Extra Credit 10

Q17.1.9

1. Why is it not possible for hydroxide ion 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 to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?

A1.)

Realize that we are basing our answer considering a balanced acidic solution. The characteristic that defines an acidic solution is having a pH below 1 x 10^-7 M. This means that hydroxide ion is out numbed by the amount of hydronium ion. Therefore, hydroxide ions would be the limiting reactant. For that reason, hydroxides wouldn't be found in the half reactions or in the overall balanced redox reaction in acidic conditions.

A2.)

This is very similar to the last question. Considering that we are balancing a basic solution (where pH is above 1 x 10^-7), hydroxide ions would out number hydronium ions. This makes the hydronium ions the limiting reactant and would only leave hydroxide ions in a balanced reaction. Thus no hydronium ions would be present in a half reaction or overall equation when balancing redox reactions in basic conditons.

Q19.1.8

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

1. \(\ce{3Fe2O3}(s)+\ce{CO}(g)⟶\ce{2Fe3O4}(s)+\ce{CO2}(g)\)
2. \(\ce{Fe3O4}(s)+\ce{CO}(g)⟶\ce{3FeO}(s)+\ce{CO2}(g)\)
3. \(\ce{FeO}(s)+\ce{CO}(g)⟶\ce{Fe}(l)+\ce{CO2}(g)\)
4. \(\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\)
5. \(\ce{C}(s)+\ce{CO2}(g)⟶\ce{2CO}(g)\)
6. \(\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)\)
7. \(\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)\)

​​​​​When considering these types of questions a helpful pneumonic to remember is OIL RIG (Oxidation is LOSS of electrons and Reduction is GAIN of electrons)

A1.) Reactants: \(O = -2, Fe = +3, C = +2, O = -2\)

Products: \(O = -2, Fe = +8/3, C = +4, O =-2\)

Iron is gaining electrons so it is reduced while carbon loses electrons and is oxidized. This is a redox reaction.

A2.) Reactants: \(O = -2, Fe =+8/3, C = +2, O = -2\)

Products: \(O = -2, Fe = +2, C = +4, O = -2\)

Iron is gaining electrons so it is reduced while carbon loses electrons and is oxidized. This is a redox reaction.

A3.) Reactants: \(O = -2, Fe = +2, C = +2 , O = -2\)

Products: \(Fe = 0, C = +4, O = -2\)

Iron is gaining electrons so it is reduced and carbon is losing electrons so it is oxidized. This is a redox reaction.

A5.) Reactants: \(C = 0, O= 0, C =+4, O=-2\)

Products: \(C = +2, O = -2\)

Carbon is losing an electron so it is getting oxidized while oxygen is gaining an electron so it is reduced. This is a redox reaction.

A6.) Reactants: \(Ca = +2, C= +4, O =-2\)

Products: \(Ca = +2, C= +4, O = -2\)

Notice that the oxidation numbers for both the reactants and the products are both the same. When that happens, the reaction is not considered a redox reaction because there is no oxidation or reduction taking place.

A7.) Reactants: \(Ca = +2, Si= +4, O =-2\)

Products: \(Ca = +2, Si= +4, O = -2\)

Similarly, the oxidation numbers for both the reactants and the products are both the same. Therefore the reaction is not an oxidation-reduction reaction.

Q19.2.10

Draw the geometric, linkage, and ionization isomers for [CoCl₅CN][CN].

*Each isomer except "A" and "The Original Structure" is an Ionization Isomer

Before we begin to answer these questions, lets first remind ourselves what geometric, linkage and ionization isomers are:

Geometric isomers are molecules that are locked into their spatial postions with respect to one another due to a double bond or a ring structure.

Linkage isomers occur with ambidentate ligands that are capable of coordinating in more than one way.

Ionization isomers are identical except for a ligand has exchanged places with an anion or a neutral molecule that was originally outside the coordination complex.

Now that we refreshed our memories, lets solve these problems:

A.) This is a linkage isomer because the CN ligand switches the linking atom from the carbon to the nitrogen.

B.) This is an ionization isomer because the ion is trading places with the Cl ligand. This change also makes this a geometric trans isomer, which is when the ligands are on the opposite side of the metal ion.

C.) From the isomer B, we see that from the ionization isomer, its ligand is switched (from being attached by the carbon atom to the nitrogen atom). This would make this isomer a linkage isomer transitioning from isomer B.

D.) From the ionization isomer, we see that two CN ligands are being linked by the nitrogen instead of the carbon in the form of a linkage isomer. These two ligands are opposite sides of the metal ion making this a geometric trans isomer

E.) This is an ionization isomer because the ion switches places with the ligand. Coincidentally this also forms a geometric cis isomer forming ligands adjacent to each other.

F.) Isomer "F" is similar to Isomer "E" however it is not a geometric cis isomer because one of the adjacent ligands swaps which atom is being linked to the central metal ion. Therefore this is a linkage isomer.

G.) Isomer "G" is similar to isomer "F". However, now both adjacent ligands are now in the same form because one of the ligands switches from carbon to nitrogen as a linkage isomer, which now makes this complex ion also a geometric cis isomer due to the linkage change.

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, Cl₂, and overall?
2. Calculate the specific rate constant for this reaction.

A1.)

From the data, you need to compare the rates of each of the molecules (Cl₂ and NO). In order to do this, you can use the data table. To solve the rate of the Cl₂ we can use the trials where the NO is a constant variable (i.e. .50 atm). Using those data points we can divide the final by the initial (1/.5) which would be 2.0. The same would be done to the initial rate (final/initial) which would give us roughly 2. We see that the rate for Cl₂ is roughly the same (2≈2) so we can conclude that this Cl₂ would be a first order reaction. The same would be done for NO. By using the trails where Cl₂ is constant (1.0) we can divide the initial over final (1.0/0.5) which would be 2.0. The same would be done for the initial reactions of the same trails (initial over final) which would equal 4.0. Since this is an increase of a factor of two (4.0/2.0), then NO is a second order rate reaction.

A2.)

\[\text{rate}=\text{k}\space\text{p}\text{(NO)}^2\text{p}\text{(Cl)}_2\]

Using the rate law equation, you can plug in any trial's value to determine "k", our rate constant.

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

By rearranging the equation you can get...

\[k=\frac{5.1*10^{-3}}{0.125}\]

Which would equal to...

\[k = 0.0408\frac{mol}{atm^2*s}\]

(The original was incorrect)

Q12.7.3

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: \[\ce{O3 \xrightarrow{sunlight} O2 + O}\\ \ce{O3 + Cl ⟶ O2 + ClO}\\ \ce{ClO + O ⟶ Cl + O2}\]

1. Explain why chlorine atoms are catalysts in the gas-phase transformation: \[\ce{2O3⟶3O2}\]
2. Nitric oxide is also involved in the decomposition of ozone by the mechanism: \[\ce{O3 \xrightarrow{sunlight} O2 + O\\ O3 + NO ⟶ NO2 + O2\\ NO2 + O ⟶ NO + O2}\]

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

Q21.4.26

Isotopes such as \(^{26}Al (\text{half life:} 7.2 x 10^5 \text{years})\) 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 \(4.7 x 10^9\) (4.7 billion) years ago. How old was the earth when 99.999999% of the \(^{26}Al\) originally present had decayed?

First lets remind ourselves what β⁺ emission or electron capture is:

β⁺ emission

• increases neutron number by one
• decreases proton number by one
• does not change mass number

electron capture

• decreases atomic number by one (one less proton)
• atomic mass number remains unchanged

Now lets answer these problems:

A1.) β⁺ emission: \(\ce{^{26}_{13}}Al = e^{-} + \ce{^{26}_{14}}Si\)

electron capture: \(\ce{^{26}_{13}}Al + e^{-} + \ce{^{26}_{12}}Mg\)

A2.) 99.999999% decayed = 0.000001% remaining.

\(t_{1/2}=(ln2/k)\)

\(k=9.627 x 10^{-7}s^{-s}\)

\[ln\tfrac{[A]_{t}}{[A]_{0}} = -kt\]

\[ln\tfrac{.000000001}{100} = -(9.627x10^{-7})t\]

Solving for time we get:

\[t=2.8701x10^{7}\]

Now in order to figure out how old the Earth was when 99.999999% decayed:

\[(4.7x10^{9})-(2.8701 x 10^{7})= 4.671299x10^{9} years\]

(The original was incorrect)

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)

When considering these types of questions two helpful pneumonics are:

• OIL RIG (Oxidation is loss of electrons and Reduction is gain of electrons)
• RED CAT & AN OX (Reduction occurs at the cathode & Oxidation occurs at the anode)

A1.) reduction: \(2H^+(aq) + 2e^- → H_2(aq); \text{cathode};\)

oxidation: \(Zn(s) → Zn^{2+}(aq) + 2e^-; \text{anode};\)

overall: \(Zn(s) + 2H^+(aq) → Zn^{2+}(aq) + H_2(aq);\)

The reduction reaction is located at the cathode region. The hydrogen is reduced because 2 electrons is being gained. These 2 electrons are from the anode, where oxidation occurs. The zinc is oxidized by losing 2 electrons and forming an aqueous zinc ion. The overall reaction includes both the oxidation and reduction reactions combined with the electrons balanced which do not show in the final equation.

A2.) reduction: \(AgCl(s) +2e^- → Ag(s) + Cl^-(aq); \text{cathode};\)

oxidation: \(H_2(g) → 2H^+(aq) + 2e^-; \text{anode};\)

overall: \(AgCl(s) + H_2(g) → 2H^+(aq) + Ag(s) + Cl^-(aq);\)

Within this reaction, we see that one of the half reactions is the decomposition of \(AgCl(s)\). The compound is decomposed through reduction, gaining an electron, and splitting into \(Ag(s)\) and \(Cl^- (aq)\). This electron is from the oxidation reaction where \(H_2\) is losing its 2 electrons to form 2 hydrogen ions.

A3.) reduction: \(Fe^{3+}(aq) + e^- → Fe^{+2}(aq); \text{cathode};\)

oxidation: \(H_2(g) → 2H^+(aq) + 2e^-; \text{anode};\)

overall: \(2Fe^{+3}(aq) + H_2(g) → 2H^+(aq) + 2Fe^{2+}(aq);\)

The reduction reaction at the cathode consisting of a \(Fe^{3+}\) ion gaining of an electron from the anode. The anode has an \(H_2\) that is oxidized which gives its electron to the \(Fe^{3+}\), creating the galvanic cell.

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?

\(\text{acetate} + {2H}^{+} + {2e}^{-} \to \text{acetaldehyde} + H_2O\;\;\;\;\;E^o = -0.58V\)

\(FAD + {2H}^{+} +{2e}^{-} \to FADH_{2}\;\;\;\;\;E^o =-0.18V\)

A1.) The question asks for the conversion of acetaldehyde to acetate, whereas the given equation is acetate to acetaldehyde. Notice when reversing the reaction, the E^o value does not change.

\(\text{acetaldehyde} + H_2O \to \text{acetate}+2H^2 + 2e^- \;\;\;E^o = -0.58V\)

(The original was incorrect)

In order to oxidize the acetaldehyde, we must have an effective oxidant. FAD must be reduced so acetaldehyde can be converted to acetate. To see if it's effective, we must look at the overall E to determine spontaneity.

\(E = E_{\text{reduction}} - E_{\text{oxidation}}\)

\(E= 0.4V\)

(The original was incorrect)

Therefore, FAD will be an effective oxidant for the conversion of acetaldehyde to acetate at pH: 4.00, since this reaction has a negative value and the reaction is spontaneous.