Extra Credit 2

Q17.1.2

For the scenario "If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?", how many electrons moved through the circuit?

S17.1.2

So here, we are asked to calculate the amount of electrons that have flowed through the circuit. However, there is not enough information currently to determine the answer. We must first determine the total charge running through the circuit before being able to determine how many electrons are moving through the circuit.

To determine the total coulombs of charge moved through the circuit, we must first understand that one coulomb is just an ampere-second, which basically means the amount of electricity, measured in amperes, transferred per second. The conversion can be seen here.

\[ 2.5\text{ A}= \frac{2.5\text{ C}}{1\text{ s}} \]

From that, we can calculate the total amount of coulombs moved through the circuit. Here, we convert seconds to the appropriate amount of minutes because remember, 1 ampere is 1 Coulomb over 1 second.

\[ 2.5\text{ A} \times \frac{60\text{ seconds}}{1\text{ min}} \times {35 \text{ minutes}} = 5250\text{ C}\]

Now that we know how many total coulombs are running through the circuit, we can determine the amount of electrons running through the circuit by using the charge per electron (1.6 x 10^-19 C).

\[ \frac {5250\text{ C}}{1.6 \times 10^{-19} \text {C} \text { per } \text {electron}}\]

\[ = 3.28 \times 10^{22} \text {electrons in the circuit}\]

Revisions are in blue

If written in mathjax/latex:

\(2.5 A = {2.5} \frac {\text {Coulombs}} {\text {s}}\)

\(1 A= \frac {\text{Coulombs}} {\text {s}}\)

\(2.5 \frac {\text {C}} {\text{s}} \times \frac { 60 \text { seconds}} {1 \text { minute}} \times {35 \text { minutes}} = 5250 { \text { Coulombs}}\)

\(5250 \text { Coulombs} \times \frac { 1\text { electron}} {1.6 \times 10^{-19} \text { Coulombs}}\)

\(= {3.28 \times 10^{22} \text { electrons}}\)

Q17.7.5

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO₃)₂ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm³. Assume the efficiency is 100%.

S17.7.5

Here, we first need to determine how much zinc there is in order to use the density of zinc to figure out the volume of zinc and then the surface area. To find the amount of zinc, we must first understand that this question refers to the process of electroplating. Electroplating is a process in which an electrical current is run through a solution of dissolved ions, in turn reducing them, and then forming a thin layer of itself onto another metal. This is very important to know because without knowledge of this being a reduction reaction, it would be near impossible to find the solution. However, because we do know that this process is rooted in reduction, we can use the reduction equation of zinc to solve this problem.

\[ \text {Zn}^\text{ 2+} + 2\text {e}^- \rightarrow \text {Zn}\]

From the reduction equation, we have to first solve for the amount of electrons. We first determine the total amount of energy like so (A or ampere-seconds are equivalent to one C)

\[ 2.599\text{ A} \times \frac{60\text{ seconds}}{1\text{ min}} \times \frac{60\text{ minutes}}{1\text{ hour}} = 9356.4\text{ C}\]

And then we can use the amount of charge carried by each electron to determine the how many electrons are running through the system.

\[ \frac {9356.4\text{ C}}{1.6 \times 10^{-19} \text {C} \text { per } \text {electron}} = 5.85 \times 10^{22} \text {electrons }\]

From that, we convert the electrons into moles using Avogadro's constant so we can use the 2:1 stoichiometric ratio between the electrons and zinc to determine how many moles of zinc have been plated.

\[ \frac {5.85 \times 10^{22} \text {electrons }}{6.022 \times 10^{23}} = .097\text { moles} \text { of } \text {electrons}\]

\[ \frac {.097 \text { moles} \text { of } \text {electrons}}{2} = .049\text { moles} \text { of } \text { Zn}\]

Now we can find the grams of zinc there are in order to see how much volume it took up.

\[ .049\text{ mol Zn} \times \frac{65.39 \text { grams}}{ 1 \text { mol Zn}} \times \frac{ 1\text { cm}^3}{7.14 \text { grams}} = .44 \text { cm}^3\]

Finally, we can use the amount of volume and use how thick the zinc was plated on to determine the total surface area of the zinc. (Hint: Convert .01123mm to .001123cm for easier calculations.)

\[ 0.01123\text{ mm} \times \frac{1 \text { cm}}{ 10 \text { mm}} = 0.001123 \text { cm}\]

\[ \frac {0.44 \text { cm}^3}{0.001123 \text { cm }} = 392 \text { cm}^2\]

Written in mathjax/latex:

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

Recall that \(1 A = {1 C \over 1 s}\)

\( 1 hour \times {60 min \over 1 hour} \times {60 seconds \over 1 min} \times {2.599 {C \over s}} = 9356.4 C\)

To find the number of electrons involved:

\({9356.4 C \over {1.6 \times 10^{-19} {C \over electron}}} = {5.85 \times 10^{22} electrons}\)

Based on the mole ratio seen from the zinc equation, we proceed by finding the moles of zinc deposited.

\({5.85 \times 10^{22} electrons} \times {1 mol \over {6.022 \times 10^{23} electrons}} \times {1 mol Zn \over 2 mol e^-} = 0.049 mol Zn\)

Now that we know the moles of zinc involved, we need to figure out the volume of the metal to determine the surface area by using the density given.

\(0.049 mol Zn \times {65.38 grams Zn \over 1 mol Zn} \times {1 cm^3 \over 7.140 grams Zn} = 0.44 cm^3\)

We also know that the layer of zinc is 0.01123 mm, which we need to convert to cm to determine the surface area of the part.

\(0.01123 mm \times {1 cm \over 10 mm} = 0.001123 cm\)

\({0.44 cm^3 \over 0.001123 cm} = 392 cm^2\)

The total surface area of the part is \(392 cm^2\).

Q19.2.2

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diaminedichloroplatinum(II)
5. potassium diaminetetrachlorochromate(III)
6. hexaaminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate

S19.2.2

Here we are told to give the coordination numbers and write the formulas for these coordination complexes. The coordination number simply refers to the number of points of attachment by ligands, or binding molecules, to the central metal. Most ligands only attach to the central metal once but some can attach at multiple points as well. The formulas for each coordination complex are written with the metal first, then the anionic in alphabetic order, then the neutral ligands in alphabetic order, and finally denoted with the proper charge, if any. The metals and ligands are written in brackets while the charge is written outside. Sometimes, a coordination complex with have an charge balancing ion like potassium or chloride written in before or after the brackets.

The naming of these coordination complexes has multiple parts. The ligands are given a shortened name such as CN^- being -cyano and the transition metals just being their original names unless the the overall charge of the complex is negative, in which the endings of their names are cut off (from chromium to chrom-) and to the endings "-ate" is added (ex. chrom- + -ate = chromate). The number of ligands are denoted using greek (mono, di, tri, tetra, etc.) with polydentate ligands such as ethylenediamine and oxalate being numbered with bis, tris, tetrakis, etc. to indicate multiple polydentate ligands.

1. [Zn(OH)₄]²⁻; the coordination number is 4 because there are 4 hydroxo ligands attached to the central zinc ion.

2. [Pd(CN)₆]^2-; the coordination number is 6 because there are 6 cyanide ligands attached to the central palladium ion.

3. [AuCl₂]^-; the coordination number is 2 because there are 2 chloro ligands attached to the central gold ion.

4. [Pt(NH₃)₂Cl₂]; the coordination number is 4 because there are 2 ammine ligands and 2 chloro ligands attached to the central platinum ion.

5. K[Cr(NH₃)₂Cl₄]; the coordination number is 6 because there are 2 ammine ligands and 4 chloro ligands attached to the central chromium ion.

6. [Co(NH₃)₆][Cr(CN)₆]; the coordination number of each of these complexes is six because there are 6 ammine ligands attached to the central cobalt ion and 6 cyano ligands attached to the central chromium ion respectively.

7. [Co(en)₂Br₂]NO₃; the coordination number is 6 because there are 2 bromo ligands and 2 ethylenediamine ligands (which attaches twice so contributed two to the coordination number) attached to the central cobalt ion.

Revision (highlighted in yellow, correction below):

[Pd(CN)₆]^2-

To add: I think it would be good to add that "bis" is only used to indicate bidentate ligands such as ethylenediamine.

Written in mathjax/latex:

1. \([Zn(OH)_4]^{2-}\)

2. \([Pd(CN)_6]^{2-}\)

3. \([AuCl_2]^{-}\)

4. \([Pt(NH_3)_2Cl_2]\)

5. \(K[Cr(NH_3)_2Cl_4]\)

6. \([Co(NH_3)_6][Cr(CN)_6]\)

7. \([Co(en)_2Br_2]NO_3\)

Q12.3.14

From the following data, determine the rate equation, the rate constant, and the order with respect to A for the reaction \[ \text{ A} \rightarrow 2\text{C}\] .

S12.3.14

We are told to find the rate equation, rate constant, and the order with respect to A for the reaction. The rate equation is quite simple to find as it is just simply rate = k[reactant]ⁿ[reactant]ⁿ[reactant]ⁿ... etc. until all of the reactants have been represented and with n being the coefficients of the reactants. In this case, the rate equation is rate = k[A] because there is the one "A" reactant.

Now, with the rate equation, k is calculated by just plugging the numbers from the chart directly into the rate equation.

3.8 x 10^-7 = k[1.33 x 10^-2]² ; k = 2.15 x 10^-3

There are many types of orders; the most common few being zero order, where the concentration of the reactant does not affect the rate; first order, where the concentration of the reactant affects the rate linearly; second order, where the concentration of the reactant affects the rate to the power of two; and third order, where the concentration of the reactant affects the rate to the power of three.To determine the order of the reaction, we must determine the relationship between [A] and rate, as k is a constant and will not change. Order here can be determined looking at patterns between the different values. When [A] doubles from 1.33 x 10^-2 to 2.66 x 10^-2, the rate quadruples. That means that the doubling of the concentration of the reactant changes the rate by 2 to the power of 2, which means the order of the reaction is second order. It can be seen when [A] goes from 1.33 x 10^-2 to 3.99 x 10^-2. The concentration of the reactant triples but the rate multiplies by 9, or 3 to the power of two. Therefore, this reaction is second order.

Revision (highlighted in yellow, correction below):

k can only be found after finding the order of the reaction, which is second order. Therefore, we can say that the rate law is: rate= k[A]². Substituting any set of values, we can then find k. For example,

3.8 x 10^-7 M/h= k[1.33 x 10^-2 M]² ; k = 2.15 x 10^-3 1/(M x h)

Written in mathjax/latex:

\(rate= k[A]^m\)

Based on the data given, we can write:

\({3.80 \times 10^{-7}} {M \over h} = k [1.33 \times 10^{-2} M]^m\)

\({1.52 \times 10^{-6}} {M \over h} = k [2.66 \times 10^{-2} M]^m\)

\({3.42 \times 10^{-6}} {M \over h} = k [3.99 \times 10^{-2} M]^m\)

Using any of these equations, we can use ratios to determine k.

Ex: \({{1.52 \times 10^{-6}} {M \over h} = k [2.66 \times 10^{-2} M]^m} \text { or } {{3.80 \times 10^{-7}} {M \over h} = k [1.33 \times 10^{-2} M]^m}\)

m=2

So, the rate law is \(rate= k[A]^2\)

Using any of the three equations above, we can now find k.

Ex: \({3.80 \times 10^{-7}} {M \over h} = k [1.33 \times 10^{-2} M]^2\)

\(k = {2.15 \times 10^{-3}} {1\over {M \times h}}\)

Q12.6.6

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?

S12.6.6

For an elementary reaction, or a one step reaction, the rate law for the reaction will be the same as the balanced equation. Only those with the rate law matching the balanced equation would possibly have the overall and elementary reaction be the same. Therefore, only b), d), and e) could be the possible answers.

(correct)

Q21.4.18

The isotope \(\ce{^{90}_{38}Sr}\) is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.

S21.4.18

To find half life, we must first know what half life means. Half life is the amount of time necessary to for an element, or radioactive isotope, to reduce by half from decay. The equation for half life is

A = A_o x (0.5)^t/h where A is the current amount, A_o as the original amount, t as in time, and h as in half life.

If all the numbers from the problem are plugged into the equation, the result is

\[.393 = .5 \times {.5}^{10/h}\]

Through some algebra, the equation turns into

\[ \text {h} = \frac {10 ln 0.5}{ln \frac{.393}{.5}}\]

h or half life is calculated to be 28.8 years, which means it takes \(\ce{^{90}_{38}Sr}\) 28.8 years to decay by half its amount.

(correct)

Could also use alternate form:

\(ln[A] = ln[A]_0 -kt\). This can be used because radioactive decay is a first-order process.

Written in mathjax/latex form:

\(ln[0.393 g] = ln[0.500 g] -k(10.0 years)\)

k = \(0.0241 \over year\)

To find the half-life, we use the equation for first-order half-life:

\(t_{1 \over 2} = {ln (2) \over k}\)

\(t_{1 \over 2} = {ln (2) \over 0.0241 {1 \over year}}\)

\(t_{1 \over 2} = 28.8 years\)

Q20.3.6

It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?

S20.3.6

Using one beaker with two electrodes and an electrolyte connecting the two is better at measuring voltage. This is because of liquid junction potential, or where two differently concentrated solutions exchange ions and the greater concentrated solution will diffuse faster into the less concentrated, thus causing a difference in charges at the "junction" or meeting place of the solutions. This normal process is disrupted when using two beakers, as there will be resistance between the high amount of electrons flowing through the salt bridge and can cause inaccuracies in determining the voltage.

(correct)

Q20.5.17

What is the standard change in free energy for the reaction between Ca²⁺ and Na(s) to give Ca(s) and Na⁺? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?

S20.5.17

Here, we need to determine the reduction equations, both cathode (reduction) and anode (oxidation), in order to calculate the proper cell potential. From that, then we can convert cell potential into free energy.

Cathode, or reduction, is the gaining of electrons is Ca²⁺ because it is gaining two electrons to become Ca.

\[ \text{Cathode:}\;\text{Ca}^2+(\mathit{aq}) + \text{2e}^- \rightarrow \text{Ca}(\mathit{l})\;\;\;\;\text{E}^°= -2.76 V \]

Anode, or oxidation, is the loss of electrons as Na is losing one electron to become Na⁺.

\[ \text{Anode:}\;\text{2Na}^{+} + \text{2e}^{-} \rightarrow \text{2Na}(\mathit{s})\;\;\;\;\text{E}^°= -2.71 V \]

Then using the equation \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

\[\text{E}^°(\mathit{Cell}) = (-2.76V) - (-2.71 V) \]

\[ = (-2.76V) + (2.71 V) \]

\[ = -0.05V \]

After solving for E^o(cell), we use the equation \[ \Delta\text{G}^° = -\text{nFE}^°_{cell} \]

n, or the moles of electrons being exchanged was determined earlier to be 2, and Faraday's constant, or F, is 96,485 J/V⋅mol.

Therefore,

\[ \Delta\text{G}^° = -(2)(96,486 J/V⋅mol)(-0.05V) \]

\[ = 9648.5 J \]

This result is expected from two elements so close to each other on the periodic table that this reaction would be nonspontaneous. It is also expected that the magnitude of E^o(cell) would not be large because their respective cell potentials were not far apart.

Revision (written in mathjax/latex):

The standard reduction potentials of calcium and sodium differ depending on the source, but based on the Libretext chart found here, the E° for Ca is -2.868 V and E° for Na is -2.71 V. To determine ΔG°, we need to first calculate \(E°_{cell}\).

\(Ca^{2+} (aq) + 2e^- → Ca(s)\) E° = -2.868 V

\(Na^{+} (aq) + e^- → Na(s)\) E° = -2.71 V

Anode: \(Na (s) → Na^{+} + e^-\) E° = -2.71 V (note that the sign of E° does not change because it is an intensive property)

Cathode: \(Ca^{2+} (aq) + 2e^- → Ca(s)\) E° = -2.868 V

\(E°_{cell} = E°_{cathode} - E°_{anode}\)

\(E°_{cell} = -2.868 V - (-2.71 V)\) = - 0.158 V

\(ΔG°= - nFE°\)

\(ΔG°= - {2 mol e^-} \times {96485 C \over mol e^-} {- 0.158 V}\)

*Note that \(1 V= 1 {J \over C}\)

\(ΔG°= 30489.26 J = 30.489 kJ\)

The sign of ΔG° is positive, which indicates that it is nonspontaneous. This makes sense because their standard reduction potentials are close in value, since they are close together on the periodic table (sodium is an alkali metal, calcium is an alkaline earth metal), so the cell voltage \(E°_{cell}\) is very small in value as well.