Extra Credit 12

Question 17.2.1

Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.

1. Mg(s) + Ni²⁺(aq) ⟶ Mg²⁺(aq) + Ni(s)

2. 2Ag⁺(aq) + Cu(s) ⟶ Cu²⁺(aq) + 2Ag(s)

3. Mn(s) + Sn(NO₃)₂(aq) ⟶ Mn(NO₃)₂(aq) + Au(s)

4. 3Cu(NO₃)(aq) + Au(NO₃)₃(aq) ⟶ 3Cu(NO₃)₂(aq) +Au(s)

Expanded Solution 17.2.1

In order to answer the question, we need to understand that we are being asked to construct a cell diagram through the two half-reactions, as well as the identification of the anode and cathode. Since we know that cellular notation is read from the left to the right, we know to place the anode (where oxidation occurs) on the left and the cathode (where reduction occurs) on the right.

There are also several symbols we need to make note of:

• || represents the salt bridge between the two half-cells
• | separates different phases
• , (a comma) differentiates between species of the same phase

1. Mg(s) + Ni²⁺(aq) ⟶ Mg²⁺(aq) + Ni(s)

We can break this down into the two half-reactions:

Mg(s) ⟶ Mg²⁺(aq)

Ni²⁺(aq) ⟶ Ni(s)

Since the oxidation state of Mg(s) increases from 0 to 2, we know that oxidation takes place here, and thus we are able to identify it as the anode. Similarly, we can say that Ni²⁺(aq) is the cathode as its oxidation state decreases from 2 to 0.

Now that we have all the relevant information, we can construct the cell diagram:

Mg(s)│Mg²⁺(aq)║Ni⁺(aq)│Ni(s)

2. 2Ag⁺(aq) + Cu(s) ⟶ Cu²⁺(aq) + 2Ag(s)

We can break this down into the two half-reactions:

2Ag⁺(aq) ⟶ 2Ag(s)

Cu(s) ⟶ Cu²⁺(aq)

Since the oxidation state of Cu(s) increases from 0 to 2, we know that oxidation takes place here, and thus we are able to identify it as the anode. Similarly, we can say that Ag⁺(aq) is the cathode as its oxidation state decreases from 1 to 0.

Now that we have all the relevant information, we can construct the cell diagram:

Cu(s)│Cu²⁺(aq)║Ag⁺(aq)│Ag(s)

3. Mn(s) + Sn(NO₃)₂(aq) ⟶ Mn(NO₃)₂(aq) + Au(s)

We can break this down into the two half-reactions:

Mn(s) ⟶ Mn(NO₃)₂(aq)

Sn(NO₃)₂(aq) ⟶ Mn(NO₃)₂(aq)

Since the oxidation state of Mn(s) increases from 0 to 2, we know that oxidation takes place here, and thus we are able to identify it as the anode. Similarly, we can say that Sn(NO₃)₂(aq) is the cathode as its oxidation state decreases from 2 to 0.

Now that we have all the relevant information, we can construct the cell diagram:

Mn(s)│Mn²⁺(aq)║Sn²⁺(aq)│Sn(s)

4. 3Cu(NO₃)(aq) + Au(NO₃)₃(aq) ⟶ 3Cu(NO₃)₂(aq) +Au(s)

We can break this down into the two half-reactions:

3CuNO₃(aq) ⟶ 3Cu(NO₃)₂(aq)

Au(NO₃)₃(aq) ⟶ Au(s)

Since the oxidation state of 3Cu(NO₃)(aq) increases from 1 to 2, we know that oxidation takes place here, and thus we are able to identify it as the anode. Similarly, we can say that Au(NO₃)3(aq) is the cathode as its oxidation state decreases from 3 to 0.

Because this reaction does not include a solid, we can use Platinum as an inert electrode. Now that we have all the relevant information, we can construct the cell diagram:

Pt(s)│Cu⁺(aq), Cu²⁺(aq)║Au³⁺(aq)│Au(s)

Question 19.1.10

Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.

Expanded Solution 19.1.10

According to the Lewis acid-base theory, a lewis acid is any substance that can accept lone pair electrons. Similarly, a Lewis base is any chemical species that can donate lone pair electrons. The overall oxidation state of the complex Mn₂O₇ is 0 as it is neutral, and we know that Oxygen has an oxidation state of -2, therefore we can conclude that the oxidation state of Manganese is equal to +7 through simple algebra. The higher the oxidation state, the less stable the compound will be due to its tendency to gain stability by completing its octet through accepting electrons. Thus, it tends to act as an electrophilic group. Electron accepting groups are also known as acids, and therefore, Mn₂O₇ will have a pH of 7.00.

Question 19.3.2

Draw the crystal field diagrams for [Fe(NO₂)₆]⁴⁻ and [FeF₆]³⁻. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δ_oct to P for each complex.

Expanded Solution 19.3.2

Both of these complexes are octahedral complexes as they have six ligands each. Since we know that the complexes are octahedral, we know that their orbitals are split into two sublevels; the higher level is called e_g (consisting of d_z² and d_x²−y² orbitals), and the lower level is the t_2g (consisting of the d_xy, d_xz, and d_yz orbitals). The difference in energy between the two levels is called the crystal field splitting otherwise known as Δ_oct. The crystal field splitting is determined by several factors such as its strength to the metal atom (can conclude from the spectrochemical series). P is the amount of energy required to pair electrons together, and if it is less than the crystal field splitting then the electrons will remain in the lower t_2g level. However, if P is greater than Δ_oct, then the electrons will be able to inhabit both levels. When unpaired electrons exist, the species is called paramagnetic and is considered high-spin. Converseley, if only paired electrons exist, the species is called diamagnetic, and is considered low-spin.

A) [Fe(NO₂)₆]⁴⁻

The overall charge of the complex is -4, and we know that the charge of NO₂ is -1, therefore we can conclude that Fe is Fe(II) and that we have six electrons. From the spectrochemical series, we know that NO₂ is fairly strong, and therefore has a Δ_oct greater than P, meaning that it occupies the lower level, has no unpaired electrons, is low-spin, and diamagnetic.

B) [FeF₆]³⁻

The overall charge of the complex is -3 and the charge of F is -1, and using algebra we can conclude that Fe is is Fe(III) and that there are 6 electrons. F is fairly weak, therefore it has a lower Δ_oct causing the elctrons to fill up each orbital as single electrons first. Since only the first orbital is paired, the other four are unpaired, meaning that the species is high-spin and paramagnetic.

Question 12.4.2

Use the data provided to graphically determine the order and rate constant of the following reaction: SO2Cl2⟶SO2+Cl2

Plotting a graph of ln[SO₂Cl₂] versus t reveals a linear trend; therefore we know this is a first-order reaction:

Expanded Solution 12.4.2

Since this is a first-order reaction, we can use the equation:

ln([A]_t/[A]₀) = kt

We simply have to plug in our values:

ln(0.0415M/0.100M) = k(4.00 x 10⁴s)

ln(0.415) = k(4.00×10⁴s)

−0.8794767588 = k(4.00×10⁴s)

k = −2.1986919×10⁵ = -2.20 x 10⁵ s^-1

Question 12.7.5

For each of the following pairs of reaction diagrams, identify which of the pairs is catalysed:

Expanded Solution 12.7.5

Catalysts increase the rate of reaction by lowering the activation energy, thus causing the transition energy state to lower.

Q1. Graph A shows a first activation energy of E_a = 45kJ − 2.5kJ = 42.5kJ and a second activation energy of E_a = 42.5kJ − 2.5kJ = 40kJ

Graph B shows a first activation energy of Ea = 45kJ − 2.5kJ = 42.5kJ and a second activation energy of E_a = 37.5kJ − 2.5kJ = 35kJ

Because the activation energy is lower in graph B it must be the graph of the catalysed reaction.

Q2. Graph A shows a first activation energy of E_a = 45kJ − 35kJ = 10kJ and a second activation energy of E_a = 35kJ − 35kJ = 0kJ

Graph B shows a first activation energy of E_a = 40kJ − 35kJ = 5kJ and a second activation energy of E_a = 40kJ − 35kJ = 5kJ

Because the activation energy is lower in graph B it must be the graph of the catalysed reaction.

Question 21.4.28

Write a balanced equation for each of the following nuclear reactions:

1. mercury-180 decays into platinum-176
2. zirconium-90 and an electron are produced by the decay of an unstable nucleus
3. thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium-228 by beta decay
4. neon-19 decays into fluorine-19

Expanded Solution 21.4.28

Before we can solve the equations, we need to be familiar with certain particles in nuclear chemistry:

1. In this reaction, mercury loses the equivalent of a Helium nucleus (4 amu and 2 protons).

¹⁸⁰Hg ⟶ ¹⁷⁶Pt + ⁴α where Hg has 80 protons, Pt has 78 protons

2. In this reaction zirconium-90 and an electron are produced by decay, through algebra it can be determined that the original nucleus was Yttrium-90.

3. Thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into Actinium-228 by beta decay.

²³²Th --> ⁴α + ²²⁸Ra --> ⁰_-1β + ²²⁸Ac

4. Neon-19 decays into fluorine-19 and a beta particle.

¹⁹Ne --> ¹⁹F + ⁰₁β

Question 20.3.14

For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.

1. Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe²⁺(aq)
2. Fe³⁺(aq) + 1/2H₂(g) → Fe²⁺(aq) + H⁺(aq)

Expanded Solution 20.3.14

1. Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe²⁺(aq)

The first step is to write out the half-reactions:

Ag(s) ⟶ Ag⁺(aq)

Fe³⁺(aq) ⟶ Fe²⁺(aq)

Next, we identify the anode and the cathode:

Since Ag(s) increases its oxidation state from 0 to 1, we know that it is the anode and that it is the negative electrode. Similarly, Fe³⁺(aq) decreases its oxidation state from 3 to 2, therefore we know that is the cathode and is the positive electrode.

Finally, we can construct the cell diagram. Since no solid exists on the cathode side we must use platinum as an inert electrode.

Ag(s)│Ag⁺(aq)║Fe³⁺(aq),Fe²⁺(aq)│Pt(s)

2. Fe³⁺(aq) + 1/2H₂(g) → Fe²⁺(aq) + H⁺(aq)

The first step is to write out the half-reactions:

Fe³⁺(s) ⟶ Fe²⁺(aq)

1/2H₂(g) ⟶ H⁺(aq)

Next, we identify the anode and the cathode:

Since 1/2H₂(g) increases in oxidations state from 0 to 1, we know that it is the anode and the negative electrode.

Since Fe³⁺(aq) decreases its oxidation state from 3 to 2, we know that this is the cathode and is the positive electrode.

Finally, we can construct the cell diagram. Since no solid exists in the equation, we need to use platinum as an inert electrode:

Pt(s)│1/2H₂(g), H⁺(aq)║Fe³⁺(aq), Fe²⁺(aq)|Pt(s)

Question 20.5.27

Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction O₂(g) + 4H⁺(aq) + 4e ⟶ 2H₂O(l)

1. 1. Will aqueous acidic KMnO₄ evolve oxygen with the formation of MnO₂?
2. At pH 14.00, what is E° for the oxidation of water by aqueous KMnO₄ (1 M) with the formation of MnO₂?
3. At pH 14.00, will water be oxidized if you are trying to form MnO₂ from 2MnO₄²⁻− via the reaction 2MnO₄²⁻(aq) + 2H₂O(l) ⟶ 2MnO₂(s) + O₂(g) + 4OH⁻(aq)?

Expanded Solution 20.5.27

1. From the table of Standard Reduction Potentials, we know that KMnO₄ oxidizes the water, therefore the water is the cathode.

Next we can use the equation:

E^o = E^o_cathode − E^o_anode to find E^o

E^o = E^o_cathode − E^o_anode

E^o = 1.68V − 1.23VE^o = 0.45V

Yes, aqueous acidic KMnO₄ will evolve oxygen with the formation of MnO₂.

2. From the table of Standard Reduction Potentials, we know that KMnO₄ oxidizes the water, therefore the water is the cathode.

Next we can use the equation:

E^o = E^o_cathode − E^o_anode to find E^o

E^o = E^o_cathode − E^o_anode

E^o = 1.424V − 1.23VE^o = 0.194V

The E^o for the oxidation of water is 0.194V.

3. From the table of Standard Reduction Potentials, we know that KMnO₄ oxidizes the water, therefore the water is the cathode.

Next we can use the equation:

E^o = E^o_cathode − E^o_anode to find E^o

E^o = E^o_cathode − E^o_anode

E^o = 1.43V − 1.23VE^o = 0.20V

Yes, water be oxidized if you are trying to form MnO₂ from 2MnO₄²⁻.