Extra Credit 10

Q17.1.9

1. 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?
2. 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?

Solution:

1. An acidic solution is any solution that has a higher concentration of hydrogen ions than water. This means that \(1 * 10^{−7} M\) < \([H^+]\). The hydroxide cannot appear as a reactant because its concentration is nearly zero.

2. A basic or alkaline solution is any solution that has lower concentration of hydrogen ions than water. In basic solution we add OH^- for every H⁺, \([OH^−]\) > \(1 * 10^{−7} M\) > \([H^+]\) . Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.

Q19.1.8

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

a. \(\ce{3Fe2O3}(s)+\ce{CO}(g)⟶\ce{2Fe3O4}(s)+\ce{CO2}(g)\)

b. \(\ce{Fe3O4}(s)+\ce{CO}(g)⟶\ce{3FeO}(s)+\ce{CO2}(g)\)

c. \(\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\)

d. \(\ce{C}(s)+\ce{CO2}(g)⟶\ce{2CO}(g)\)

e.\(\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)\)

f.\(\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)\)

Solution:

To identify redox reaction, we have to determine if have to see if the equation is an oxidation-reduction reaction-meaning that the species are changing oxidation states during the reaction, which involves the transfer of electrons between two species. If a species is losing electrons, then that species is being oxidized. If a species is gaining electrons, then that species is being reduced. A way to remember this is using the acronyms OIL RIG. Oxidation Is Loss, and Reduction Is Gain, referring to electrons. Both of these must occur for an equation to be a redox reaction. Let's see if these equations are redox reactions or not:

a. In the reactants side \(\ce{Fe2O3}\) , Fe is has an oxidation number of +3. In the product \(\ce{Fe3O4}\) , Fe has an oxidation number of +2.67. Since Fe changed from +3 to +2.67, we can say that Fe had gained electrons and therefore reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.

b. In the reactant, \(\ce{Fe3O4}\) , Fe has an oxidation number of +2.67. In the product, FeO, Fe has an oxidation number of +2. Since the oxidation of Fe has changed from +2.67 to +2, electrons have been added therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.

c. In the reactant side, in FeO, Fe has an oxidation number of +2 and in the products side Fe has an oxidation number of 0. Since the oxidation number of Fe changed from +2 to 0, electrons have been gained and therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.

d. In the reactants C has an oxidation number of 0, and in the products side in \(\ce{CO2}\) , C has an oxidation number of +4. Since the oxidation number of C has changed from 0 to +4, we can say that C has been oxidized. In the reactants, in \(\ce{O2}\) oxygen has an oxidation number of 0, and in the products CO2, oxygen has an oxidation number of -2. Since the oxidation number of oxygen has changed from 0 to -2, oxygen has been reduced. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.

e. In the reactants \(\ce{CO2 }\) has an oxidation number of +4, and in the products side in CO, C has an oxidation number of +2. Since carbon went from +4 to +2, carbon has been reduced. In the reactants, in \(\ce{CO2 }\) oxygen has an oxidation number of -4 and in the products CO carbon has an oxidation number of -2. Since oxygen went from -4 to -2, it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.

f. In the reactants, \(\ce{CaCO3}\) Ca has an oxidation number of +2, and in products CaO Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.

g. In the products CaO Ca has an oxidation number of +2, and in the products \(\ce{CaSiO3}\) Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.

Q19.2.10

Draw the geometric, linkage, and ionization isomers for \(\ce{[CoCl5CN][CN]}\).

Solution:

Geometric Isomers: Geometric isomers come from restricted rotation. There are two types cis and trans, cis isomers are isomers with the same species being next to each other and trans isomers are isomers with different species or elements across from each other. Mer and Fac are also geometric isomers. Fac isomers are all angles B-M-B angles being 90^o. Mer isomers have one B-M-B angle being 180^o, and two B-M-B angles being 90^o. "B" being ions or ligands.

Linkage Isomers: Isomers in which two or more coordination compounds from the donor atom of at least one of the ligands are different, meaning that the connectivity between the atoms are different. There isomers can only exist when the compound had a ligand that bonds to the metal atoms in different ways. It is important to know that not all coordination compounds have linkage isomers

Ionization Isomers: There isomers are almost identical. Meaning that the ligand has to exchange places with the anion or the neutral molecule that was originally outside the coordination complex.

Q12.3.23

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

1. From these data, write the rate equation for this gas reaction. What order is the reaction in NO, \(Cl_2\) , and overall?
2. Calculate the specific rate constant for this reaction.

Solution:

1. To find the order of reaction for NO, we have to keep the pressures of \(Cl_2\)constant. Therefore to figure out the order of NO, we use the last two rows of the table:

\(\dfrac{rate_2}{rate_1}\) = k \(\dfrac{pNO_2}{pNO_1}^x\)

x= order of the reaction, k= rate of reaction

Plug in the numbers: \(\dfrac{1*10^{-2}}{4.0*10^{-2}}\) = k \(\dfrac{0.5}{1.0}^x\)

x= 2, therefore NO is a second order reaction

To find the order of the reaction for \(Cl_2\) , we have to keep the pressures of NO constant. Therefore, to figure out the order of \(Cl_2\) , we use the first and last rows of the table:

\(\dfrac{rate_2}{rate_1}\) = k \(\dfrac{(pCl_2)_2}{(pCl_2)_1}^y\)

y= order of the reaction, k= rate of reaction

Plug in the numbers: \(\dfrac{5.1*10^{-3}}{1.0*10^{-2}}\) = k \(\dfrac{0.5}{1.0}^y\)

y= 1, therefore \(Cl_2\) is a first order reaction

The overall rate equation for this gas reaction: rate= k \([{NO}]^2[{Cl_2}]\)

To figure out the overall order of the reaction, we have to add up the orders of both species which is 3 (2+1).

2. To find the rate constant of the reaction (k), we just pick any row in the table and plug the numbers into the rate equation from (#1), I will use the first row:

rate= k \([{NO}]^2[{Cl_2}]\)

\(5.1*10^{-3}\) = k \([{0.5}]^2[{0.5}]\)

k= 0.0408 \(\dfrac{mol}{atm^2*s}\)

To figure out the units for k, we plug in the units in the rate equation

The initial rate is atm/s, and the initial pressures are atm. So when we divide the atm over in the following equation leaving the K, we would be left with atm² in the denominator with seconds.

\(\dfrac{atm*mol}{s}\) = k \([{atm}]^2[{atm}]\)

k= \(\dfrac{mol}{atm^2*s}\)

Q12.7.3

Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:

\[O_{3} \xrightarrow{\text{sunlight}} O_{2} + O\]

\[O_{3} + Cl \xrightarrow{\text{sunlight}} O_{2} + ClO\]

\[ClO + O ⟶ O_{2} + Cl\]

1. Explain why chlorine atoms are catalysts in the gas-phase transformation

\[2O_{3} ⟶ 3O_{2}\]

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

\[O_{3} \xrightarrow{\text{sunlight}} O_{2} + O\]

\[O_{3} +NO ⟶ NO_{2} + ClO\]

\[NO_{2} + O ⟶ O_{2} + NO\]

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

Solution:

1. A catalysts works by lowering the energy of activation, and this enhances the rate of forward and backward reactions. Catalysts usually form an intermediate with the reactants in the initial step of the reaction and it is also released in the product formation. Chlorine atoms are a catalyst because they react in the second step but are regenerated in the third step. Thus, they are not used up, which is a characteristic of catalysts. Chlorine atoms are generated from CFCs, which are known to destroy ozone. The UV radiation break down the CFCs into chlorine atoms, and these chlorine atoms break the ozone down. As you can see, more chlorine is made in the last step to break more ozone molecules down. So overall, since chlorine is a reactant in the second step, it is being used to create products. Also Cl is a product, meaning it is produced and cancels out with the reactant.
2. NO is a catalyst for the same reason as in part (a), they act as intermediates on the reactant side in step two and are produced as products in step three. NO comes from nitrous oxide, usually man-made, has similarities to CFCs, the only difference is that nitrous oxide works to break the ozone layer through NO catalyzed reactions and CFCs work through chlorine catalyzed reactions.

Q21.4.26

Isotopes such as \(^{26}Al\) (half-life: \(7.2 × 10^5\) years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.

1. \(^{26}Al\) decays by β+ emission or electron capture. Write the equations for these two nuclear transformations.
2. The earth was formed about \(4.7 × 10^9\) (4.7 billion) years ago. How old was the earth when 99.999999% of the \(^{26}Al\) originally present had decayed?

Solution:

1. β+ emission is the decay of an isotope by a fast moving beta particle or an electron, which is represented by \(^{0}_{-1}e\). The mass number is 0, and the atomic number is -1. Since this is an emission or a decay, we put the electron on the products, and reduce the atomic number of particle by 1. This results in an atomic number of 14, which is the atomic number of silicon. If we put all this information together, we get an equation like so:

\(^{26}_{13}Al\) ⟶ \(^{0}_{-1}e\) + \(^{26}_{14}Si\)

In an electron capture, the electron \(^{0}_{-1}e\) is used, but it is used in the reactant side of the equation because it is capture. In addition, a neutrino is involved which is represented by \(^{0}_{0}v\) because it is ejected from the nucleus where the electron reacts with the isotope, so it is represented in the product side of the equation. Putting it all together, you get an equation like so:

\(^{26}_{13}Al\) + \(^{0}_{-1}e\) ⟶ \(^{0}_{0}v\) + \(^{26}_{12}Mg\)

2. Half life is represented by this equation, where λ=constant:

λ = \(\dfrac{0.693}{t_{1/2}}\)

Plugging in numbers: λ = \(\dfrac{0.693}{7.2 × 10^5}\) = \(9.63 × 10^{-7}\) years

If 99.999999% decayed, then about 0.000001% remains

To solve this problem, we use this equation:

ln \(\dfrac{n_{t}}{n_{0}}\) = -λt

λ= decay constant, given by the constant we calculated above \(9.63 × 10^{-7}\) years

\(n_{t}\) = concentration of isotope at time t (0.000001)

\(n_{0}\) = initial concentration of isotope (100)

We plug all the numbers in and calculate for t

ln \(\dfrac{0.000001}{100}\) = -\(9.63 × 10^{-7}\)*t

t= 19128432.76 years

We can round this to about 19,128,433 years

To find out how old the earth was when 99.999999% decayed we subtract this number from the age of earth: 4,700,000,000 years - 19,128,433 years= 4,680,871,567 years

Q20.3.13

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.

1. Zn(s)∣Zn2+(aq) ∥ H+(aq)∣H2(g), Pt(s)
2. Ag(s)∣AgCl(s)∣Cl−(aq) ∥ H+(aq)∣H2(g)∣Pt(s)
3. Pt(s)∣H2(g)∣H+(aq) ∥ Fe2+(aq), Fe3+(aq)∣Pt(s)

Solution:

In a galvanic cell for a half reaction to be spontaneous, the E^o_Cell must be positive. Also oxidation occurs at the anode, and reduction occurs at the cathode.

1. Half-reactions:

\(Zn_{(s)} ⟶ Zn^{2+}_{(aq)} + 2e^-\) occurs at the Anode (oxidation- electrons in the product side of the equation)

\(2H^{+}_{(aq)} + 2e^- ⟶ H_{2(g)}\) occurs at the Cathode (reduction- electrons in the reactant side of the equation )

Overall Reaction:

\(Zn_{(s)} + 2H^{+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + H_{2(aq)}\)

2. Half Reactions:

\(AgCl_{(s)} + e^- ⟶ Ag_{(s)} + Cl^-_{(aq)}\) occurs at the Cathode (reduction)

\(H_{2(g)} ⟶ 2H^+_{(aq)} + 2e^-\) occurs at the Anode (oxidation)

To get the overall reaction, we need to multiply the first equation by two- so that we can make the electrons in two equations equal, by doing so we get this:

\(2AgCl_{(s)} + 2e^- ⟶ 2Ag_{(s)} + 2Cl^-_{(aq)}\)

\(H_{2(g)} ⟶ 2H^+_{(aq)} + 2e^-\)

Since we have two electrons on either side of the equation, we can cancel those electrons out and add the equations to get the overall reaction.

Overall Reaction:

\(2AgCl_{(s)} + H_{2(g)} ⟶ 2H^+_{(aq)} + 2Ag_{(s)} + Cl^-_{(aq)}\)

3. Half Reactions:

\(Fe^{3+}_{(aq)} + e^- ⟶ Fe^{2+}_{(aq)}\) occurs at the Cathode (reduction)

\(H_{2(g)} ⟶ 2H^+_{(aq)} + 2e^-\) occurs at the Anode (oxidation)

Overall Reaction:

\(2Fe^{3+}_{(aq)} + H_{2(g)} ⟶ 2H^+_{(aq)} + 2Fe^{2+}_{(aq)}\)

Q20.5.25

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?

\(acetate + 2H^+ + 2e^− → acetaldehyde + H_{2}O\) E° = −0.58 V

\(FAD + 2H^+ +2e^− → FADH_2\) E° = −0.18 V

Solution:

By the E cell voltage of -0.18, we can tell that FAD is a fairly good reducing agent since its E cell is a negative number. However, for an effect oxidant for the conversion of acetaldehyde to acetate we need a good reducing agent. We can look at the overall E^o. Since we are looking at FAD as a oxidant, we can see it as the anode. So, -0.18-(-0.58)=+0.4V. Since it is positive it would be an effective oxidant.