^-+4H^+\rightarrow 2NO+2H_2O\)

--------------------------------------------------------

\(4H^+ +2NO_2^-+2I^-\rightarrow 2NO+2H_2O+I_2\)

Step 6 (for basic solutions): Add \(OH^-\) and cancel \(H_2O\)

\(4OH^-+4H^++2NO_2^-+2I^-\rightarrow 2NO+2H_2O+I_2+4OH^-\Longrightarrow\)

\(4H_2O +2NO_2^-+2I^-\rightarrow 2NO+2H_2O+I_2+4OH^- \Longrightarrow\)

\(2H_2O+2NO_2^-+2I^-\rightarrow 2NO+I_2+4OH^-\)

This is the balanced net reaction. As shown, \(H^+\) reacts with the \(OH^-\) to form \(H_2O\).

Q.19.1.8

The following reactions all occur in a blast furnace. Which of these are redox reactions?

a. \(3Fe_2O_3(s)+CO(g)\rightarrow 2Fe_3O_4(s)+CO_2(g)\)

b. \(Fe_3O_4(s)+CO(g)\rightarrow 3FeO(s)+CO_2(g)\)

c. \(FeO(s)+CO(g)\rightarrow Fe(l)+CO_2(g)\)

d. \(C(s)+O_2(g)\rightarrow CO_2(g)\)

e. \(C(s)+CO_2(g)\rightarrow 2CO(g)\)

f. \(CaCO_3(s)\rightarrow CaO(s)+CO_2(g)\)

g. \(CaO(s)+SiO_2(s)\rightarrow CaSiO_3(l)\)

S.19.1.8

A redox reaction is a reaction in which both an oxidation and reduction take place at the same time. A reduction involves a decrease in oxidation state and oxidation involves an increase in oxidation state. To determine which reactions are redox, first split the reaction into two half-reactions. Then, determine the oxidation state of each element. A redox reaction will have one thing reduced and one thing oxidized.

The rules for assigning oxidation states are listed here.

Following the rules for oxidation states: a, b, c, d,and e are all redox reactions.

F is not a redox reaction because the oxidation states remain the same.

G is not a redox reaction because the oxidation states remain the same.

Q.19.2.10

Draw the geometric, linkage, and ionization isomers for \([CoCl_5CN][CN]\)

S.19.2.10

Isomers are compounds with same formula but different atom arrangement. There are two subcategories: structural isomers, which are isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another, and stereoisomers, isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space.

There are three subcategories under structural isomers: ionization isomers, which are isomers that are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex; coordination isomers, isomers that have an interchange of some ligands from the cationic part to the anionic part; and linkage isomers, in two or more coordination compounds in which the donor atom of at least one of the ligands is different.

There are also two main kinds of stereoisomers: geometric isomers, metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal, and optical isomers, which occurs when the mirror image of an object is non-superimposable on the original object.

Some of the isomers look almost identical, but that is because the CN ligand can be attached by both (but not at the same time) the C or N.

Q.12.3.23

In the reaction: \(2NO+Cl_2\rightarrow 2NOCl\) the reactants and products are gases at the temperature of the reaction. The following rate data were measured for three experiments:

a. From these data, write the rate equation for this gas reaction. What order is the reaction in NO, Cl₂, and overall?

b. Calculate the specific rate constant for this reaction.

S.12.3.23

a. The rate equation can be determined by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction \(A+B\rightarrow products\), for example, we need to determine k and the exponents m and n in the following equation:
\[rate=k[A]^m[B]^n\]
To do this, the initial concentration of B can be kept constant while varying the initial concentration of A and calculating the initial reaction rate. This information would deduce the reaction order with respect to A. The same process can be done to find the reaction order with respect to B.
In this particular example,
\[\frac{rate_2}{rate_3}=\frac{k[A_2]^m[B_2]^n}{k[A_3]^m[B_3]^n}\]
So taking the values from the table,
\[\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{k[1.0]^m[1.0]^n}{k[0.5]^m[1.0]^n}\]
and by canceling like terms, you are left with
\[\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{[1.0]^m}{[0.5]^m}\]
Now, solve for m

\(4=2^m\Longrightarrow m=2\) Because m=2, the reaction with respect to \(NO\) is 2. \(NO\) is second order.

You can repeat the same process to find n.

\[\frac{rate_3}{rate_1}=\frac{k[A_3]^m[B_3]^n}{k[A_1]^m[B_1]^n}\]
Taking the values from the table,
\[\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{k[0.5]^m[1.0]^n}{k[0.5]^m[0.5]^n}\]
and by canceling like terms, you are left with
\[\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{[1.0]^n}{[0.5]^n}\]

Now this time, solve for n

\(2=2^n\Longrightarrow n=1\)

Because n=1, the reaction with respect to \(Cl_2\) is 1. \(Cl_2\) is first order.

So the rate equation is\[rate=k[NO]^2[Cl_2]^1\]

To find the overall rate order, you simply add the orders together. Second order + first order makes the overall reaction third order.

b. The rate constant is calculated by inserting the data from any row of the table into the experimentally determined rate law and solving for k. For a third order reaction, the units of k are \(frac{1}{atm^2*sec}\). Using Experiment 1,
\[rate=k[NO]^2[Cl_2]^1\Longrightarrow 5.1*10^{-3} \frac{atm}{sec}=k[0.5m atm]^2[0.5 atm]^1\]
\[k=0.0408 \frac{1}{atm^2*sec}\]

Q.12.7.3

Consider this scenario and answer the following questions:

Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl₂F₂, catalyze the decomposition of ozone in the atmosphere.

One simplified mechanism for the decomposition is:

\(O_3\rightarrow O_2+O\)

\(O_3+Cl\rightarrow O_2+ClO\)

\(ClO+O\rightarrow Cl+O_2\)

a. Explain why chlorine atoms are catalysts in the gas-phase transformation: \(2O_3\rightarrow 3O_2\)

b. Nitric oxide is also involved in the decomposition of ozone by the mechanism:

\(O_3\rightarrow O_2+O\)

\(O_3+NO\rightarrow NO_2+O_2\)

\(NO_2+O\rightarrow NO+O_2\)

Is NO a catalyst for the decomposition? Explain your answer.

S.12.7.3

a. By definition, a catalyst is a substance that speeds up a reaction without being consumed by it. Chlorine atoms are a catalyst as they react in the second step with \(O_3\) but are regenerated in the third step. As a result, they are not used up. It sped up the reaction but was not consumed by it.

b. Yes, NO is a catalyst for the decomposition. This is very similar to Part A. By definition, a catalyst is a substance that speeds up a reaction without being consumed by it. Nitric oxide is a catalyst as it reacts in the second step with \(O_3\) but is regenerated in the third step. As a result, it is not used up. It sped up the reaction but was not consumed by it.

Q.21.4.26

Isotopes such as ²⁶Al (half-life: 7.2 × 10⁵ years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.

a. ²⁶Al decays by β⁺ emission or electron capture. Write the equations for these two nuclear transformations.

b. The earth was formed about 4.7 × 10⁹ (4.7 billion) years ago. How old was the earth when 99.999999% of the ²⁶Al originally present had decayed?

S.21.4.26

a. In beta emission, a positron (the antiparticle of an electron) is emitted from the nucleus. The positron is very similar to an electron, but has a positive charge. Because there are no positrons in the nucleus, a positron must be created before emission. As a result, in a beta decay, the atomic number Z decreases by one because of the loss of the proton. The mass number A stays the same because even though a proton is lost, a neutron is created to keep the same number of nucleons.

In this example, \({^{26}_{13}Al}\rightarrow {^{26}_{12}Mg}+{^{0}_{+1}e}+{\nu}{}{}\)

In electron capture, an orbital electron gets absorbed by the nucleus. Since the electron cannot remain as an electron within the nucleus, it combines with a proton and in the process creates a neutron and a neutrino. As a result, in electron capture, the number of protons in the nucleus decreases by one hence changing Z while the number of nucleons A stays constant.

In this example, \({^{0}_{-1}e}+{^{26}_{13}Al}\rightarrow {^{26}_{12}Mg}+{\nu}{}{}\)

b. Radioactive decay is a first-order process. The formula in terms of radioactive half-life is as followed:
\[T_{1/2}=\frac{ln(2)}{k}\]
in which \(T_{1/2}\) refers to the half-life and k is the radioactive decay constant.

Radioactive decay can be described in terms of either the differential rate law or the integrated rate law:

\[N=N_0e^{-kt}\Longrightarrow t=\frac{-1}{k}ln\frac{N}{N_0}\]

In which t refers to the time, \(N_0\) refers to the initial amount, and N refers to the final amount.

Looking at this specific example, \(T_{1/2}=(4.7*10^9\). Plugging it into the first equation,
\[(4.7*10^9)=\frac{ln(2)}{k}\Longrightarrow k=9.627*10^{-7}\]

Then plugging k into the integrated rate law,

\[t=\frac{-1}{9.627*10^{-7}}ln\frac{0.00000001}{1.0}\Longrightarrow t=19,134,305.83\]

The earth was approximately 19.1 million years.

Q.20.3.13

For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.

S.20.13

In order to make it easier to describe a given electrochemical cell, a special cell notation has been adopted. The half-cells are separated by two bars or a slash, which reprsents a salt brige. Single lines indicates different phases. In cell notation, the anode (oxidation) is written on the left and the cathode (reduction) is written on the right. Concentrations can be written in parentheses along with the state of each phase. When writing the half-reactions, it is important to balance the charges by putting electrons on either side. In these reactions, platinum is not included because it is simply what the electrode is made of.

1. \(Zn(s)|Zn^{2+}(aq)||H^+(aq)|H_2(g),Pt(s)\)
   
   Reduction (Cathode): \(2H+(aq)+2e^-\rightarrow H_2(aq)\)
   Oxidation (Anode): \(Zn(s)\rightarrow Zn^{2+}(aq)+2e^-\)
   Overall: \(Zn(s)+2H^+(aq)\rightarrow Zn^{2+}(aq)+H_2(aq)\)

2. \(Ag(s)|AgCl(s)|Cl^-(aq)||H^+(aq)|H_2(g),Pt(s)\)
   Reduction (Cathode): \(AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq)\)
   Oxidation (Anode): \(H_2(g)\rightarrow 2H^+(aq)+2e^-\)
   Overall: \(AgCl(s)+H_2(s)\rightarrow 2H^+(aq)+Ag(s)+Cl^-(aq)\)

3. \(Pt(s)|H_2(aq)||Fe^{2+}(aq),Fe^{3+}(aq)|Pt(s)\)
   
   Reduction (Cathode): \(Fe^{3+}(aq)+e^-\rightarrow Fe^{2+}(aq)\)
   Oxidation (Anode): \(H_2(g)\rightarrow 2H^+(g)+2e^-\)
   Overall: \(2Fe^{3+}(aq)+H_2(g)\rightarrow 2H^+(aq)+2Fe^{2+}(aq)\)

Q.20.5.25

Given the following biologically relevant half-reactions, will FAD (flavin adenine dinucleotide), a molecule used to transfer electrons whose reduced form is FADH₂, be an effective oxidant for the conversion of acetaldehyde to acetate at pH 4.00?

\(acetate+2H^++2e^-\rightarrow acetaldehyde+H_2O, E^{\circ}=-0.58V\)

\(FAD+2H^++2e^-\rightarrow FADH_2, E^{\circ}=-0.18V\)

S.20.5.25

Yes, FAD is an effective oxidant (oxidizing agent). In general, the bigger the E° is, the more of an oxidizing agent it is. Conversely, the smaller the E°, the more of a reducing agent it is. Because \(FAD+2H^++2e^-\rightarrow FADH_2\) has a bigger E° than \(acetate+2H^++2e^-\rightarrow acetaldehyde+H_2O\) (-0.58 < -0.18), FAD is a bigger oxidizing agent.