Extra Credit 19

Q17.2.8 (My Question Changed Since I Started)

Comments: Very good work! All the solutions seem to have correct answers! :)

An aqueous solution containing the following ions has 2.15 A of current passed through for 60 min, how much metal (in g) is formed at the cathode in each solution?

1. \(\mathrm{Zn^{2+}}\)
2. \(\mathrm{Al^{3+}}\)
3. \(\mathrm{Ag^{+}}\)
4. \(\mathrm{Ni^{2+}}\)

Answer:

Explanation (Using Zinc)

Step 1: The first step is to find the half-reaction for zinc. You can do this by looking at an oxidation or reduction chart and find Zn on there. Since zinc has a +2 charge attached to it, we need to add two electrons to it to make it a neutral Zn, which means it undergoes a reduction reaction. Once you have the half reaction, you can make the statement that there are 2 moles of electrons for every 1 mole of Zn, using the mole to mole ratio concept.

Step 2: You need to convert the minutes into seconds in order to plug it into the Faraday's Law equation. You do this by using dimensional analysis and cancelling out the minutes to get seconds with the ratio that there are 60 seconds for every 1 minute.

Step 3: You use Faraday's Law in order to find the number of electrons that pass through the circuit. You have your seconds and your current, and Faraday's constant is given, so all you do is plug it into the equation to find out how many moles of electrons flow through the circuit.

Step 4: Now that you have the number of moles of electrons that flow through the circuit, you can use that number and the half-reaction from step 1 to find the moles of Zn. You do this by multiplying your number from step 3 by the ratio we created in step 1. This would cause the moles of electrons to cancel out leaving you with moles of Zn.

Step 5: The final step is to find the mass of zinc that is associated with the mole value found. To do this, you find the molar mass of zinc on the periodic table and multiply it by the moles you found in step 4, leaving you with the mass, in grams, of Zinc.

You follow these steps to solve for the remaining metals, but instead using a different half-reaction and molar mass based upon the respective metal.

Work

1.

Step 1:

\[\mathrm{Zn^{2+}}(aq)+2\mathrm{e^{-}}\rightarrow \mathrm{Zn}(s)\]

which means that for every 1 mol Zn there is 2 mol e^-.

Step 2:

\[60\: min\: \times \: \frac{60\: sec}{1\: min}\: =\: 3600\: sec\]

Step 3:

\[n_{e}= \frac{It}{F}\]

where I is the current in A, t is the time it takes in sec, and F is Faraday's constant 96485 C/mol

\[n_{e}= \frac{2.15\: A\: \times 3600\: sec}{96285\: C/mol}= 0.08022\: \mathrm{e^{-}}\]

Step 4:

\[0.08022\: \mathrm{e^{-}}\: \times \frac{\mathrm{1\: mol\: Zn}}{\mathrm{2\: mol\: e^{-}}}= 0.04011\: \mathrm{mol\: Zn}\]

Step 5:

\[0.04011\: \mathrm{mol\: Zn}\: \times \: \frac{65.38\: \mathrm{g\: Zn}}{1\: \mathrm{mol\: Zn}}= \: \mathbf{2.62\: \mathrm{\mathbf{g\: Zn}}}\]

2.

Step 1:

\[\mathrm{Al^{3+}}(aq)+3\mathrm{e^{-}}\rightarrow \mathrm{Al}(s)\]

which means that for every 1 mol Al there is 3 mol e^-.

Step 2:

\[60\: min\: \times \: \frac{60\: sec}{1\: min}\: =\: 3600\: sec\]

Step 3:

\[n_{e}= \frac{It}{F}\]

where I is the current in A, t is the time it takes in sec, and F is Faraday's constant 96485 C/mol

\[n_{e}= \frac{2.15\: A\: \times 3600\: sec}{96285\: C/mol}= 0.08022\: \mathrm{e^{-}}\]

Step 4:

\[0.08022\: \mathrm{e^{-}}\: \times \frac{\mathrm{1\: mol\: Al}}{\mathrm{3\: mol\: e^{-}}}= 0.02674\: \mathrm{mol\: Al}\]

Step 5:

\[0.02674\: \mathrm{mol\: Al}\: \times \: \frac{26.98\: \mathrm{g\: Al}}{1\: \mathrm{mol\: Al}}= \: \mathbf{0.721\: \mathrm{\mathbf{g\: Al}}}\]

3.

Step 1:

\[\mathrm{Ag^{+}}(aq)+\mathrm{e^{-}}\rightarrow \mathrm{Ag}(s)\]

which means that for every 1 mol Ag there is 1 mol e^-.

Step 2:

\[60\: min\: \times \: \frac{60\: sec}{1\: min}\: =\: 3600\: sec\]

Step 3:

\[n_{e}= \frac{It}{F}\]

where I is the current in A, t is the time it takes in sec, and F is Faraday's constant 96485 C/mol

\[n_{e}= \frac{2.15\: A\: \times 3600\: sec}{96285\: C/mol}= 0.08022\: \mathrm{e^{-}}\]

Step 4:

\[0.08022\: \mathrm{e^{-}}\: \times \frac{\mathrm{1\: mol\: Ag}}{\mathrm{1\: mol\: e^{-}}}= 0.08022\: \mathrm{mol\: Ag}\]

Step 5:

\[0.08022\: \mathrm{mol\: Ag}\: \times \: \frac{107.87\: \mathrm{g\: Ag}}{1\: \mathrm{mol\: Ag}}= \: \mathbf{8.65\: \mathrm{\mathbf{g\: Ag}}}\]

4.

Step 1:

\[\mathrm{Ni^{2+}}(aq)+2\mathrm{e^{-}}\rightarrow \mathrm{Ni}(s)\]

which means that for every 1 mol Ni there is 2 mol e^-.

Step 2:

\[60\: min\: \times \: \frac{60\: sec}{1\: min}\: =\: 3600\: sec\]

Step 3:

\[n_{e}= \frac{It}{F}\]

where I is the current in A, t is the time it takes in sec, and F is Faraday's constant 96485 C/mol

\[n_{e}= \frac{2.15\: A\: \times 3600\: sec}{96285\: C/mol}= 0.08022\: \mathrm{e^{-}}\]

Step 4:

\[0.08022\: \mathrm{e^{-}}\: \times \frac{\mathrm{1\: mol\: Ni}}{\mathrm{2\: mol\: e^{-}}}= 0.04011\: \mathrm{mol\: Ni}\]

Step 5:

\[0.04011\: \mathrm{mol\: Ni}\: \times \: \frac{58.69\: \mathrm{g\: Ni}}{1\: \mathrm{mol\: Ni}}= \: \mathbf{2.35\: \mathrm{\mathbf{g\: Ni}}}\]

Q19.1.17

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

1. \(\mathrm{Fe}(s)\: +\:\mathrm{ H_{2}SO_{4}}(aq)\rightarrow\)
2. \(\mathrm{FeCl_{3}}(aq)\: +\:\mathrm{ NaOH}(aq)\rightarrow\)
3. \(\mathrm{Mn(OH)_{2}}(s)\: +\:\mathrm{HBr}(aq)\rightarrow\)
4. \(\mathrm{Cr}(s)\: +\:\mathrm{O_{2}}(g)\rightarrow\)
5. \(\mathrm{Mn_{2}O_{3}}(s)\: +\:\mathrm{HCl}(aq)\rightarrow\)
6. \(\mathrm{Ti}(s)\: +\:xs\mathrm{F_{2}}(aq)\rightarrow\)

Answer:

Explanation (Using #1)

\(\mathrm{Fe}(s)\: +\:\mathrm{ H_{2}SO_{4}}(aq)\rightarrow\) ?

Step 1: Determine the type of reaction

This one is a precipitation reaction, so we must check the solubility rules. This is to see whether a precipitate was formed or not, thus seeing if the reaction went to completion.

Step 2: Find the Products

We must find the products in order to see if a precipitate forms or not. To do this, we need to check the charge of all of the molecules involved and see which can replace which. In this reaction, we have Fe with a charge of 0, H₂ with a charge of +2 since it is attached to the polyatomic ion SO₄, which has a charge of -2. Seeing this, we now can replace H₂ by Fe and have H₂ by itself.

\[\mathrm{Fe}(s)\: +\:\mathrm{ H_{2}SO_{4}}(aq)\rightarrow\: \mathrm{H_{2}}(g)\: +\: \mathrm{FeSO_{4}}\]

Step 3: Determine if the products are soluble or insoluble

By checking the solubility chart, we find that FeSO₄ is soluble meaning it will have an aqueous attached to it. The final result is given below.

\[\mathrm{Fe}(s)\: +\:\mathrm{ H_{2}SO_{4}}(aq)\rightarrow\: \mathrm{H_{2}}(g)\: +\: \mathrm{FeSO_{4}}(aq)\]

Step 4:

Break up the reaction into its soluble particles and nonsoluble particles to see which are the spectator ions and which are essential for the reaction. Make sure to balance out all the charges and ions in the reaction from both the product and reactant side.

\[\mathrm{Fe(s)}\: +\: \mathrm{2H_{3}O^{+}}(aq)\: +\: \mathrm{SO_{4}\, ^{2-}}(aq)\rightarrow \mathbf{\mathrm{\mathbf{H_{2}}(g)}\: +\: \mathrm{\mathbf{Fe^{2+}}}(aq)\: +\: \mathrm{\mathbf{SO_{4}\, ^{2-}}}(aq)\: +\: \mathrm{\mathbf{2H_{2}O}}(l)}\]

This would be the final breakdown of the reactants of the equation.

Work

2. \(\mathrm{FeCl_{3}}(aq)\: +\:\mathrm{ NaOH}(aq)\rightarrow\)

Charges of Each Element in the Compounds:

Fe: +3

Cl: -3

Na: +1

OH: -1

Products Formed:

\(\mathrm{3NaCl}(aq)\: +\: \mathrm{Fe(OH)_{3}}(s)\)

Forms the precipitate Fe(OH)₃

Breakdown into Ions:

\[\mathrm{FeCl_{3}}(aq)\: +\: \mathrm{3Na^{+}}(aq)\: +\: \mathrm{3OH^{-}}(aq)\rightarrow \mathrm{\mathbf{3Na^{+}}}(\mathbf{aq})\: +\: \mathrm{\mathbf{3Cl^{-}}}(\mathbf{aq})\: +\: \mathrm{\mathbf{Fe(OH)_{3}}}(\mathbf{s})\]

3. \(\mathrm{Mn(OH)_{2}}(s)\: +\:\mathrm{HBr}(aq)\rightarrow\)

Charges of Each Element in the Compounds:

Mn: +2

OH: -2

H: +1

Br: -1

Products Formed:

\(\mathrm{MnBr_{2}}(aq)\: +\: \mathrm{2H_{2}O}(l)\)

MnBr₂ is soluble

Breakdown into Ions:

\[\mathrm{Mn(OH)_{2}}(s)\: +\: \mathrm{2H_{3}O^{+}}(aq)\: +\: \mathrm{2Br^{-}}\rightarrow \mathrm{\mathbf{Mn^{2+}}}(\mathbf{aq})\: +\: \mathrm{\mathbf{2Br^{-}}}(\mathbf{aq})\: +\: \mathrm{\mathbf{4H_{2}O}}(\mathbf{l})\]

4. \(\mathrm{Cr}(s)\: +\:\mathrm{O_{2}}(g)\rightarrow\)

Charges of Each Element in the Compounds:

Cr: 0

O₂: 0

Products Formed:

\(\mathrm{\mathbf{Cr_{2}O_{3}}}(\mathbf{s})\)

This was a synthesis reaction that created chromium (III) oxide based on their respective ionic charges on the periodic table, taking note that Cr is a transition metal

Final Reaction:

\[\mathbf{4Cr(s)\: +\: 3O_{2}(g)\rightarrow 2Cr_{2}O_{3}(s)}\]

5. \(\mathrm{Mn_{2}O_{3}}(s)\: +\:\mathrm{HCl}(aq)\rightarrow\)

Charges of Each Element in the Compounds:

Mn: +3

O: -2

H: +1

Cl: -1

Products Formed:

\(\mathrm{MnCl_{3}}(s)\: +\: \mathrm{H_{2}O}(l)\)

MnCl₃ is insoluble based on the solubility rules

Breakdown into Ions:

\[\mathrm{Mn_{2}O_{3}}(s)\: +\: \mathrm{6H_{3}O^{+}}(aq)\: +\: \mathrm{6Cl^{-}}(aq)\rightarrow \mathrm{\mathbf{2MnCl_{3}}}(\mathbf{s})\: +\: \mathrm{\mathbf{9H_{2}O}}(\mathbf{l})\]

6. \(\mathrm{Ti}(s)\: +\:xs\mathrm{F_{2}}(aq)\rightarrow\)

Charges of Each Element in the Compounds:

Ti: 0

F₂: 0

Products Formed:

\[\mathrm{Ti}(s)\: +\:xs\mathrm{F_{2}}(aq)\rightarrow \mathrm{\mathbf{TiF_{4}}}(\mathbf{g})\]

Ti has a +4 charge as an ion, meaning F, which has a -1 charge, would need four to bind on to Ti and create a 0 overall charge

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?

Answer:

Work

Step 1: Assume the percentage is the grams of each element out of 100 grams.

\[\frac{\mathrm{x\: Percentage}}{100}=\mathrm{x\: grams}\]

\(\frac{21.5\: \mathrm{g}\: \mathrm{Ni}}{58.69\:\mathrm{ g\: Ni}}=0.366\: \mathrm{mol\: Ni}\)

\(\frac{26\: \mathrm{g}\: \mathrm{Cl}}{35.45\:\mathrm{ g\: Cl}}=0.733\: \mathrm{mol\: Cl}\)

\(\frac{52.5\: \mathrm{g}\: \mathrm{P(CH_{3})_{3}}}{76.08\:\mathrm{ g\: P(CH_{3})_{3}}}=0.69\: \mathrm{mol\: P(CH_{3})_{3}}\)

Step 2: The lowest one is 0.366 moles of Ni, so divide each mole by that number.

\(\frac{0.366\: \mathrm{mol\: Ni}}{0.366\: \mathrm{mol}}=1\)

\(\frac{0.733\: \mathrm{mol\: Cl}}{0.366\: \mathrm{mol}}=2.00\)

\(\frac{0.69\: \mathrm{mol\: P(CH_{3})_{3}}}{0.366\: \mathrm{mol}}=1.88\mathrm{\: \: which\: \: rounds\: \: up\: \: to\: \: 2}\)

Step 3: The values calculated creates the empirical formula of NiCl₂(P(CH₃)₃)₂.

Step 4: Since it states that the molecular mass is approximately 270 g, we have to calculate of the empirical formula and find the ration between the two.

Ni - 58.69 g

Cl - 35.45 g

P(CH₃)₃ - 76.08 g

So 58.69 + 2(35.45) + 2(76.08) = 281.75

\(\frac{270\: \mathrm{g}}{281.75\: \mathrm{g}}=0.958\) which rounds up to 1, meaning the empirical formula we found is the molecular formula the question asked for, which is NiCl₂(P(CH₃)₃)₂.

Step 5: The problem contains a metal Ni, two ligands P(CH₃)₃, and 2 Cl. Ni is the metal cation that has the two ligands and chlorine attached to it. Since in the formula there are two P(CH₃)₃ and Cl, that means that there are four attachments to Ni. The shape is tetrahedral, with the center being Ni.

Explanation

Step 1: In order to calculate the moles of each species, assume that the percentage means the grams out of 100 g, in order to make it easier to compute the moles. Once you have done that, you need to find the molar mass of each species using the periodic table. Divide the grams of each element by its molar mass to obtain the moles of each element.

Step 2: To find the number that each element is present in, you take the lowest mole number out of each element involved and it divide each element's mole number by it. In this case, it was Ni that had the lowest, so the moles of Cl and P(CH₃)₃ were divided by the mole number of Ni, which was 0.366. Ni divided by itself yields 1, Cl yielded a value very close to 2, and P(CH₃)₃ did not give an exact whole number, but it was relatively close to 2. We can round 1.88 up to 2 in this case.

Step 3: Now using the values we obtained, this is the subscript for each element in the empirical formula. There is 1 Ni, 2 Cl, and 2 P(CH₃)_3.

Step 4: To solve for the molecular formula, we take the ratio of the given molecular mass and the empirical molecular mass and the number it yields is what we would multiply each subscript with. Once we calculated the empirical molecular mass, we divided the molecular mass by it and got a number close to 1, which means we do not need to change the subscripts.

Step 5: In order to find the structure, the easiest way would be to draw it out, but another way would be to think of the bonds each has with the center metal cation Ni. Since there are four bonds in the molecule, we can rule out octahedral and linear, leaving us with either tetrahedral or square planar as the geometry of the molecule. The reason it is tetrahedral is because if it were square planar, the molecule would possess cis/trans isomers, which are different ways to construct the molecule. This one does not, leading to the conclusion that it is in fact tetrahedral.

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.

Answer:

Work

Step 1: For a second-order reaction, the formula for half life is t_1/2= 1/k[A]₀ where k is the rate constant in L/mol/s and [A]₀ is the initial concentration of the compound.

Given:

\(\mathrm{k}=50.4 \frac{\mathrm{L}}{\mathrm{mol}\cdot \mathrm{h}}\)

\([\mathrm{A}]_{0}=2.35\times 10^{-6}M\)

Step 2: We then plug in our unknowns into the equation:

\[\mathrm{t}_{1/2}=\frac{1}{(50.4\frac{\mathrm{L}}{\mathrm{mol}\cdot \mathrm{h}})(2.35\times 10^{-6}M)} =8443.09\: h=\: \mathbf{8443\: h}\]

Overall, it takes 8433 hours for the concentration to decay by half. We use a second-order reaction in order to solve for the half life which has different equations than other reactions to solve for the half life, rate constant, and more. It is also important to remember the units for each kind of reaction equation.

Explanation

Step 1: The first step is to recognize the half life equation for a second-order reaction. After we have figured that out, we are able to have the basis of how to solve for the half life. Another way to solve for the half life is to use the integrated form of second-order reactions and isolate t. This would look like t_1/2 = (1/[A] - 1/[A]₀)/k, where [A] is just half the concentration of [A]₀, since we are solving for half life. Either method is acceptable, but method 1 is easier since it involves a simpler equation.

Step 2: We have to convert k into units that are able to be plugged into the equation. In the equation, the half life is in seconds, so our rate constant must have seconds in there as well. But we have L/mol/h, and in order to convert it to seconds, we must cancel out the hours by multiplying it by the ratio 1 h/3600 sec. Once we do this, we are able to obtain the k in the proper units, L/mol/s.

Step 3: Plug the new value of k and the initial concentration given in the problem into the second-order reaction equation for half life to obtain the time in seconds.

Q21.2.4

For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.

Isotopes in Question 21.2.3:

(a) \(_{14}^{34}\textrm{Si}\); (b) \(_{15}^{36}\textrm{P}\); (c) \(_{25}^{57}\textrm{Mn}\); (d) \(_{56}^{121}\textrm{Ba}\)

Answer:

Explanation (Using Si)

Step 1: In order to find the amount of protons, electrons, and neutrons, we must first find the atomic number, which is Z. Z tells us the amount of protons in the element. This number for each element can be found on the periodic element chart. Since all the elements in this problem are neutral, the number of electrons are going to equal the number of protons. Commonly elements are written as:

\[_{\mathrm{Z}}^{\mathrm{A}}\: \: \ {\mathrm{Element}}\: \: \mathrm{Symbol}\]

Step 2: The next piece of information is the mass number, A. The mass number is the sum of the number of protons and neutrons. An equation to use to find the number of neutrons is A = Z + N, and isolate N to get N = A - Z.

Basically, Neutron Number = Mass Number - Atomic Number

Step 3: Using the information provided, you can calculate the number of protons, electrons, and neutrons in the element.

Use the steps given to solve the remaining elements.

Work

(a) \(_{14}^{34}\textrm{Si}\)

Step 1: Z = 14

Step 2: A = 34

Step 3:

Protons = 14

Electrons = 14

Neutrons = 34 - 14 = 20

(b) \(_{15}^{36}\textrm{P}\)

Z = 15

A = 36

Protons = 15

Electrons = 15

Neutrons = 36 - 15 = 21

(c) \(_{25}^{57}\textrm{Mn}\)

Z = 25

A = 57

Protons = 25

Electrons = 25

Neutrons = 57 - 25 = 32

(d) \(_{56}^{121}\textrm{Ba}\)

Z = 56

A = 121

Protons = 56

Electrons = 56

Neutrons = 121 - 56 = 65

Q21.5.7

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

Answer:

Work and Explanation

Nuclear Reactors produce electricity, and unlike conventional plants that burn coal and fuel and produce electricity through boiling water into steam, nuclear reactors use uranium fuel through the process of fission to create steam. The potential of uranium at the beginning is stored in ceramic pellets that are placed inside the nuclear reactor. Through fission, the atoms in uranium are split, which creates the heat needed to create the steam. The heat that is created through the splitting of the uranium atoms creates the steam necessary to drive the turbine. The driving of the turbine spins the generator, and that in turn creates the electricity. Through the process of fission within the nuclear power plant, the potential energy in uranium turns into heat, which then creates electricity.

In nuclear fission, the nucleus of a bigger and/or heavier atom splits into lighter atoms.

.

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?

Answer:

Work

The SHE refers to Standard Hydrogen Electrode.

Here is the Standard Reduction Potentials Chart which is used to determine the reduction potential in V and whether it is an oxidizing or reducing agent.

Cell diagram involves two species, a reduction and an oxidation, and two metals on the outsides that act as a transport for the electrons. Assuming this is in standard condition, the molarity will be 1 M for aqueous solutions and 1 atm for gases. Since we only have one metal, Ni, we need to use Platinum, Pt, which will be used to allow the electrons to transport through.

Step 1: Half Reactions:

\(\mathrm{Ni^{2+}}(aq)\: +\: \mathrm{2e^{-}}\rightarrow \mathrm{Ni}(s)\: \: \: \: \mathrm{E^{0}}=-0.25\: \mathrm{V}\)

\(\mathrm{2H^{+}}(aq)\: +\: \mathrm{2e^{-}}\rightarrow \mathrm{H_{2}}(g)\: \: \: \: \mathrm{E^{0}}=0.00\: \mathrm{V}\)

The Hydrogen Half-Reaction is a reference electrode, hence why it has 0.00 V.

Step 2: Now we assign which is the reduction and oxidation half-reaction. Since Ni is a reducing agent based on the chart, it is given the oxidation title, and we call its E⁰ the E anode. Hence H would be the E cathode.

\[\mathbf{E_{Anode}=-0.25\: V}\]

\[\mathbf{E_{Cathode}=0.0\: V}\]

Step 3: \(\mathrm{E_{Cell}= E_{Cathode}-E_{Anode}}\)

*(Reminder: Anode is where Oxidation occurs, Cathode is where Reduction occurs)

\[\mathrm{2H^{+}}(aq)\: +\: \mathrm{Ni}(s)\rightarrow \mathrm{H_{2}}(g)\: +\: \mathrm{Ni^{2+}}(aq)\]

\[\mathbf{E_{Cell}= 0.00-(-0.025)=+0.25 }\: \mathbf{V}\]

Step 4: Knowing this, we can construct our cell diagram:

\[\mathbf{Ni}\mathbf{(s)}|\mathbf{Ni^{2+}(aq)}||\mathbf{H^{+}(aq, 1\: M)}|\mathbf{H_{2}(g, 1\: atm)}|\mathbf{Pt(s)}\]

Step 1: In order to begin, we must construct our half-reactions with the two species, Ni and H. We find them on the Reduction Half-Reaction Chart. Once we have found those values, we can construct the equation with the corresponding E⁰.

Step 2: Even though they are both gaining 2e^- based off of the half-reactions, Ni is the oxidation reaction since it is below H and thus has a higher reducing agent strengh. This would mean that H has a higher oxidizing agent strength, making it the reduced half-reaction.

Step 3: Using the E cathode and E anode values, you can find the E cell by subtracting the E cathode by the E anode and obtaining a value.

Step 4: The final step is creating the cell diagram using standard conditions. The oxidized reaction go on the left side, and the reduction on the right side, and the metals go on the outside as they allow for the electrons to flow through them and cause the half reactions to occur to generate a voltage. The double lines represent the salt bridge of the cell, meanwhile the single lines separate the products from the reactants. In parenthesis one must put the state of the species and the amount if given.

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

Answer:

Work and Explanation

There are three main battery types, Primary, Secondary, and Fuel Cells. Primary batteries cannot be recharged, and common examples are the button battery and lithium-iodine battery. Secondary batteries are rechargeable, and a common example is the lead-acid (storage) battery. Fuel Cells are not necessarily batteries but they are very efficient and at times better than most combustion engines in creating electricity, so are a bit more expensive.

Primary batteries are portable voltaic cells that cannot be recharged since they come already fully charged. Secondary batteries are rechargeable though do deteriorate after awhile. Fuel cells are electrochemical cells that function through converting chemical energy in fuel into electricity through an oxidation reaction.

1. In order to power an electric motor scooter, you need the battery to be rechargeable, and that would be a secondary battery, specifically a Lead-Acid (storage) battery so you can constantly charge and reuse the battery cell.

2. A backup battery for a smartphone does not need to be rechargeable, it just needs to be able to charge the smartphone. For this case, it only needs to be a primary battery, specifically lithium-iodine batteries.

3. In order to power an iPod, it needs to have a rechargeable batter, since the iPod comes with a charger. For this case, we would use a secondary battery. Common secondary batteries are lead-acid, NiCd, NiMH, and Li-ion batteries.