Extra Credit 27
- Page ID
- 82734
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q 17.3.6
Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br2(l) is the same as for Br2(aq).
A 17.3.6
Pt(s)│H2(g)│H+(aq)║Br2(aq)│Br-(aq)│Pt(s)
Using the cell diagram, we know that the first equation is
Br2(aq) + 2e- \(\rightleftharpoons\) 2Br-(aq) with E˚cell: 1.06 V
From this equation, we can tell that \(Br_2\) is being reduced- meaning that it is gaining electrons (electrons on the reactant side).
Since the species is being reduced, we can confirm that this half reaction is the cathode.
The second equation is
H2(g) \(\rightleftharpoons\) 2H+(aq)+ 2e- with E˚cell: 0V
Similarly, we can see from the half reaction that this species is being oxidized- meaning that the species is losing electrons (electrons on the product side).
From this, we can confirm that this half reaction is the anode.
NOTE: To get the E˚cell for each of the half reactions shown above, refer to the standard reduction potential table.
Knowing that the cathode is where reduction occurs and that the anode is where oxidation occurs we can set up
E˚cell = Ecathode - Eanode
1.06V = 1.06V - 0V
E˚cell = 1.06 V
The reaction is spontaneous at standard conditions with a cell potential of 1.06V. We can say that it is spontaneous since the E˚cell value is a positive value.
Q 19.1.25
Give the oxidation state of the metal for each of the following oxides of the first transition series.
A 19.1.25
For all of these oxides we know that oxygen has an oxidation state of -2 and that the total charge is equal to 0.
a. Sc2O3 Sc: +3 O: -2; 0 = 2(x) + 3(-2), x = \(Sc^{+2}\) Wrong x= 3, \(Sc^{+3}\)
b. TiO2 Ti: +4 O: -2 0 = (x) + 2(-2), x = 4, \(Ti^{+4}\)
c. V2O5 V: +5 O: -2 0 = 2(x) + 5(-2), x = 5, \(V^{+5}\)
d. CrO3 Cr: +6 O: -2 0 = (x) + 3(-2), x = 6, \(Cr^{+6}\)
e. MnO2 Mn: +4 O: -2 0 = (x) + 2(-2), x =4, \(Mn^{+4}\)
f. Fe3O4, These are known as mixed valence compounds, and since Fe has more than one oxidation state this compound can be rewritten as FeO and Fe2O3.
For FeO: Fe: +2 O: -2 0 = (x) + (-2), x = 2, \(Fe^{+2}\)
For Fe2O3: Fe: +3 O: -2 0 = 2(x) + 3(-2), x = 3, \(Fe^{+3}\)
g. Co3O4 These are also known as mixed valence compounds, and since Co has more than one oxidation state this compound can also be rewritten as CoO and Co2O3.
For CoO: Co: +2 O: -2 0 = (x) + (-2), x = 2, \(Co^{+2}\)
For Co2O3: Co: +3 O: -2 0 = 2(x) + 3(-2), x = 3, \(Co^{+3}\)
h. NiO Ni: +2 O: -2 0 = (x) + -2, x = 2, \(Ni^{+2}\)
i. Cu2O Cu: +1 O: -2 0 = 2(x) + -2, x = 1, \(Cu^{+1}\)
Q 12.4.18
Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.
A 12.4.18
In order to find out what year King Richard III died, set [A]/[A0] (the percent of carbon-14 still contained) equal to 0.5time(t)/half life (t1/2) or use the equation N(t) = N0e-rt.
Using the first equation:
\(A/A_{0}\) = \(0.5^{t/t_{1/2}}\) plug in the given numbers \(.9379 = 0.5^{t/5730}\) and solve for t.
\(ln.9379\) = \((t/5730)(ln0.5)\) (using the rule of logs)
\(-.0641\) = \((t/5730)(-.693)\)
\(-367.36\) = \(-.693t\)
\(t = 530.1 years\)
Using \(N(t) = N_{0}e^{-rt}\) this problem is solved by the following:
\(1/2 = e^{-5730r}\)
\(r = 0.000121\)
Now that we know what r is, we can use this value in our original formula and solve for t, the amount of years that have passed.
This time, we use 93.78, the percent of the carbon-14 remaining as N(t) and 100 as the original, N0.
\(93.78 = 100e^{-0.000121t}\)
\(t = 530.7\) years
Another way of doing this is by using these two equations:
λ = \(\dfrac{0.693}{t_{1/2}}\) and \(\dfrac{n_{t}}{n_{0}}\) = -λt
\(n_{t}\) = concentration at time t (93.79)
\(n_{0}\) = initial concentration (100)
First solve for lambda or the decay constant by plugging in the half life.
Then plug in lambda and the other numbers into the second equation, and solve for t- which should equal to 530.1 years as well.
If we want to find out what year King Richard III died, we take the current year, 2017, and subtract 530 years. Doing this, we find that King Richard III died in the year 1487.
Q 21.3.2
Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?
A 21.3.2
Nuclear reactions in general are reactions in which the nucleus of one element is changed into the nucleus of another elements by the emission of energetic particles- this changes the characteristics of the atomic nucleus itself. Nuclei are made up of protons and neutrons. Alpha particles (α2+) are helium nuclei consisting of two protons and two neutrons, symbolized by \(^{4}_{2}He\). Also, the proton particle is a nuclei of hydrogen atoms, symbolized by \(^{1}_{1}H\) or \(^{1}_{1}p\). Additional information and reading can be found here: Nuclear Radiation
Q 21.7.4
A scientist is studying a 2.234 g sample of thorium-229 (t1/2 = 7340 y) in a laboratory.
- What is its activity in Bq?
- What is its activity in Ci?
A 21.7.4
In order to find the activity we know the formula is
\(activity\) = \(0.693/{t_{1/2}}\) x mass
Now we plug our values in
\(activity\) = \((0.693/7340)(2.234)\)
We get that \(activity\) = \((9.44x10^{-5})\) grams/year
We want our answer to be in Bq which is decay per second. So we need to convert our answer using the following:
(9.44x10-5g/year) x (1year/365 days) x (1 day/24 hours) x (1 hour/60 min) x (1 min/60 sec) x (1 mol/229 g) x (6.022x1023 molecules/1 mol)* 1 decay/1 molecule
= \(7.87x10^{9} Bq\) Bq= decay/s
to get this in curies we use the conversion factor \(1 curie\)= \(3.7X10^{10} Bq\)
= \(7.87X10^{9} Bq\) x \(1 Ci/3.7x10^{10} Bq\) \( = 0.2128 Ci\)
Q 20.4.17
The standard cell potential for the oxidation of Pb to Pb2+ with the concomitant reduction of Cu+ to Cu is 0.39 V. You know that E° for the Pb2+/Pb couple is −0.13 V. What is E° for the Cu+/Cu couple?
A 20.4.17
We know that Pb2+/Pb is -0.13V and that the overall reaction (Ecell= Cathode-Anode) is equal to 0.39V.
We also know that Pb2+/Pb is the anode (Pb is being oxidized) and Cu+/Cu is the cathode (Cu is being reduced).
So knowing that E˚cell is cathode - anode we can set up an equation and solve for X = cathode.
\(0.39V\) = \(X - -0.13V\)
\(X\) = \(cathode\) = \(0.26V\)
The E° for the Cu+/Cu half reaction is 0.26V
Q 20.9.2
How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?
A 20.9.2
One could use an electrolytic cell to compare the strengths of various oxidants and reductants by computing the cell potential. Cell potential is defined as Ecell = Ecathode - Eanode, and the individual strengths of the cathode and the anode can be determined using the standard electrode potentials. The more positive the number, the stronger of an oxidizing agent it is, and the more negative the number, the stronger of a reducing agent it is. It can also be determined that the more positive the total Ecell, the more spontaneous the cell is, however this does not tell us much more than that. In addition, we can also use the stoichiometric ratios for a given electrochemical reaction, the length of time, and the current passed to calculate the quantity of material that is oxidized or reduced in an electrolytic cell. This information can also help with the electroplating processes, which is plating one metal onto another through hydrolysis. For more information, check Comparing Strengths of Oxidants and Reactants.
Q 14.1.2
If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?
A 14.1.2
Studying the chemical kinetics of the reaction that would be taking place would be very beneficial since in doing so, some important information could be determined. One of the most beneficial things that could be determined is the time it would take for the reaction to occur. In an industrial facility, many tasks are time dependent and knowing how long a reaction would need to take place could save money and resources- wasteful usage of resources can be avoided since through these calculates we can determine the concentration and amounts needed.