Extra Credit 24

Q17.3.3

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

Cu(s)│Cu²⁺(aq)║Au³⁺(aq)│Au(s)

Compounds on the left side of the cell notation refers to where oxidation (a loss of electrons) occurs at the anode. The double lines indicate the presence of a salt bridge. To the right lies the cathode, where reduction (a gain of electrons) occurs. The solid copper and gold represent electrodes. A single bar identifies individual solid, liquid, or aqueous phases within a half-cell.

Cu and Cu²⁺ will be the anode where the oxidation reaction occurs because it is on the left. Au and Au³⁺ will be the cathode and thus where reduction reaction occurs. These equations must be balanced using electrons and coefficients.

Anode (oxidation): \[3(Cu\rightarrow Cu^{2+}+2e^-) \]

Cathode (reduction): \[2(Au^{3+}+3e^-\rightarrow Au) \]

Overall reaction: \[3Cu(s)+2Au^{3+}(aq)\rightarrow 3Cu^{2+}(aq)+2Au(s) \]

The standard cell potential (E°_cell ) can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half reaction at the cathode. If E°_cell is greater than zero, then the reaction is considered spontaneous at standard conditions. If E°_cell is less than zero, then the reaction is considered non-spontaneous at standard conditions.Find the reduction potential of each electrode:

\[E°_{cell} = E°_{cathode} − E°_{anode} \]

E°_cathode or E°_Au³⁺/Au= 1.498 V

E°_anode or E°_Cu²⁺/Cu= 0.337 V

\[E°_{cell} = E°_{Au^3{^+}/Au} − E°_{Cu^2{^+}/Cu} \]

\[E°_{cell} = 1.498V − .337V \]

\[E°_{cell} = 1.16V \]

Yes, this reaction is spontaneous at standard conditions because E°_cell is greater than zero.

Solution: 3Cu(s)+2Au³⁺(aq) ⟶3Cu²⁺(aq)+2Au(s); +1.16 V; spontaneous

Q19.1.22

Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.

\[Co(NO_3)_2(s)\rightarrow Co_2O_3(s)+NO_2(g)+O_2(g)\]

Split the equation into its constituent oxidation and reduction half-reactions. Add H₂O molecules to balance oxygen and H⁺ ions to balance hydrogen. If there is a an imbalance of charge, add electrons as e^- accordingly. If this equation were to be balanced in basic solution, an OH^- molecule would be added for every H⁺ ion.

Oxidation: \[2Co(NO_3)_2(s) \rightarrow Co_2O_3(s) + 4NO_2(g)+H_2O(l) \]

\[2H^+(aq)+2Co(NO_3)_2(s) \rightarrow Co_2O_3(s) + 4NO_2(g)+H_2O(l) \]

\[2H^+(aq)+2e^-+2Co(NO_3)_2(s) \rightarrow Co_2O_3(s) + 4NO_2(g)+H_2O(l) \]

Reduction: \[2H_2O(l) \rightarrow O_2(g) \]

\[2H_2O(l) \rightarrow O_2(g)+4H^+(aq) \]

\[2H_2O(l) \rightarrow O_2(g)+4H^+(aq)+4e^- \]

In this reaction, cobalt changes from an oxidation state of +1 to +3 by oxidation, nitrogen from +5 to +4 by oxidation, and oxygen from -2 to 0 by reduction. In order to combine the two half-reactions, the coefficients must be balanced so that there are no electrons remaining and like terms are eliminated. The oxidation half-reaction must be multiplied by a factor of two. All H₂O, H⁺

Total Reaction: \[2H_2O(l)+4Co(NO_3)_2(s)+4e^-+4H^+ \rightarrow O_2(g)+2Co_2O_3(s)+8NO_2(g)+4H^++4e^-+2H_2O(l) \]

Final Balanced Reaction: \[4Co(NO_3)_2(s)\rightarrow O_2(g)+2Co_2O_3(s)+8NO_2(g) \]

Solution: \[4Co(NO_3)_2(s)\rightarrow O_2(g)+2Co_2O_3(s)+8NO_2(g) \]

Q12.4.14

There are two molecules with the formula C₃H₆. Propene, CH₃CH=CH₂, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:

When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of 5.95 × 10⁻⁴ s⁻¹. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5 °C?

A rate constant relates the rates of a reaction with the product of the concentration of its reactants in a constant proportion. Because the reaction of cyclopropane being converted into propene has the general form A → products, it is a first order reaction, so rate=k[A]. K is our rate constant. The half-life, t_1/2 equation that relates to the rate constant for a first order reaction is:

$$t_\frac{1}{2}=\frac{.693}{k}$$

$$t_\frac{1}{2}=\frac{.693}{.000595s^{-1}}$$

Thus, t_1/2= 1164.71 seconds or .324 hours.

We must convert the seconds into hours because the question is asking for the answer in terms of hours.

The equation below is a general equation for first order reactions. If it were zero or second order, the equation would be different, so it is very important to pay attention to the order of the reaction.

$$\frac{[A_t]}{[A_o]}=e^{(-kt)}$$

\[ e^{-(.000595)(2700)}=.20059(100)=20.06\]

20.06% remaining

Solution: 1.16 x 10³ s; 20% remains

Q21.2.9

Which of the following nuclei lie within the band of stability?

1. chlorine-37
2. calcium-40
3. ²⁰⁴Bi
4. ⁵⁶Fe
5. ²⁰⁶Pb
6. ²¹¹Pb
7. ²²²Rn
8. carbon-14

Belt (or Band) of Stability. Graph of isotopes by type of nuclear decay. Orange and blue nuclides are unstable, with the black squares between these regions representing stable nuclides. The solid black line represents the theoretical position on the graph of nuclides for which proton number is the same as neutron number (N=ZN=Z. Elements with more than 20 protons require more neutrons than protons to be stable. Figure used with permission from Wikipedia and transferred from Stability of the Atomic Nucleus.

If a nucleus is to the left or to the right of the band of stability, it will be unstable and radioactive. Beta emission is preferred to the left of the belt of stability, positron emission is preferred to the right of the belt of stability, and alpha decay is most common for elements whose atomic number is greater than 83. The nucleus will decay into other nuclei that are closer to or on the band of stability. During a decay reaction, an unstable isotope (or radioisotope) is converted into a more stable isotope. The closer in number the neutrons and protons are, the more likely an isotope is stable. Also, even numbers are typically more stable than odd numbers. The "magic" numbers of stability for protons are 2, 8, 20, 28, 50, 82, and 114. The "magic" numbers of stability for neutrons are 2, 8, 20, 28, 50, 82, 126, and 184. The optimal range of stability of the ratio of neutrons to protons is approximately 1 to 1 for light isotopes and 1 to 1.52 for heavy isotopes.

1. chlorine-37 has 20 neutrons and 17 protons. The ratio is 1.17 and 20 is an even number, so it will be stable.
2. calcium-40 has 20 neutrons and 20 protons. The ratio is 1 and both numbers are even, so it will be stable.
3. ²⁰⁴Bi has 121 neutrons and 83 protons. The ratio is 1.46, so it will be stable.
4. ⁵⁶Fe has 30 neutrons and 26 protons. The ratio is 1.15, and both numbers are even, so it will be stable.
5. ²⁰⁶Pb has 124 neutrons and 82 protons. The ratio is 1.51, and both numbers are even, so it will be stable.
6. ²¹¹Pb has 129 neutrons and 82 protons. The ratio is 1.57, so it will be unstable.
7. ²²²Rn has 136 neutrons and 86 protons. The ratio is 1.58, so it will be unstable.
8. carbon-14 has 8 neutrons and 6 protons. Although the ratio 1.33 is within the heavy isotope range, carbon is a light isotope, so it will be unstable.

Solution: (a), (b), (c), (d), and (e)

Q21.7.1

If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:

1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)

Radioactive Emmission Penetration

Alpha (α), beta (β), and gamma (γ) particles can be emitted if an unstable nucleus undergoes radioactive decay in order to achieve a more stable configuration. Alpha decay is the emission of a helium-4 nucleus as an alpha particle; produces a daughter nucleus with a mass number four fewer and a nuclear charge two fewer the parent nucleus. An alpha ray could be stopped by a single sheet of paper. Beta decay is the release of when a neutron is converted into a proton and a high energy electron as a beta particle, which has moderate penetration and ionization capabilities. There are two kinds of Beta decay: beta emission, and positron emission. Beta emission is coupled with the release of an anti-neutrino, and Positron emission is coupled with the release of a neutrino. Gamma rays are high-energy photons that a nucleus releases in an excited state when it returns to the ground state. A gamma ray is capable of passing through 1 cm of aluminum and is more penetrating than either alpha or beta radiation.

1. Since cobalt-60 releases powerful γ-rays, a thick piece of lead is needed for protection.

2. A thin shield of metal would protect against the beta particles emitted by molybdenum-99.

Solution: 1) thick piece of lead 2) thin shield of metal

Q20.4.10

For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?

1. measuring the potential of a Cl⁻/Cl₂ couple
2. measuring the pH of a solution
3. measuring the potential of a MnO₄⁻/Mn²⁺ couple

An electrode should not create an unwanted reaction. It should allow ion flow between the anode and the cathode so a reaction can continue. It should also not form a precipitate.

1. Platinum (Pt) should be used because it is an inert electrode yet a good conductor that does not participate in redox chemistry.

\[2Cl^-\rightarrow Cl_2+2e^-\], \[E°_{cell} = -1.36V \]

Since platinum has a standard reduction potential of +1.2, the difference between the measured potential and the corresponding E° is .16V.

2. A standard hydrogen electrode (SHE) should be used to measure the pH. The metal in a typical pH electrode is silver because it comes from the silver chloride that makes up the electrode.

\[H_2\rightarrow 2H^++2e^-\], \[E°_{cell} = 0V \]

Silver has a standard reduction potential of +0.7994V, so the difference between the measured potential and the corresponding E° is +.7994V.

3. Any electrode with a more negative standard reduction potential could be a potential reference electrode for a MnO₄⁻/Mn²⁺ couple.

\[MnO^{4-}+8H^++5e^-\rightarrow Mn^{2+}+4H_2O\], \[E°_{cell} = 1.5V \]

Solid copper has a standard reduction potential of +.337, so the difference between the measured potenetial and the corresponding E° is 1.163V

\[E°_{cell} = E°_{cathode} − E°_{anode} \]

\[E°_{cell} = 1.5V − .337V=1.163V \]

Q20.4.11

Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction:

\[H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)]\

In order to draw a cell diagram, one must determine which half-reaction occurs at the anode and cathode. Copper is reduced from an oxidation state of +2 to 0 and hydrogen changes from an oxidation state of 0 to +1 so it will be oxidized. No coefficients or spectator ions should be included. A single line identifies individual solid, liquid, or aqueous phases within a half-cell. Double lines indicate the presence of a salt bridge. To the right lies the cathode, where reduction (a gain of electrons) occurs. Electrodes are placed at the ends. Platinum (Pt) is often used as an electrode because it is inert and its potential determines the standard hydrogen electrode (SHE).

Cathode (reduction): \[Cu^{2+}(aq)+2e^-\rightarrow Cu(s)\]

Anode (oxidation): \[H_2(g)\rightarrow 2H^+(aq)+2e^-\]

\[ Pt(s) \,|H_2(g)|H^+(aq)||Cu^{2+}(aq)\,|Cu(s)\]

Solution: \[ Pt(s) \,|H_2(g)|H^+(aq)||Cu^{2+}(aq)\,|Cu(s)\]

Q20.8.2

What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?

When a metal is coated with a sacrificial layer, another more reactive metal that has a more negative electrode potential is placed over the metal to protect from corrosion, because it will be preferentially reacted. The more reactive metal acts as an anode and will corrode before the metal that is meant to be protected. Corrosion is irreversible damage caused by a chemical or an electrochemical reaction. Metals spontaneously corrode if they undergo oxidation. On earth, the atmosphere has plenty of oxygen to supply this type of reaction spontaneously. Sacrificial layering provides a cathodic protection. Galvanic protection is another name for a sacrificial layer, so it is not different.