Extra Credit 47
- Page ID
- 82858
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The texts in blue are or additional notes from Uyen Ngo.
Q17.7.1
Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Table P1 are the same as those at each of the melting points. Assume the efficiency is 100%.
- CaCl2
- LiH
- AlCl3
- CrBr3
A17.7.1
1. CaCl2
First, we need to determine the corresponding half reactions, which are:
\[Cl_2\ +2e^−⇌2Cl^{−}(aq),\ E°=1.396V\]
\[Ca^{2+}(aq) +2e^−⇌Ca(s),\ E°=−2.84V\]
To determine which reaction is at the cathode and which reaction is at the anode, we have to figure which element will be reduced and which will be oxidized. To identify the reaction at the anode and cathode, we can use the Standard Reduction Potential Table, that is how we got the values 1.396 V and -2.84V. In electrolysis, the half reaction with a more negative standard reduction potential will be reduced (Calcium since it has a SRP of -2.84V), and will therefore be the cathode.
\[Cathode: Ca^{2+}(aq) +2e^−⇌Ca(s),\ E°=−2.84V\]
\[Anode: 2Cl^{−}(aq)⇌Cl_2\ +2e^−,\ E°=1.396V\]
We have to reverse the anode equation first so that there are electrons on both side so they can cancel out when the two reaction are added together. That is why the anode equation above is reversed.
To find the overall reaction:
Since the electrons are balanced on both sides already, we simply add the reactions together and cancel out the like terms to obtain
\[Ca^{2+}(aq)+2Cl^− (aq)⇌Ca(s)+Cl_2\]
To calculate the potential required for electrolysis, we use the equation
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
\[E°_{cell}=−2.84V-(1.396V)=-4.236\]
2.LiH
Again, in electrolysis, we know that the reaction with the more negative standard reduction potential will be reduced. So given the half reactions, we know that lithium will be reduced and hydrogen will be oxidized. Therefore, Lithium will be at the cathode and hydrogen will be at the anode.
\[Cathode: Li^{+}(aq) +e^−⇌Li(s),\ E°=−3.040V\]
\[Anode: H_{2}(g)⇌2H^{+}(aq)+2e^−,\ E°=0.000V\]
To determine the over all reaction, we need to balance the electrons. We do so by multiplying the lithium half reaction by 2:
\[(Li^{+}(aq) +e^−⇌Li(s))x2\]
\[2Li^{+}(aq) +2e^−⇌2Li(s)\]
Next we reverse the oxidation reaction
\[H_{2}(g)\rightleftharpoons 2H^{+}(aq)+2e^{-}\]
Next, we can add the two reactions together:
\[2Li^{+}(aq) +2e^{−}+H_{2}(g)⇌2Li(s)2H^{+}(aq)+2e^−\]
Cancel out the electrons on both sides ot obtain the overall reaction:
\[2Li^{+}(aq) +H_{2}(g)⇌2Li(s)2H^{+}(aq)\]
The potential required for the electrolysis reaction is:
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
\[E°_{cell}=−3.040V-(0.000V)=-3.040V\]
3.AlCl3
Based on the half reactions, we can say that the aluminum reaction will be reduced at the cathode since its potential is more negative. The chlorine reaction will be oxidized at the anode since it reduction potential is more positive.
\[Cathode: Al^{3+} (aq)+ 3e^-⇌Al(s),\ E°= -1.676 V\]
\[Anode: 2Cl^- (aq)⇌Cl_2 (g)\ +2e^-,\ E°= 1.396V\]
To balance the reactions, we need to multiply each equation by 2 and 3, respectively, to obtain the same number of electrons in each equation.
\[( Al^{3+} (aq)+ 3e^-⇌Al(s))x2\]
\[(2Cl^- (aq)⇌Cl_2 (g)\ +2e^-)x3\]
This gives us:
\[2Al^{3+} (aq)+ 6e^-⇌2Al(s)\]
\[6Cl^- (aq)⇌3Cl_2 (g)\ +6e^-\]
We can add the equations together and cancel out the electrons to get the overall reaction:
\[2Al^{3+} (aq)+6Cl^- (aq)⇌3Cl_2 (g)+2Al(s)\ +\]
To find the potential:
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
\[E°_{cell}=−1.676-(1.396V)=-3.072V\]
4.CrBr3
Based on the half reactions and standard potentials, we can see that chromium will be reduced at the cathode and bromine will oxidized at the anode.
\[Anode:2Br^-(aq) ⇌Br_2 (l)+ 2e^-,\ E°=1.087 V\]
\[Cathode: Cr^{3+} (aq)+ e^-⇌ Cr(s), E°= -0.74 V\]
To find the overall reaction, we need to multiply the cathode reaction by 2:
\[(Cr^{3+} (aq)+ e^-⇌ Cr(s))x2\]
\[2Cr^{3+} (aq)+ 2e^-⇌ 2Cr(s)\]
Next, we add the reactions and cancel out the electrons to get the overall reaction:
\[2Cr^{3+}+2Br^-⇌Br_2+2Cr^{2+}\]
The potential is:
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
\[E°_{cell}=−0.424V-(1.087V)=-1.511V\]
I'm not sure where the value -0.424V came from but I believe that the reaction is actually -0.74V for the cathode so it should be:
\[E^{\circ }_{cell} = -0.74V-(1.087V)= -1.827V\]
Q12.3.10
The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 × 10−8 L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 × 10−4 M?
A12.3.10
The instantaneous rate of is the rate at which the concentration is changing at a specific point. Therefore that's how we got the equation below.
The equation is a derivative of the equation:
\[\frac{1}{[A]}t=kt+\frac{1}{[A^\circ ]}\]
Since this is a second order reaction, we know the rate law will be Rate=k[A]2
We are given k (the rate constant) which is 4.7x10-8 L/mol/s and are asked to solve the equation with the concentration 5.55x10-4 M. We simply plug in the values into Rate=k[A]2
\[Rate=k[A]^2=4.7x10^{-8}\frac{L}{mol\ \times\ s}[5.55x10^{-4} M]^2=1.45x10^{-14} M/s \]
Q12.6.2
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A+B⟶C ? Can we predict the effect if the reaction is known to be an elementary reaction?
A12.6.2
In general, we can't predict the effect of doubling the concentration of A on the rate of the overall reaction. It is impossible to determine the rate law and reaction order with the balanced equation. This has to be determined experimentally. However, if it is known to be an elementary reaction, we know that the rate law will be rate=k[A][B]. This is because it is known at in an elementary reaction, the reaction order of the reactants depend of the coefficent of the reactant in the reaction equation. The rate law came from the reaction \[A+B\rightarrow C\], which each reactant has a coefficient of 1.
The doubling of the concentration of A will double the rate of the reaction. This is of A being directly proportional to rate.
Q21.4.14
A 1.00 × 10⁻⁶ g sample of nobelium, No-254, has a half-life of 55 seconds after it is formed. What is the percentage of No-254 remaining at the following times?
- 5.0 min after it forms
- 1.0 h after it forms
A21.4.14
First, we have to determine the rate constant. Since this is a first order reaction, we can use the equation t1/2 = 0.693/k.
Rearrange the first-order half- life equation to solve for k:
\[t_{1/2}=\frac{ln2}{\lambda }\]
\[k=\frac{0.693}{t_{1/2}} \]
\[k=\frac{0.693}{55 s}=0.0126 s^{-1} \]
1. Convert 5.0 min to seconds
\[(5.0min) (60 \frac{s}{min})=300 s \]
Using \[ln \frac {[A_t]}{[A_0]}=-kt \], where [At] is the amount at time t and [A0] is the initial amount:
\[ln \frac {[A_t]}{[A_0]}=-(0.0126s^{-1})(300s) \]
\[\frac {[A_t]}{[A_0]}=e^{-(0.0126s^{-1})(300s)} \]
\[\frac {[A_t]}{[A_0]}=0.0228 \]
Multiply by 100% to make this into a percentage
0.00228X100=2.28%
Therefore, the percentage left at 5 minutes is 2.28%
2. Convert 1.0 h to seconds
\[60 min) 60 \frac{s}{min}=3600 s \]
This equation is the integrated rate law for reaction rate of a first order reaction. Using \[ln \frac {[A_t]}{[A_0]}=-kt \], where \[[A_t]\] is the amount at time t and \[[A_0]\] is the initial amount:
\[ln \frac {[A_t]}{[A_0]}=-(0.0126s^{-1})(3600s) \]
\[\frac {[A_t]}{[A_0]}=e^{-(0.0126s^{-1})(3600s)} \]
\[\frac {[A_t]}{[A_0]}=2.00x10^{-20} \]
The percentage remaining after 1 hour is 2.00x10-18 %
Q20.3.2
If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?
A20.3.2
If two half reactions are physically separated, it possible for a redox reaction to occur if there is a complete circuit and an external electrical connection between the two circuits. The electrons can flow through this electrical connection. The name of the apparatus in which two half-reactions are carried out simultaneously is called a Galvanic cell., in which two half reactions are carried our simultaneously. There is a anode, which is where the oxidize half reaction occurs, and the cathode, where the reduced half reaction occurs. The electrons would flow from the anode to the cathode. There is also a salt bride so that spectator ions can migrate in between the two electrode. Though they are not a part of the reaction, they are used to maintain electrical neutrality.
This is what a Galvanic cell would generally look like.
Q20.5.13
For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?
A20.5.13
We can figure out how many electrons are transferred by determining the balanced equation. First, we have to separate the reaction into two half reactions.
\[Al(s) \rightarrow Al(aq)^{3+} \]
\[Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) \]
Next, we balance the charges by adding the appropriate number of electrons.
\[Al(s) \rightarrow Al(aq)^{3+} +3e^- \]
\[Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq) \]
Multiply the top equation by 2 and the bottom equation by 3 to obtain the same amount of electrons in each half reaction.
\[2x(Al(s) \rightarrow Al(aq)^{3+} +3e^-) \] = \[2Al(s) \rightarrow2 Al(aq)^{3+} +6e^- \]
\[3x(Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq)) \] =\[3Sn^{2+}(aq)+6e^- \rightarrow 3Sn^{4+}(aq) \]
Add the resulting equations together and cancel out the like terms.
\[3Sn^{2+}(aq)+2Al(s)+6e^- \rightarrow 3Sn^{4+}(aq)+Al(aq)^{3+}+6e^-\]
The electrons cancel out and we get
\[3Sn^{2+}(aq)+2Al(s) \rightarrow 3Sn^{4+}(aq)+Al(aq)^{3+}\]
Therefore, the number of electrons transferred in the redox reaction is 6.
The equation used to calculate standard cell potentials is
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
We have to determine which of the reactions are the cathode and anode.
Looking back at our half reactions, we see that we have:
\[Al(s) \rightarrow Al(aq)^{3+} +3e^- \]
Since Al is losing electrons, it is being oxidized. This means it is the anode.
\[Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq) \]
Sn is gaining electrons which means it has been reduced- therefore it is the cathode.
We now have to refer to a standard reduction potential table to obtain E°cathode and E°anode.
E°cathode=0.154 E°anode=-1.676
\[E°_{cell}=E°_{cathode}-E°_{anode}=0.154-(-1.676)= 1.83V\]
Since E°cell is positive, this is a spontaneous reaction. If the Standard Cell potential was negative, it would be non-spontaneous.
To calculate ΔG°, we use the equation \[ΔG°=−nFE°cell\]
n=moles of electrons transferred=6
F=Faraday's constant=96,468 J/(V⋅mol e-)
E°cell=1.83 V
\[ΔG°=−nFE°cell=-(6 mol\, e^-)(96,468 \frac {J}{(V)(mol)})=-1,059,416\, J\]
Q24.6.8
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
- [Cu(NH3)4]2+
- [Ni(CN)4]2−
A24.6.8
- [Cu(NH3)4]2+
To determine the structure, we have to look at the coordination number of the complex. By looking at the number of ligands, we see that we have a coordination number of 4, so the structure will either be tetrahedral or square planar. To decide, we need to find the electron configuration of Cu. Copper's configuration is [Ar]4s13d10. The charge on the complex is 2+, and since NH3 has a neutral charge, the we know that it is Cu2+ in the complex. The electron configuration of Cu2+ is [Ar]3d9. We do this by removing 1 electron from the 4s subshell and 1 from the 3d subshell. Therefore, copper has 9 electrons. This is a square planar complex.
From this spectrum, we can see that NH3 is a strong ligand, therefore making this low spin.
Next, we have to look at the energy diagram. Since this is low spin, we fill up the bottom row, which is the t2g level, before we move to the top, the eg level. A strong spin would mean that we place one electron in both eg and t2g levels before pairing them up. This gives:
There is one unpaired electron.
2.[Ni(CN)4]2−
The coordination number of this complex is also 4, which means it will have a square planar or tetrahedral structure. CN has a -1 charge so we know Ni has a +2 charge. The electron configuration of Ni2+ is [Ar]3d8. Since the outermost orbital has a d8 configuration, this is square planar. From the spectrum, we can see that CN is a strong field ligand, so this is a low spin complex. This is because strong field ligands have large splitting energy because it takes less energy to pair electrons in lower energy orbitals than in high energy orbital.
The energy diagram of Ni2+ is
There are two unpaired electrons.
Q14.4.6
Iodide reduces Fe(III) according to the following reaction:
2Fe3+(soln) + 2I−(soln) → 2Fe2+(soln) + I2(soln)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?
A14.4.6
Since doubling the concentration of Fe(III) doubles the reaction rate, the rate of reaction is proportional to the concentration of Fe3+, so it is first order. If doubling the iodide concentration quadruples the rate, the rate of the reaction is proportional to the square of I-, therefore, it is second order.
The overall rate law will be:
\[Rate=k[Fe^{3+}][I^-]^2 \]
To determine the overall reaction order, we simply add up the reaction order of each species in the rate law. The reaction order is given by the exponents:
\[1+2=3 \]