Extra Credit 19

Q.17.2.8

(No question was provided so I made up the question that would suit the answers) What happens within the electrochemical cell with and without a salt bridge?

• Without a salt bridge

The circuit will be open where no current can flow from the anode to the cathode. The salt bridge is involved with maintaining the charge neutrality of the cell. Without the salt bridge there would be an accumulation of charge on either side of the cell and the current will not flow.

• With a salt bridge
  • Each half-cell remains electrically neutral and current can flow through the circuit.
  • Salt charge is needed to maintain the charge balance in the cell as the electrons are flowing from one half-cell to the other
    • electrons flow from the anode (-) to the cathode (+) where oxidation occurs at the anode and generates electrons and positively charged ions. The electrons move through the wire and leave unbalanced positive charge in the half-cell. To keep this charge balance, the salt bridge is added so that anions (-) and cations (+) will flow across the bridge.
    • NOTE: Anions (-) will migrate to the anode (-) and cations (+) will migrate to the cathode (+)

Q19.1.17

Predict the products of each of the following reactions.

• Fe(s)+H₂SO₄(aq)⟶ ?

Whenever a metal reacts with an acid, the products are salt and hydrogen. Because Fe is lower on the activity series, we know that when it reacts with an acid it will result in the formation of Hydrogen gas. To simplify the equation is:

\[Metal + Acid ⟶ Salt + Hydrogen\]

The salt produced will depend on the metal and in this case, the metal is iron (Fe) so the resulting equation would be:

\[\ce{Fe}(s)+\ce{H2SO4}(aq)⟶ \ce{FeSO4}(aq) + \ce{H2}(g)\]

This equation works out as the H2 is removed from H₂SO₂, resulting in a SO₄^2- ion where Fe will take on an oxidation state of Fe⁺² to form FeSO₄ which will be the salt in this example.

But since FeSO₄ and H2SO₄ are aqueous, the reactants and products can also be written as its ions where the overall equation can be:

\[\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO2^{−4}}(aq)⟶\ce{Fe2+}(aq)+\ce{SO2^{−4}}(aq)+\ce{H2}(g)+\ce{2H2O}(l)\]

• FeCl₃(aq)+NaOH(aq)⟶ ?

In this case, adding a metal hydroxide (NaOH) to a solution with a transition metal ion (Fe) will form a transition metal hydroxide (XOH). As iron is bonded to three chlorine atoms in the reactants side, it has the oxidation state of +3 where three hydroxide ions (OH^-) are needed to balance out the charges when they are bonded in the products. The remaining ions are Na⁺ and Cl^- where they bond together in a 1:1 ratio where there are 3 molecules of NaCl once the reaction is balanced.

The overall reaction will be:

\[\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶ \ce{Fe(OH)3}(s) + \ce{3NaCl}(aq)\]

NOTE: Fe(OH)₃(s) is a solid as it is a rule that all all transition metal hydroxides are insoluble and a precipitate is formed.

Since NaOH(aq) and NaCl(aq) are aqueous, we can write them out in their ion forms:

\[\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH^{−}}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)\]

• Mn(OH)2(s)+HBr(aq)⟶ ?

This is an example of a metal hydroxide reacting with an acid where a metal salt and water will always be formed:

\[Metal Hydroxide + Acid ⟶ Metal Salt + Water\]

When this rule is applied to this equation, we will get the following:

\[\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶ \ce{MnBr2}(aq)+\ce{2H2O}(l)\]

But to follow through with this question, the aqueous solutions such as HBr(aq) and MnBr₂(aq) can be re-written as:

\[\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)\]

• Cr(s)+O₂(g)⟶ ?

This is the general reaction of a metal reacting with oxygen which will always result in a metal oxide. However, the metal oxide is determined by the oxidation state of the metal so there may be several outcomes of this reaction such as:

\[\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)\]

\[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{2CrO}(s)\]

\[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{CrO2}(s)\]

\[\ce{2Cr}(s)+\ce{3O2}(g)⟶\ce{CrO3}(s)\]

However, Cr₂O₃ is the main oxide of chromium so it can be assumed that this is the general product of this reaction.

• Mn₂O₃(s)+HCl(aq)⟶?

This follows the general reaction of a metal oxide and an acid will always result in a salt and water

\[Metal Oxide + Acid ⟶ Salt + Water\]

Using this general reaction, similar to the general reactions above, the reaction will result in:

\[\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\]

However, since HCl is an aqueous solution, the overall equation can also be re-written as:

\[\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\]

• Ti(s)+xsF₂(g)⟶?

Titanium is able to react with the halogens where there are two oxidation state that titanium can be: +3 and +4. The following reactions follow each oxidation state accordingly:

\[\ce{2Ti}(s)+\ce{3F2}(g)⟶ \ce{2TiF3}(s)\]

\[\ce{Ti}(s)+\ce{2F2}(g)⟶\ce{TiF4}(s)\]

However, since there is the symbol "xs", this indicates that F₂ is added in excess so the second reaction is favored more as it drives the reaction to completion.

OVERALL:

\[\ce{Ti}(s)+\ce{xsF2}(g)⟶\ce{TiF4}(g)\]

Q19.3.9

Trimethylphosphine, P(CH₃)₃, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH₃)₃ can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?

First let's find out the molecular formula and then figure out the geometry. Using CHE 2A knowledge, we should calculate the mass of each individual element/compound, the amount of moles of the individual element/compound and then the mole ratio. In order to derive the molecular formula we first obtain the empirical formula and then divide the molecular mass by the empirical formula mass.

270 x (percentage of the individual element/compound in decimal form) = mass of the individual element/compound

Ni: 270 * 0.215 = 58.05g

Cl: 270 * 0.260 = 70.20g

P(CH₃)₃: 270 * 0.525 = 141.75g

Mass of the individual element/compound / Molar Mass = Moles of individual element/compound

Using the smallest mole value, we divide the other elements/compounds by that value to get the empirical formula of: NiCl₂(P(CH₃)₃)₂

With this empirical formula, the mass of this compound is ≈ 281 g which is close to 270 g so this is also the molecular formula

As there are a total of 4 ligands, the compound could have a tetrahedral or square planar shape. However, since it was given that the compound does not have any isomeric forms, the compound must have a tetrahedral form and not a square planar shape. Square planar compounds are able to have isomeric forms in the cis/trans form. Tetrahedral compounds can only have isomeric forms if all of the ligands are different, unlike this particular compound. In order to determine whether the shape is tetrahedral or square planar, we must use a gui balance to determine whether the compound is paramagnetic or diamagnetic.Square planar geometries are by nature low spin configurations and will therefore be diamagnetic. A tetrahedral geometry could either be a low or high spin configuration, and this can only be determined by checking ligand field strength withe respect to the spectrochemical series.

Q12.4.9

What is the half-life for the decomposition of O₃ when the concentration of O₃ is 2.35 × 10⁻⁶ M? The rate constant for this second-order reaction is 50.4 L/mol/h.

This can be calculated by using the calculation for a half life for a second order reaction:

\[t_{1/2}=\dfrac{1}{k[A]o}\]

So:

\[t_{1/2}=\dfrac{1}{(50.4M^{-1}/h)[2.35×10^{-6}M]} = \dfrac{1}{1.1844×10^{-4}h^{-1}} = 8443.1hr\]

Q21.2.4

For each of the isotopes in Question 21.5.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.

Since there may have been a typo in the question, I will do this question based on the isotopes in Question 21.5.1

• berkelium-244 (²⁴⁴Bk)

Since we know that the mass number (A) of Bk is 244, we can find that the unique atomic number (Z) for berkelium is 97 which is the number of protons in this isotope.

NOTE: The atomic number for each element is unique to it because if the atomic number changes, then the element will also change as well. (This is what occurs during nuclear transmutation)

We can use math to calculate the number of neutrons by doing:

Mass number (A) - Atomic Number (Z) = No. of neutrons

\[244 - 97 = 147\]

Since this is a neutral atom, the number of electrons is equal to the atomic number (Z)

FINAL: Protons = 97, Neutrons = 174, Electrons = 97

• fermium-254 (²⁵⁴Fm)

The known mass number (A) of Fm is 254 where the atomic number (Z) is 100. Math can be used to calculate the number of neutrons

\[Neutrons = 254 - 100 = 154\]

FINAL: Protons = 100, Neutrons = 154, Electrons = 100

• lawrencium-257 (²⁵⁷Lr)

The known mass number (A) of Lr is 257 where the atomic number (Z) is 103. Math can be used to calculate the number of neutrons

\[Neutrons = 257 - 103 = 154\]

FINAL: Protons = 103, Neutrons = 154, Electrons = 103

• dubnium-260 (²⁶⁰Db)

The known mass number (A) of Db is 260 where the atomic number (Z) is 105. Math can be used to calculate the number of neutrons

\[Neutrons = 260 - 105 = 155\]

FINAL: Protons = 105, Neutrons = 155, Electrons = 105

Q21.5.7

Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.

Based on knowledge of nuclear chemistry, uranium is traditionally used in a nuclear power plant as a fuel where it would undergo nuclear fission. This is a process where a heavier atom is split into two lighter ones where a large amount of energy is released. The energy that is required for the nucleons to be held together in the nucleus is referred to as binding energy.This is the energy that is released as heat energy which can be harnessed to be converted into electrical energy. Similar to the typical methods to generate electricity, an external steam generator is used where the generated heat energy is used to heat up water to turn into steam. This steam rises and travels through pipes where it would then turn turbines to make electricity.

Q20.4.5

Write a cell diagram representing a cell that contains the Ni/Ni²⁺ couple in one compartment and the SHE in the other compartment. What are the values of E°_cathode, E°_anode, and E°_cell?

To write the cell diagram for this cell, we need to find out the E°_cathode, E°_anode, and E°_cell which can be worked out based off the standard electrode potentials. We need to find out the half cell reactions and potentials for Ni/Ni²⁺ and SHE.

NOTE: SHE is the Standard Hydrogen Electrode where hydrogen gas is used to compare all other values to. These reactions are always carried out at standard conditions of 1M H⁺, 1 atm and at 273K. Hydrogen in this case as a cell potential of 0.00V where H/H⁺ can be oxidized or reduced depending on the reaction.

As Nickel has the more negative standard reduction potential (SRP) value than H₂, Ni is less likely to be reduced than H₂. Following the OIL RIG mnemonic, oxidation is loss and reduction is gain. We can see the H₂ will be reduced and gain electrons whereas Ni will be oxidized and gain electrons. So the redox equations should be:

\[Ni^{2+}(aq)+2e^-→Ni(s)\]

\[H+(aq)+2e^-→H_{2}(g)\]

We should also remember that oxidation ALWAYS occurs at the anode and reduction ALWAYS occurs at the cathode so we can conclude that:

Anode: Ni²⁺(aq)+2e⁻→ Ni(s) (-0.257V)

Cathode: H⁺(aq)+2e⁻→H₂(g) (0.000V)

NOTE: You can try to remember the mnemonic RED CAT where REDucition always occurs at the CAThode.

Since we know which reaction is at the cathode and anode, we can use their SRP values to calculate the Standard Electrode Potential with the equation:

E°_cell = E°_cathode - E°_anode

E°_cell = (0.000V) - (-0.257V) = 0.257V

The overall reaction will be: H⁺(aq)+Ni(s)→H₂(g)+Ni²⁺(aq)

As we know everything about this reaction, we can construct the cell diagram which will always illustrate everything about the reaction, including what is being reduced and oxidized.

NOTE: In a cell diagram, the left side always indicates the half-cell reaction that is at the cathode (being reduced) and the right side always indicates the half-cell reaction at the anode (being oxidized)

• At the cathode, solid Ni is being converted into its Ni2+ ions (since oxidation occurs at the anode) where Ni will be used as the electrode so the phase boundary will be in between the solid and aqueous ions
• At the anode, hydrogen gas is formed however (since reduction occurs at the cathode) , the solid that is used in the SHE is always platinum (Pt)

NOTE: When writing down cell diagrams with hydrogen, always separate H⁺(aq) and H₂(g) with a phase boundary before the Pt electrode

NOTE: Always write down all of the states of all of the ions and elements/ compounds

Ni(s)∣Ni²⁺(aq)∥H⁺(aq, 1 M)∣H₂(g, 1 atm)∣Pt(s)

Q20.7.3

What type of battery would you use for each application and why?

1. powering an electric motor scooter
2. a backup battery for a smartphone
3. powering an iPod

SOLUTION:

1. For an electric motor scooter, you would want a battery that you can recharge so you that you can use up the electricity in the scooter and recharge it for further use. Thus, you would want a secondary battery for this purpose. An example that we've learnt for this purpose is a Lead-Acid (storage) battery which can also be found in cars but it can also be used for an electric motor scooter as well. Secondary batteries can be recharged easillly unlike primary batteries and is therefore cost and energy efficient. Moreover it is not as harmful to the environment as primary cells.
2. Assuming this is not for iPhone users (would be hard to replace the battery yourself...), you would want to have a primary battery that cannot be recharged as this would have a single use. However, there are three primary batteries that could be used:
   • Lechanche dry cell (alkali battery)
     • although this dry cell may be inexpensive to manufacture, it only produces about 1.55V and is not efficient in producing electrical energy as only a small amount is produced. Moreover, there is a limited shelf life where the dry cell may corrode and have its contents leak out. You would want a battery that is able to be stored until needed so the dry cell does not fulfill this category.
   • Button battery
     • these button batteries tend to be used for devices that are smaller such as power watches, calculators and cameras. In the case of a smartphone, a larger battery is preferred. These batteries are also energy and cost efficient since they can be used repeatedly after charging.
   • Lithium iodine battery
     • this cell type is the best for this purpose as these batteries can be long-lived and are reliable. Thus, they are the most common type of battery where you would not want to replace the battery. They are also portable yet powerful and can therefore be used to power such devices.
3. For the use of an iPod, you would want a rechargeable battery as the battery would be drained after each use, however, you would want to recharge it. Thus, having a secondary battery would be the most useful for this purpose. Examples that we have learnt could be the Nickel-Cadmium (NiCad) battery, Nickel-metal hydride battery (NiMH), and Lead-Acid (Lead storage) battery. Howeover, the Lead-Acid (lead storage) battery is most commonly used in cars and is far too large for a small ipod so the NiCad and NiMH would be preferred for thus purpose.