Extra Credit 20

Q17.2.9

• An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

Answer17.2.9

The active electrode are electrode in which mass are found to be changed in the chemical reaction process in electrolytic solution. Metal form cation in oxidation, so metal electrode will lose weight metal ions under go oxidation reaction. However, if metal ion undergo reduction reaction, then metal electrode will gain weight.

When electricity pass through a solution of an electrolyte, chemical reaction occurs at the electrode. The electrode connect to negative terminal of battery is the cathode, and the electorode connect to the positive terminal is called anode.

Since the electrode was found to gain mass then it must be the reduction part of the reaction also known as the cathode. The reason that reduction/the cathode must have the gain in mass is that when you reduce anything you are giving the electrons to the cathode and since you have a metal which tend to be positively charged when you add electrons its oxidation state will change and it will bind to the metal and solidify increasing the mass of the cathode.

When there is loss of mass the opposite occurs (oxidation/anode), since oxidation and reduction come hand and hand you need to remember that the electrons you are giving the cathode need to come from some where (because mass is neither created nor destroyed). When oxidation occurs you are loosing electrons and you are reducing the cathode. This loss of electrons has the same concept that the electron gain has on the the cathode, EXCEPT since in the anode you are loosing electrons you are becoming positively charged and this will create aqueous metal ions that will detach to from the anode and will cause it to decrease in mass.

Q19.1.18

Describe the electrolytic process for refining copper.

Answer19.1.18

Electrolytic refining is a process of obtaining pure metal by the process of electrolysis (first it is treated to remove any remaining sulfur and then it is cast into anodes so that it can be refined through electrolysis). To refine cooper using the electrolytic process, you take an impure cooper rod as an anode and a thin sheet of pure cooper as the cathode and place it in an electrolytic cell containing an acidifying solution of cooper sulfate as electrolyte. In this method (Electrolytic refining), an electric current is passed through a solution that copper(II) sulfate solution. When passing a current through this copper(II) sulfate solution we can observe the cooper in the impure cooper rod getting dissolved as Cu²⁺. On the cathode we see a deposit of copper ions (Cu²⁺) on to the negatively charged cathode, the pure copper. These copper ions were deposited on to the negative pure copper cathode because the copper ions were attracted to the electrons as they will become neutral when they deposit onto the cathode. However these high purity copper ions were stripped from the anode (the impure copper) and migrate to the cathode they were not provided by the solution.This reaction causes the anode to decrease in size, gradually disappearing and the cathode to increase in size because the of the transfer of copper from the anode to the cathode. Meanwhile, the sulfate ions from the impure cooper (anode) and other impurities goes into the electrolyte.

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Q19.3.10

Would you expect the complex [Co(en)₃]Cl₃ to have any unpaired electrons? Any isomers?

Answer19.3.10

The complex does not have any unpaired electrons. In the ligand, the central metal ion is bonded with three [en], which are bidentates. Since there is no charge for [en], Co shows -3 charge and the negative charge is balanced when it pairs up with three Cl-.

The complex does not have any geometric isomers, but the mirror image is non-superimposable, so it has two optical isomers.

The first thing to notice about [Co(en)3]Cl₃ is that it has a coordination number of 6 so it in an octahedral geometry. The second thing to notice is that it has a strong field ligand (en), this will affect the spin. Since the ligand is a strong ligand we know that it will have a low spin. Now we need to know how many d-shell electrons it would have by looking at the oxidation number. Based off of that we see that Co has a 2+ charge. meaning it will have 6 electrons. Based on the way that electrons are arranged in a he octahedral crystal field diagram 6 electrons would fill the bottom 3 rows leaving it as diamagnetic (no unpaired electrons). [Co(en)3]Cl₃ has no geometric isomers but it it does have optical isomers/enantiomers because the mirror image is not superimposable on its self.

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Q12.4.10

The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t_1/2?

Answer12.4.10

As mentioned in the question the reaction of compound A will result in the formation of compounds C and D. This reaction was found to be second-order in A. Therefore, we should use the second order equation for half-life which relates the rate constant and initial concentrations to the half-life:

\[t_{\frac{1}{2}}=\frac{1}{k[A]_{0}}\]

Since we were given k (rate constant) and Initial concentration of A, we have everything needed to calculate the half life of A.

\[k=0.5\frac{\frac{L}{mol}}{s}\]

\[[A]_{0}=2.42\frac{mol}{L}\]

When we plug in the given information notice that the units cancel out to seconds.

\[t_{\frac{1}{2}}=\frac{1}{\frac{2.42Lmol^{-}}{s}[0.500\frac{mol}{L}]}=0.826 s\]

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Q21.2.5

Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:

1. 25 protons, 20 neutrons, 24 electrons
2. 45 protons, 24 neutrons, 43 electrons
3. 53 protons, 89 neutrons, 54 electrons
4. 97 protons, 146 neutrons, 97 electrons

S21.2.5

Since the format for nuclide notation is:

\[_{protons}^{mass}\textrm{Element symbol}^{charge}\]

\[mass = protons+neutrons\]

\[charge= protons-electrons\]

\[neutrons =mass-protons\]

1. 25 protons, 20 neutrons, 24 electrons\[_{25}^{45}\textrm{Mn}^{1+}\]

2. 45 protons, 24 neutrons, 43 electrons \[_{45}^{69}\textrm{Rh}^{2+}\]

3.53 protons, 89 neutrons, 54 electrons\[_{53}^{142}\textrm{I}^{-}\]

4. 97 protons, 146 neutrons, 97 electrons\[_{97}^{243}\textrm{Bk}\]

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Q21.5.8

The mass of a hydrogen atom is 1.007825 amu; that of a tritium atom (31H)(13H) is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of He24He24 produced is released by the following fusion reaction:\[_{1}^{1}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{4}^{2}\textrm{He}\]

Answer 21.5.8

Since we were asked for the energy in kilojoules released by the reaction we need to use the equation:\[\Delta E=\Delta mc^{2}\]

We need to find ▵m

Step1: Total mass before the reaction\[_{1}^{1}\textrm{H}+_{1}^{3}\textrm{H}\]

\[1.007825+3.01605=4.023875 amu\]

Step2: total mass after the reaction\[_{4}^{2}\textrm{He}\]

\[4.00150amu(\alpha particle)\]

Step 3: Mass defect (▵m)

\[\Delta m = mass(reactants) - mass(products)\]

\[\Delta m =4.023875amu-4.0015amu=0.022375 amu\]

Since\[1amu=1.6605 *10^{^{-27}}kg.\]

The mass defect equals to:\[0.022375amu*1.6605*10^{^{-27}}\frac{kg}{amu}=3.715*10^{^{-29}} kg\]

Energy released:\[\Delta E=\Delta mc^{2}\]

Therefore,\[\Delta E=3.715*10^{-29}× (3*10^{8})^2= 3.3438*10^{-12}J\]

since \[1000J =1 kJ\]

Binding Energy \[\Delta E=3.3438*10^{-15} \frac{kJ}{atom}\]

Since there are\[Na=6.022*10^{23}\frac{atoms}{mol He}\]

The binding energy for one mole of He will be \[\Delta E= 3.3438*10^{-15}\frac{kJ}{atom} *6.022*10^{23} \frac{atom}{mole} =2.0129*10^{9} \frac{kJ}{mole}\]

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Q20.4.6

Explain why E° values are independent of the stoichiometric coefficients in the corresponding half-reaction.

Answer 20.4.6

E° is defined as the potential of a cell measured under standard conditions. Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties (meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system) and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.

Q20.7.4

Why are galvanic cells used as batteries and fuel cells? What is the difference between a battery and a fuel cell? What is the advantage to using highly concentrated or solid reactants in a battery?

Answer 20.7.4

Since galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells.

A battery is a galvanic cell that contains all the reactants needed to produce electricity. In contrast, a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. The major difference is that a fuel cell does not store chemical or electrical energy. However it is able to extract energy directly from a chemical reaction a battery is a contained unit that produces electricity. Another major difference between batteries and the fuel cells is that commercial batteries typically use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. This is the advantage of using solid reactants in a battery that it maximizes the electrical output per unit mass.

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