Extra Credit 29
- Page ID
- 83538
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q19.10B
\(Pb (s) | Pb^{2+} (aq) (0.07M)||Fe^{3+} (aq) (0.64M), Fe^{2+} (aq) (0.85M)|Pt (s)\)
a. Will the cell reaction be spontaneous?
b. What ratio of \(\dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\) will reaction will not be spontaneous in either direction?
S19.10B
a. In order to determine the spontaneity of the cell, you have to use the equation \(E_{cell}^\circ = E_{reduction}^\circ - E_{oxidation}^\circ\) in order to find the difference in the standard reduction potentials and the Nernst equation in order to find the cell potential of this particular reaction.
First, you'd determine in which cell the oxidation occurs and in which cell the reduction occurs. A species is oxidized when it loses electrons and a species is getting reduced when it gains electrons.
Pb (s) → \(Pb^{2+} (aq) + 2e^{-}\)
\(Fe^{3+}\) (aq) + \(e^{-}\) → \(Fe^{2+}\) (aq)
In this case, Pb is the species being oxidized to \(Pb^{2+}\) and \(Fe^{3+}\) is the species being reduced to \(Fe^{2+}\). Now that we've figured out which reaction is the reduction and which is the oxidation, we can look up the standard cell potentials for each reaction.
\(E_{oxidation}^\circ\) = +0.771 V
and
\(E_{reduction}^\circ\) = -0.125 V
so
\(E_{cell}^\circ\) = +0.771 V - (-0.125 V) = 0.896 V
Then, you use the Nernst equation, \(E_{cell} = E_{cell}^\circ - \dfrac{0.0592}{2} \times log K\) and plug everything in. In order to find K, you must use the balanced redox reaction.
Pb (s) → \(Pb^{2+} (aq) + 2e^{-}\)
2 (\(Fe^{3+}\) (aq) + \(e^{-}\) → \(Fe^{2+}\) (aq))
Pb (s) + \(2Fe^{3+}\) (aq) → \(Pb^{2+} (aq) + 2Fe^{2+}\) (aq)
Once you have the balanced redox reaction, you plug in the concentrations of each species raised to their respective powers. All the products go in the numerator and are multiplied by each other and all the reactants go in the denominator and are multiplied by each other. In this case, you only have one reactant to plug in because Pb is a solid in the reactant.
\(E_{cell} = E_{cell}^\circ - \dfrac{0.0592}{2} \times log \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
\(E_{cell} = 0.896 V - \dfrac{0.0592}{2} \times log \dfrac{[.07][0.85]^2}{[0.64]^2}\)
\(E_{cell}\) = 0.869 V
Because this value is positive, the cell reaction is spontaneous.
b. In order to make the reaction not spontaneous, \(E_{cell}\) must be negative. In order to make it negative, we'd have to alter the equation so that we end up with a negative value overall and we must do so by altering the value of K.
We start with the inequality:
0 > \(E_{cell}^\circ - \dfrac{0.0592}{2} \times log \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Then we add \(E_{cell}\) to both sides so that the inequality now looks like:
\(-E_{cell}\) > \(-\dfrac{0.0592}{2} \times log \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Divide both sides by \(-\dfrac{0.0592}{2}\), which is the equivalent of multiplying by \(-\dfrac{2}{0.0592}\). When you divide or multiply by a negative, the inequality is flipped.
\(-E_{cell} \times -\dfrac{2}{0.0592}\) < log \(\dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Then raise everything to the power of 10 on both sides to cancel out the log function.
\(10^{(-\dfrac{2}{0.0592})(-E_{cell})} < \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Now plug in all your values.
\(10^{(-\dfrac{2}{0.0592})(-0.869)} < \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Then simplify to get:
\(10^{29} < \dfrac{[Pb^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2}\)
Q19.75B
Using the reactions below, calculate \(E_{cell}\) for \(X^{3+} (aq) +e^- →X^{2+} (aq)\).
\(X^{2+} (aq) + XO^{2+} (aq) + 2H^+ (aq) → 2X^{3+} (aq) +H_{2}O (l)\) \(E_{cell}^\circ\) = 0.789 V
\(X^{3+} (aq) + Fe^{3+} (aq) + H_{2}O (l) → XO^{2+} (aq) + 2H^+ (aq) + Fe^{2+} (aq)\) \(E_{cell}^\circ\) = 0.243 V
S19.75B
Using the second equation, we can split them into their respective reduction and oxidation reactions.
Reduction: \(Fe^{3+} (aq) +e^- → Fe^{2+} (aq)\) \(E_{reduction}^\circ\) = 0.771 V
Oxidation: \(X^{3+} (aq) + e^- + H_{2}O (l)→ XO^{2+} (aq) +2H^+ (aq)\) \(E_{oxidation}^\circ\) = ?
When the two equations are combined, the overall \(E_{cell}\) = 0.243 V. We know that \(E_{cell}\) = E_{reduction}^\circ - E_{oxidation}^\circ\) so we can simply plug in the values we do know and solve for \(E_{oxidation}^\circ\).
\(0.243 V = 0.771 V - E_{oxidation}^\circ\)
\(0.528 V = E_{oxidation}^\circ\)
Then we use the first equation and split them into their respective oxidation and reduction reactions.
Reduction: \(XO^{2+} (aq) + 2H^{+} (aq)→ X^{3+} (aq) + H_{2}O (l) +e^-\) \(E_{reduction}^\circ\) = 0.528 V
\(E_{reduction}^\circ\) is the oxidation half reaction from above but flipped so we know that the standard reduction potential of this reduction half reaction is the same as the standard reduction potential of the oxidation equation above.
Oxidation: \(X^{2+} (aq) → X^{3+} (aq) + e^- (aq)\) \(E_{oxidation}^\circ\) = ?
After splitting them, use the overall cell potential to calculate \(E_{oxidation}^\circ\).
\(0.789 V = 0.528 V - E_{oxidation}^\circ\)
\(-0.261 V = E_{oxidation}^\circ\)
The standard reduction potential for \(X^{2+} (aq) → X^{3+} (aq) + e^- (aq)\) is the same as \(X^{3+} (aq) +e^- (aq) → X^{2+} (aq)\) since they are the same equation but flipped.
Q21.5F
Of the following complexes, name whether each pair is identical, geometric, or enantiomers.
S21.5F
- These two complexes are identical.
- These two complexes are geometric isomers. The bidentate ligand ethylenediamene is bonded to the metal in the center differently in the second complex, switching positions with the chloro ligand so that the ligand now has both bonds in the same plane. The first complex has the two chloro ligands in a trans position, meaning they are across from each other in the same plane, whereas the second complex has them in a cis position, meaning they form a \(90^\circ\) angle.
- These two complexes are enantiomers. Enantiomers are complexes that are mirror images of each other and can't be superimposed on one another no matter how you rotate the molecule.
Q24.14A
The following was obtained for the initial rates of reaction in the reaction A + 2B + C → 2D +E.
| Experiment | Initial [A], M | Initial [B], M | [C], M | Initial Rate |
|---|---|---|---|---|
| 1 | 1.20 | 1.20 | 1.00 | \(R_1\) |
| 2 | 0.60 | 1.20 | 1.00 | \(R_2 = \dfrac{1}{2}\times R_1\) |
| 3 | 0.60 | 0.60 | 1.00 | \(R_3 = \dfrac{1}{4}\times R_2\) |
| 4 | 1.20 | 1.20 | 0.50 | \(R_4 = 16 \times R_3\) |
| 5 | 0.60 | 0.60 | 0.50 | \(R_5\) = ? |
a. What are the reaction orders with respect to A, B, and C?
b. What is the value of \(R_5\) in terms of \(R_1\)?
S24.14A
a. The rate of a reaction is determined by the equation \(rate = k[A]^m[B]^n[C]^o\). The order of a reaction is represented by the variables m, n, and o with respect to A, B, and C respectively. In order to determine the order of a reaction, we set up a proportion of the rate laws for two trials. In order to find the reaction order with respect to A, we must find the two experiments where all else is held equal except for the concentration of A. Experiments 1 and 4 are the two that fit that criteria.
\(\dfrac{k[1.20]^m[1.20]^n[1.00]^o}{k[0.60]^m[1.20]^n[1.00]^o} = \dfrac{R_1}{\dfrac{1}{2}R_1}\)
After canceling everything out that can be canceled out, you should be left with
\(\dfrac{[1.20]^m}{[0.60]^m} = \dfrac{1}{\dfrac{1}{2}}\)
Then you simplify the ratios and solve for m.
\(2^m = 2\)
m = 1
Then you do the same to solve for the reaction order with respect to B, this time using experiments 2 and 3.
\(\dfrac{k[0.60]^m[1.20]^n[1.00]^o}{k[0.60]^m[0.60]^n[1.00]^o} = \dfrac{\dfrac{1}{2}R_1}{\dfrac{1}{4}R_2}\)
After canceling out everything that can be canceled out, you should be left with:
\(\dfrac{[1.20]^n}{[0.60]^n} = \dfrac{\dfrac{1}{2}R_1}{\dfrac{1}{4}R_2}\)
However, this time there is an extra step. Because the reaction rates don't have the same variables, you must change the reaction rate for experiment 3 so that it is in terms of \(R_1\) like the reaction rate for experiment 2.
\(\dfrac{1}{4}R_2 = \dfrac{1}{4}\dfrac{1}{2}R_1 = \dfrac{1}{8}R_1\)
After plugging the reaction rate for experiment 3 back into the equation, you can simplify the ratios and solve for n.
\(\dfrac{[1.20]^n}{[0.60]^n} = \dfrac{\dfrac{1}{2}R_1}{\dfrac{1}{8}R_1}\)
\(2^n = 4\)
n = 2
For the reaction order with respect to C, you repeat the same steps but with experiments 1 and 4.
\(\dfrac{k[1.20]^m[1.20]^n[1.00]^o}{k[1.20]^m[1.20]^n[0.50]^o} = \dfrac{R_1}{16R_3}\)
\(\dfrac{[1.00]^o}{[0.50]^o} = \dfrac{R_1}{16R_3}\)
\(16R_3 = 16 \times \dfrac{1}{4}R_2 = 4 \times R_2 = 4 \times \dfrac{1}{2}R_1 = 2R_1\)
\(\dfrac{[1.00]^o}{[0.50]^o} = \dfrac{R_1}{2R_1}\)
\(2^o = \dfrac{1}{2}\)
o = -1
b.