Extra Credit 47
- Page ID
- 82909
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Q 17.7.1
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%.
a. CaCl2
b. LiH
c. AlCl3
d. CrBr3
S 17.7.1
a. CaCl2
To identify which reactions occur at the anode and cathode, use the standard reduction potential (SRP) table. In electrolysis, the species with the larger, more positive SRP is oxidized at the anode and the species with the smaller, more negative SRP is reduced at the cathode. Since the SRP of calcium is -2.84 V and the SRP of chlorine is 1.396 V, calcium will be reduced at the cathode and chlorine will be oxidized at the anode.
\[Anode: Cl_2(g)\ +2e^-⇌ 2Cl^-(aq),\ E°= 1.396V\]
\[Cathode: Ca^{2+} (aq)\ +2e^-⇌Ca(s),\ E°= -2.84 V\]
To obtain the overall reaction:
1. Balance the number of electrons transferred
\[Cl_2(g)\ +2e^-⇌ 2Cl^-(aq)\]
\[Ca^{2+} (aq)\ +2e^-⇌Ca(s)\]
2. Reverse the oxidation reaction
\[2Cl^-(aq) ⇌ Cl_2(g)\ +2e^-\]
3. Sum the reactions
\[2Cl^-(aq) ⇌ Cl_2(g)\ +2e^-\]
\[Ca^{2+} (aq)\ +2e^-⇌Ca(s)\]
---------------------------------------------------------------------
\[2Cl^-(aq)\ +Ca^{2+} (aq)\ +2e^{-}⇌Cl_2(g)\ +Ca(s)\ +2e^{-}\]
4. Cancel out the electrons on both sides
\[2Cl^-(aq)\ +Ca^{2+} (aq)+ \require{cancel} \bcancel{2e^{-}} ⇌Cl_2(g)\ +Ca(s)+ \require{cancel} \bcancel{2e^{-}}\]
5. Overall reaction:
\[2Cl^-(aq)\ +Ca^{2+} (aq)⇌Cl_2(g)\ +Ca(s)\]
To approximate the potential required for the electrolysis of CaCl2, you must first calculate the standard cell potential by using the equation:
\[E°_{cell}= E°_{cathode}- E°_{anode}\]
\[E°_{cell}= -2.84 V - (1.396 V) = -4.236 V\]
The approximate potential required for the electrolysis of CaCl2 must be greater than the potential required for the reverse (spontaneous) reaction. Therefore, the potential must be greater than 4.236V, *indicating that the reaction is spontaneous because E cell is positive.
b. LiH
To identify which reactions occur at the anode and cathode, use the standard reduction potential (SRP) table. In electrolysis, the species with the larger, more positive SRP is oxidized at the anode and the species with the smaller, more negative SRP is reduced at the cathode. Since the SRP of lithium is -3.040 V and the SRP of hydrogen is 0.000 V, lithium will be reduced at the cathode and hydrogen will be oxidized at the anode.
\[Anode: 2H^+(aq)+2e^-⇌H_2(g),\ E°= 0.000 V\]
\[Cathode: Li^+ (aq)\ + e^-⇌Li (s),\ E°= -3.040 V\]
To obtain the overall reaction:
1. Balance the number of electrons transferred
\[2H^+(aq)+2e^-⇌H_2(g)\]
\[2\ (Li^+ (aq)\ + e^-⇌Li (s))\]
2. Reverse the oxidation reaction
\[H_2(g)⇌2H^+(aq)+2e^-\]
3. Sum the reactions
\[H_2(g)⇌2H^+(aq)+2e^-\]
\[2Li^+ (aq)\ + 2e^-⇌2Li (s)\]
---------------------------------------------------------------------
\[H_2(g)+2\ Li^+ (aq)+ 2e^-⇌2\ Li(s)+ 2H^+(aq)+2e^-\]
4. Cancel out the electrons on both sides
\[H_2(g)+2\ Li^+ (aq)+ \require{cancel} \bcancel{2e^{-}} ⇌2\ Li(s)+ 2H^+(aq)+ \require{cancel} \bcancel{2e^{-}}\]
5. Overall reaction:
\[H_2(g)+2\ Li^+ (aq)⇌2\ Li(s)+ 2H^+(aq)\]
To approximate the potential required for the electrolysis of LiH, you must first calculate the standard cell potential by using the equation:
\[E°_{cell}= E°_{cathode}- E°_{anode}\]
\[E°_{cell}= -3.040 V - (0.000 V)= -3.040 V\]
The approximate potential required for the electrolysis of LiH must be greater than the potential required for the reverse (spontaneous) reaction. Therefore, the potential must be greater than 3.040 V,*indicating that the reaction is spontaneous because the Ecell is positive.
c. AlCl3
To identify which reactions occur at the anode and cathode, use the standard reduction potential (SRP) table. In electrolysis, the species with the larger, more positive SRP is oxidized at the anode and the species with the smaller, more negative SRP is reduced at the cathode. Since the SRP of aluminum is -1.676 V and the SRP of chlorine is 1.396 V, aluminum will be reduced at the cathode and chlorine will be oxidized at the anode.
\[Anode: Cl_2 (g)\ +2e^-⇌2Cl^- (aq),\ E°= 1.396V\]
\[Cathode: Al^{3+} (aq)+ 3e^-⇌Al(s),\ E°= -1.676 V\]
To obtain the overall reaction:
1. Balance the number of electrons transferred
\[3\ (Cl_2 (g)\ +2e^-⇌2Cl^- (aq))\]
\[2\ (Al^{3+} (aq)+ 3e^-⇌Al(s))\]
2. Reverse the oxidation reaction
\[6Cl^- (aq)⇌ 3Cl_2 (g)\ +6e^-\]
3. Sum the reactions
\[6Cl^- (aq)⇌ 3Cl_2 (g)\ +6e^-\]
\[2Al^{3+} (aq)+ 6e^-⇌2Al(s))\]
--------------------------------------------------------------------
\[6Cl^- (aq)+2Al^{3+} (aq)+ 6e^-⇌2Al(s)+ 3Cl_2 (g)+6e^-\]
4. Cancel out the electrons on both sides
\[6Cl^- (aq)+2Al^{3+} (aq) + \require{cancel} \bcancel{6e^{-}}⇌2Al(s)+ 3Cl_2 (g)+ \require{cancel} \bcancel{6e^{-}}\]
5. Overall reaction:
\[6Cl^- (aq)+2Al^{3+} (aq)⇌2Al(s)+ 3Cl_2 (g)\]
To approximate the potential required for the electrolysis of AlCl3, you must first calculate the standard cell potential by using the equation E°cell= E°cathode- E°anode.
\[E°_{cell}= E°_{cathode}- E°_{anode}\]
\[E°_{cell}= -1.676 V - (1.396 V) = -3.072 V\]
The approximate potential required for the electrolysis of AlCl3 must be greater than the potential required for the reverse (spontaneous) reaction. Therefore, the potential must be greater than 3.072 V, *indicating that the reaction is spontaneous since Ecell is positive.
d. CrBr3
To identify which reactions occur at the anode and cathode, use the standard reduction potential (SRP) table. In electrolysis, the species with the larger, more positive SRP is oxidized at the anode and the species with the smaller, more negative SRP is reduced at the cathode. Since the SRP of chromium is –0.74 V and the SRP of bromine is 1.087 V, chromium will be reduced at the cathode and bromine will be oxidized at the anode.
\[Anode: Br_2 (l)+ 2e^-⇌2Br^-(aq),\ E°=1.087 V\]
\[Cathode: Cr^{3+} (aq)+ 3e^-⇌ Cr(s),\ E°= -0.74 V\]
To obtain the overall reaction:
1. Balance the number of electrons transferred
\[3\ (Br_2 (l)+ 2e^-⇌2Br^-(aq))\]
\[2\ (Cr^{3+} (aq)+ 3e^-⇌ Cr(s))\]
2. Reverse the oxidation reaction
\[6Br^-(aq)⇌ 3Br_2 (l)+6e^-\]
3. Sum the reactions
\[6Br^-(aq)⇌ 3Br_2 (l)+6e^-\]
\[2Cr^{3+} (aq)+ 6e^-⇌ 2 Cr(s)\]
--------------------------------------------------------------------
\[6Br^-(aq)+2Cr^{3+} (aq)+ 6e^-⇌ 2 Cr(s)+3Br_2 (l)+6e^-\]
4. Cancel out the electrons on both sides
\[6Br^-(aq)+2Cr^{3+} (aq)+ \require{cancel} \bcancel{6e^{-}}⇌ 2 Cr(s)+3Br_2 (l)+ \require{cancel} \bcancel{6e^{-}}\]
5. Overall reaction:
\[6Br^-(aq)+2Cr^{3+} (aq)⇌ 2 Cr(s)+3Br_2 (l)\]
To approximate the potential required for the electrolysis of CrBr3, you must first calculate the standard cell potential by using the equation E°cell= E°cathode- E°anode.
\[E°_{cell}= E°_{cathode}- E°_{anode}\]
\[E°_{cell}= -0.74 V - 1.087 V= -1.827 V\]
The approximate potential required for the electrolysis of CrBr3 must be greater than the potential required for the reverse (spontaneous) reaction. Therefore, the potential must be greater than 1.827 V, *since Ecell is positive the reaction is spontaneous.
Q 12.3.10
The decomposition of acetaldehyde is a second order reaction with a rate constant of \(4.71\times10^{-8} \frac{L}{mol\ \times\ s}\). What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of \(5.55\times10^{-4} M\)?
S 12.3.10
Since the decomposition of acetaldehyde is a second order reaction, the rate of the reaction would be proportional to the rate constant times the square of the concentration of the reactant, acetaldehyde.
\[rate=k\left[A \right]^2\]
\[rate= (4.71\ \times10^{-8}\frac{L}{mol\ \times\ s})\left[5.55\times10^{-4}M \right]^2\]
\[rate= 1.45\times10^{-14} \frac{M}{s}\]
*The constant of the order is determined by -(x-1), then -(2-1), it will be -1. So that is why it is M-1s-1
Q 12.6.2
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?
S 12.6.2
We cannot predict the effect of doubling the concentration of A on the rate of the overall reaction, \(A+B⟶ C\), since the rate law and the reaction order cannot be derived from the stoichiometric equation. The reaction order and the rate law must be determined by experiment. However, we can predict the effect if the reaction is known to be an elementary reaction, since the rate law can be determined from the balanced reaction equation. In an elementary reaction, the reaction order of each reactant is indicated by its coefficient in the reaction equation. If \(A+B⟶ C\) were an elementary reaction, the rate law would be \(rate=k\left[ A\right]\left[B \right]\) since each of the reactants has a coefficient of 1. This means that A, with a first reaction order, is directly proportional to the rate, and doubling the concentration of A would double the rate of the reaction. *Since the reaction is in the first oder the units would be s-1.
Q 21.4.14.
A \(1.00\times10^{-6} g\) 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?
a. 5.0 min after it forms
b. 1.0 h after it forms
S 21.4.14.
a. 5.0 min after it forms
Since radioactive decay is a first-order process, you can use the integrated rate law for the reaction rate of a first-order reaction, in terms of the number of atoms instead of concentration, to calculate the amount of No-254 remaining.
\[ ln\left(\frac{N_t}{N_0} \right)=-\lambda\ t\]
To calculate the percentage of No-254 remaining 5.0 minutes after it forms,
Use the first-order half-life equation to find the value of the radioactive decay constant
\[t_\frac{1}{2}=\frac{ln \ 2}{\lambda}\]
\[\lambda=\frac{ln \ 2}{t_{\frac{1}{2}}}\]
\[\lambda=\frac{ln\ 2}{55\ seconds}\]
\[\lambda≈0.0126 \ s^{-1}\]
Rearrange the integrated rate law to solve for the fraction remaining
\[\frac{N_t}{N_0}=e^{-\lambda\ t}\]
Since the radioactive decay constant is in terms of seconds, convert 5.0 minutes to seconds
\[5.0\ minutes\;\times\frac{60\ seconds}{1\ minute}= 300\ seconds\]
Calculate the fraction remaining by plugging in 300 seconds for time and 0.0126 s-1 for the radioactive decay constant
\[\frac{N_t}{N_0}=e^{-\lambda\ t}\]
\[\frac{N_t}{N_0}=e^{-\ (0.0126\ s^{-1})(300\ s)}=0.0228\]
Multiply by 100% to obtain a percentage
\[0.0228\ \times100\%=2.28\%\ remaining\]
*This answer is completely right, I could not find any errors.
b. 1.0 h after it forms
Since radioactive decay is a first-order process, you can use the integrated rate law for the reaction rate of a first-order reaction, in terms of the number of atoms instead of concentration, to calculate the amount of No-254 remaining.
\[ ln\left(\frac{N_t}{N_0} \right)=-\lambda\ t\]
To calculate the percentage of No-254 remaining 1.0 hour after it forms,
Use the first-order half-life equation to find the value of the radioactive decay constant
\[t_\frac{1}{2}=\frac{ln \ 2}{\lambda}\]
\[\lambda=\frac{ln \ 2}{t_{\frac{1}{2}}}\]
\[\lambda=\frac{ln\ 2}{55\ seconds}\]
\[\lambda≈0.0126 \ s^{-1}\]
Rearrange the integrated rate law to solve for the fraction remaining
\[\frac{N_t}{N_0}=e^{-\lambda\ t}\]
Since the radioactive decay constant is in terms of seconds, convert 1.0 hour to seconds
\[1.0\ hour\ \times\frac{60\ minutes}{1\ hour}\ \times\frac{60\ seconds}{1\ minute}=3,600\ seconds\]
Calculate the fraction remaining by plugging in 3,600 seconds for time and 0.0126 s-1 for the radioactive decay constant
\[\frac{N_t}{N_0}=e^{-\lambda\ t}\]
\[\frac{N_t}{N_0}=e^{-\ (0.0126\ s^{-1})(3,600\ s)}=2.00\ \times10^{-20}\]
Multiply by 100% to obtain a percentage
\[(2.00\times10^{-20})\ \times100\%=2.00\ \times10^{-18}\%\ remaining\]
*This answer is completely right, I could not find any errors.
Q 20.3.2
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?
S 20.3.2
If two half-reactions are physically separated, a redox reaction can occur if there is a complete circuit that allows for the flow of electrons from the oxidation reaction to the reduction reaction. An electrochemical cell is an apparatus in which two half-reactions are carried out simultaneously. In an electrochemical cell, there is a solution with an anode, where the oxidative half-reaction occurs, and a separate solution with a cathode, where the reductive half-reaction occurs. *The electrons flow from oxidation (loss of electrons) to reduction (gain of electrons). The electrodes are connected by a wire and establish a complete circuit by providing an external electrical connection between the two separate solutions. The cell also contains either a porous membrane or a salt bridge that allows charged spectator ions to migrate between the electrode compartments. These spectator ions do not participate in the electrochemical reaction, and are used to maintain electrical neutrality in the cell.
Q 20.5.13
For the cell represented as \(Al(s)\ |\ Al^{3+}(aq)\ ||\ Sn^{2+}(aq),\ Sn^{4+}(aq)\ |Pt(s)\), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is \(\Delta\ G°\)?
S 20.5.13
To calculate the standard cell potential, use the equation \(E°_{cell}= E°_{cathode}- E°_{anode}\) and the values from the standard reduction potential table. Based on the cell diagram, aluminum is being oxidized since it was written on the left side and tin is being reduced since it was written on the right side.
\[Anode: Al^{3+} (aq)+3e^{-}⇌Al(s),\ E°=-1.676\ V\]
\[Cathode: Sn^{4+} (aq)+2e^{-}⇌ Sn^{2+} (aq),\ E°=0.154\ V\]
\[E°_{cell}=E°_{cathode}-E°_{anode}\]
\[E°_{cell}=0.154\ V- (-1.676\ V)= 1.83\ V\]
To determine how many electrons are transferred in the redox reaction, you must balance the reaction. Since aluminum is losing 3 electrons and tin is gaining 2 electrons, the least common factor for the number of electrons transferred is 6.
\[2\ (Al (s)⇌Al^{3+} (aq)+3e^{-})\]
\[3\ (Sn^{4+} (aq)+2e^{-}⇌Sn^{2+} (aq))\]
--------------------------------------------------------------------
\[2Al (s)+3Sn^{4+} (aq)+\require{cancel} \bcancel{6e^{-}}⇌ 3Sn^{2+} (aq)+ 2Al^{3+} (aq)+\require{cancel} \bcancel{6e^{-}}\]
To calculate \(\Delta\ G°\), use the equation \(\Delta\ G°=-nFE°_{cell}\).
- n= number of electrons transferred
- F= Faraday’s constant= \(96,485 \frac{J}{V\ \times\ mol\ e^{-}}\)
\[\Delta\ G°= -6\ moles\ e^{-}\ \times 96,485\ \frac{J}{V\ \times\ mole\ e^{-}}\ \times 1.83\ V\ \times\frac{1\ kJ}{1000\ J}\]
\[\Delta\ G°=-1,059\ kJ\]
Since \(\Delta\ G°\) is negative and \(E°_{cell}\) is positive, this process is spontaneous, *favoring the products.
Q 24.6.8
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
a. [Cu(NH3)4]2+
b. [Ni(CN)₄]²⁻
S 24.6.8
a. [Cu(NH3)4]2+
To predict the structure of this complex ion, you must look at the metal center with respect to crystal field theory. There are four monodentate, NH3 ligands that are attached to the copper atom, so the complex has a coordination number of four. Complexes with a coordination number of four have either a tetrahedral or square planar structure.
Then, you must determine the number of d-electrons in the transition metal, copper. The complex ion has a charge of 2+ and there are four neutral NH3 ligands, so the charge of the copper ion is 2+. Copper normally has an electron configuration of [Ar] 4s13d10. To obtain the electron configuration of Cu2⁺, you would remove 1 electron from the 4s subshell and 1 electron from the 3d subshell. Thus, copper would have 9 d-electrons.
Transition metal complexes with a coordination number of four and a d⁸ configuration tend to have a square planar geometry and a low spin, since the energy gap between the dxy and dx2-y2 orbitals is very large. However, since Cu2+ has a d9 configuration, determining whether the complex has a tetrahedral or square planar structure would require the use of information that hasn't been discussed yet in this course. But in this case, [Cu(NH3)4]2+ has a square planar structure.
To determine whether the complex has a high spin or low spin, you would use the spectrochemical series to determine the strength of the ligands. Strong field ligands have a large splitting energy, so it would take less energy to pair the electrons in a lower energy orbital than to singly place the electron in a higher energy orbital. This explains why a complex with a strong field ligand tends to have a low spin and a complex with a weak field ligand that produces a small splitting energy tends to have a high spin. Based on the series, NH3 is neither a highly strong field nor weak field ligand and its strength would depend on the metal atom that it is coordinated to. However, since high vs. low spin possibilities only occur for complexes with d4 to d7 electron configurations, this complex with a d9 configuration has neither a high nor low spin.
With a d9 electron configuration and a square planar structure, there is only one possible energy diagram. Eight of the nine d-electrons would occupy the four lowest energy orbitals, and there would be one unpaired electron in the dx2-y2 orbital. *It is rare to have one electron on the last level, since it requires a lot of energy to reach dx2-y2
b. [Ni(CN)₄]²⁻
To predict the structure of this complex ion, you must look at the metal center with respect to crystal field theory. There are four monodentate, CN ligands that are attached to the nickel, so the complex has a coordination number of four. Complexes with a coordination number of four have either a tetrahedral or square planar structure.
Then, you must determine the number of d-electrons in the transition metal, nickel.The complex ion has a charge of 2- and there are four CN ligands with a charge of 1- each, so the charge of the nickel ion is 2+. Nickel normally has an electron configuration of [Ar] 4s²3d⁸. To obtain the electron configuration of Ni²⁺, you would remove 2 electrons from the outermost shell, the 4s subshell. Thus, nickel would have 8 d-electrons.
Based on the spectrochemical series, CN is a strong field ligand. Since strong field ligands have a large splitting energy, it would take less energy to pair the electrons in a lower energy orbital than a higher energy orbital, and they tend to induce low spin complexes.
[Ni(CN)₄]²⁻ has a square planar geometry. Transition metal complexes with the d⁸ configuration tend to have a square planar geometry and a low spin, since the energy gap between the dxy and dx2-y2 orbitals is very large. Therefore, it would require less energy for all 8 of the electrons to occupy the lowest four energy orbitals.
There are zero unpaired electrons, since the complex has a square planar structure and a low spin. The 8 electrons are all paired in the 4 lowest energy orbitals: dyz, dxz, dz2, and dxy. *For tetrahedral it is very hard to reach dx2-y2 because it requires high energy.
Q 14.4.6
Iodide reduces Fe(III) according to the following reaction:
\[2Fe^{3+}(soln)+ 2I^- (soln)→2Fe^{2+} (soln)+I_2(soln)\]
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?
S 14.4.6
Since doubling the concentration of Fe³⁺ doubled the reaction rate, the rate of the reaction is proportional to the concentration of Fe³⁺ and Fe³⁺ has a first reaction order. Since doubling the concentration of I⁻ increased the reaction rate by a factor of 4, the rate of the reaction is proportional to the square of the concentration of I⁻and I⁻ has a second reaction order.
The overall rate law would be the product of the rate constant times the concentration of each reactant raised to its respective reaction order.
\[rate=k\left[Fe^{3+} \right]\left[I^- \right]^2\]
This reaction has a third reaction order overall, based on the sum of the reaction order for each species in the rate law, *which shows (1+2)=3.
\[1\ (reaction\ order\ of Fe^{3+})\ + 2\ (reaction\ order\ of\ I^-)= 3\]