Extra Credit 21
- Page ID
- 83415
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Calculate \(\Delta G^\circ\) for the following reactions and determine if spontaneous:
- \(NO^{−3}\)(aq)+Al(s)+\(4H^+\)(aq)→NO(g)+\(Al^3+\)(aq)+\(2H_2O\)(l)
- \(F_2\)(g)+2Li(s)→\(2F^−\)(aq)+\(2Li^+\)(aq)
A 19.7A
PART A
To answer this question we must first find the energy potential of the cell using the equation:
\[\mathrm{E^\circ_{cell} = (E^\circ_{cathode}) - (E^\circ_{anode})}\]
Using the oxidation states, we find that the \(NO_3^-/NO^-\) reaction is the reduction reaction and therefore occurs at the cathode. Meanwhile, the \(Al^{3+}/Al\) reaction is an oxidation reaction and occurs at the anode. Therefore, we know the equation for the standard potential of the cell is:
\[\mathrm{E^\circ_{cell} = (E^\circ_{NO_3^-/NO^-}) - (E^\circ_{Al^{3+}/Al})}\]
Next, we use the Standard Reduction Potential Table below to find the standard reduction potentials (SRP) of the individual half-reactions we are looking at.
| Reduction Half-Reaction Eo, V | |
| Acidic Solution | |
| F2(g) + 2e- → 2 F-(aq) | +2.866 |
| O3(g) + 2H+(aq) + 2e- → O2(g) + H2O(l) | +2.075 |
| S2O82-(aq) + 2e- → 2SO42-(aq) | +2.01 |
| H2O2(aq) + 2H+(aq) +2e- → 2H2O(l) | +1.763 |
| MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) | +1.51 |
| PbO2(s) + 4H+(aq) + 2e- → Pb2+(aq) + 4H2O(l) | +1.455 |
| Cl2(g) + 2e- → 2Cl-(aq) | +1.358 |
| Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) | +1.33 |
| MnO2(s) + 4H+(aq) +2e- -> Mn2+(aq) + 2H2O(l) | +1.23 |
| O2(g) + 4H+(aq) + 4e- → 2H2O(l) | +1.229 |
| 2IO3-(aq) + 12H+(aq) + 10e- → I2(s) + 6H2O(l) | +1.20 |
| Br2(l) + 2e- → 2Br-(aq) | +1.065 |
| NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2 H2O(l) | +0.956 |
| Ag+(aq) + e- → Ag(s) | +0.800 |
| Fe3+(aq) + e- → Fe2+(aq) | +0.771 |
| O2(g) + 2H+(ag) + 2e- → H2O2(aq) | +0.695 |
| I2(s) + 2e- → 2I-(aq) | +0.535 |
| Cu2+(aq) + 2e- → Cu(s) | +0.340 |
| SO42-(aq) + 4H+(aq) + 2e- → 2H2O(l) + SO2(g) | +0.17 |
| Sn4+(aq) + 2e- → Sn2+(aq) | +0.154 |
| S(s) + 2H+(aq) + 2e- → H2S(g) | +0.14 |
| 2H+(aq) + 2e- → H2(g) | 0 |
| Pb2+(aq) + 2e- → Pb | -0.125 |
| Sn2+(aq) + 2e- → Sn(s) | -0.137 |
| Fe2+(aq) + 2e- → Fe(s) | -0.440 |
| Zn2+ + 2e- → Zn(s) | -0.763 |
| Al3+(aq) + 3e- → Al(s) | -1.676 |
| Mg2+(aq) + 2e- → Mg(s) | -2.356 |
| Na+(aq) + e- → Na(s) | -2.713 |
| Ca2+(aq) + 2e- → Ca(s) | -2.84 |
| K+(aq) + + e- → K(s) | -2.924 |
| Li+(aq) + e- → Li(s) | -3.040 |
From the table, we find that the SRP of the \(NO_3^-/NO^-\) reaction is 0.956 and the SRP of the \(Al^{3+}/Al\) reaction is -1.676. We plug these values into the equation for standard cell potential equation.
\(\mathrm{E^\circ_{cell} = (0.956) -(-1.676)}\)
\(\mathrm{E^\circ_{cell} = 2.63\,V}\)
Using the oxidation states, we also find that 3 electrons are transferred during this reaction because this is the amount that is needed to make the charges equal on each side of the equation.
\(\ce{N}\) oxidation: reactant = +5 product= +2
\(\ce{Al}\) oxidation: reactant = 0 product= +3
And therefore \(\mathrm{N = 3}\). You can also see evidence of this in the reactions in the table above.
Lastly, we know \(\mathrm{F=96,485\,C/mol\,e^-}\) because this is Faraday's constant.
We plug all of this information into the Gibbs Free Energy equation and solve.
\(\mathrm{\Delta G^\circ= -nFE^\circ_{cell}}\)
\(\mathrm{\Delta G^\circ=-[(3)(96,485\,C/mol\,e^-)(2.63\,V)]}\)
\(\mathrm{\Delta G^\circ= -761\, kJ\, Spontaneous}\)
Because the change in Gibbs Free Energy is negative and the \(\mathrm{E^\circ_{cell}}\) is positive the reaction is spontaneous.
The change in Gibbs Free Energy is the change in free energy during a reaction. If the change is negative, the reaction is positive as the reactants have higher energy than the products do, so energy is released in the reaction. This correlates with a spontaneous reaction because the reactants move towards a more stable product and lower energy implies a more stable state.
PART B
Using the same method as above, we first need to find the standard cell potential, knowing that \(F_2/F^-\) occurs at the cathode because it is reduced and \(Li^+/Li\) occurs at the anode because it is oxidized. Therefore, we use the equation \(\mathrm{E^\circ_{cell} = (E^\circ_{F_2/F^-})- (E^\circ_{Li^+/Li})}\).
We find the individual half-reactions in the Standard Reduction Potential table (above) and use these values to plug into the standard cell potential equation and solve.
\(\mathrm{E^\circ_{cell} = 2.866 + 3.040 = 5.906\,V}\)
Next, we find the number of electrons transferred (N) by looking at the half reactions and balancing their charges. From this we find that \(\mathrm{N = 2}\).
\(\mathrm{F_2 + 2e^- \rightarrow 2F^-}\)
\(\mathrm{2Li^+ + 2e^- \rightarrow 2Li(s)}\)
Lastly, we plug in all of these values, in addition to Faraday's constant and solve for the change in Gibbs Free Energy.
\(\mathrm{\Delta G^\circ= -nFE^\circ_{cell}}\)
\(\mathrm{\Delta G^\circ= -[(2\,mol\,e^-)(96,485\,C/mol\,e^-)(5.906\,V)]}\)
\(\mathrm{\Delta G^\circ=-1,140\,kJ}\)
Because the change in Gibbs Free Energy is negative and the \(\mathrm{E^\circ_{cell}}\) is positivethe reaction is spontaneous.
Q19.59C
Imagine an iron nail that is corroding in a solution. Predict the appearance of the nail under the given conditions using your knowledge of corrosion in voltaic cells. Assume standard temperature conditions.
- The head and tip of the nail are covered in magnesium.
- The nail is galvanized, but there is a break in plating.
- The nail is plated in copper, but there is a break in plating.
A 19.59C
1. The \(\mathrm{E^\circ_{cell}}\) of magnesium is smaller than the \(\mathrm{E^\circ_{cell}}\) for iron and, therefore, magnesium oxidizes much more readily than oxygen. A higher \(\mathrm{E^\circ_{cell}}\) means that the metal will be more easily reduced than those with lower \(\mathrm{E^\circ_{cell}}\). Oxidation of metals causes the metal to corrode. Because magnesium is much more active than iron, it acts as a sacrificial coating for the nail. Sacrificial coatings are applied to metals to reduce oxidation and thus corrosion. Therefore, the head and tip of the nail will not corrode while the rest of the nail will.
2. A steel nail is galvanized when it is coated with a layer of zinc to protect it from corrosion. Zinc a more active metal and it leaves a negative charge on iron, thus which inhibiting the further corrosion of Fe2+. Zinc has a lower \(\mathrm{E^\circ_{cell}}\) than iron. Therefore, although there is a break in the plating, the nail will not corrode due to the addition of a negative charge.
3. If there is a break in the copper plating of a nail, the nail will begin to oxidize because iron is more active than copper and therefore corrosion will occur and spread through the nail until the whole thing corrodes.
Q21.4A
- How many unpaired electrons would you expect to find in the complex \([Fe(CN)_6]^{3+}\)?
- How many unpaired electrons would you expect to find in the tetrahedral complex \([MnBr_4]^{2−}\)? Would you expect more, fewer, or the same number of unpaired electrons as in the octahedral complex \([Mn(NH_3)_6]^{2+}\)?
A21.4A
1. We know that Cyanide (CN) has a -1 charge and there's a total of 6 Cyanides in this complex so the overall charge of these ligands is -6 . We also know that the total charge of the complex is 3+. Therefore, we can find that Iron (Fe) must have a charge of 9+ in order to make the complex's charges comprehensive. We use this knowledge to find the total number of electrons in the complex. The electron configuration of neutral Iron is \([Ar]4s^23d^6\) and the neutral iron atom has however when Iron has a 9+ charge it must lose 9 electrons from its configuration. Iron loses electrons from its s orbital first and then from its d orbital, therefore, in the end, the configuration of \(Fe^{+9}\) looks like \([Ar]3d^1\). Thus there is only one unpaired electron left in the d orbital of \(Fe^{+9}\), which is depictive of the complex as a whole.
There would be 1 unpaired electron in the \([Fe(CN)_6]^{3+}\) complex.
2. The Bromo (Br) ligand has a charge of -1 and there are 4 Bromo ligands in this complex, therefore there is a charge of 4- on the ligands. There is an overall charge of 2- on the complex and therefore we know that Manganese (Mn) has a charge of 2+. When Manganese has a 2+ charge it loses 2 electrons from the 4s orbital, making its electron configuration equal to \([Ar]3d^5\). Therefore, there are 5 electrons in the tetrahedral complex, which are high spin because \(Br^-\) is a weak-field ligand with a smaller \(\Delta\) splitting and thus us more likely to be high-spin. Therefore all 5 electrons are unpaired in the \([MnBr_4]^{2−}\) complex.
We would expect fewer unpaired electrons in the octahedral complex \([Mn(NH_3)_6]^{2+}\) because \(NH_3\) is a strong-field ligand with a larger \(\Delta\) splitting and thus us more likely to be low-spin. Low spin electrons tend to pair up and therefore there's bound to be less unpaired electrons. If you do the math, its true, \([Mn(NH_3)_6]^{2+}\) has only one unpaired electron.
Phase II:
1. I disagree with this solution. I believe there is an error in the question and the compound that is shown should be \([Fe(CN)_6]^{3-}\) instead of \([Fe(CN)_6]^{3+}\) . In this case, we know that Cyanide (CN) has a -1 charge and there's a total of 6 Cyanides in this complex so the overall charge of these ligands is -6 . We also know that the total charge of the complex is 3-. Therefore, we can find that Iron (Fe) must have a charge of 3+ in order to make the complex's charges comprehensive. We use this knowledge to find the total number of electrons in the complex. The electron configuration of neutral Iron is \([Ar]4s^23d^6\) and the neutral iron atom has however when Iron has a 3+ charge it must lose 3 electrons from its configuration. Iron loses electrons from its s orbital first and then from its d orbital, therefore, in the end, the configuration of \(Fe^{+3}\) looks like \([Ar]3d^5\). Because CN- is a strong field electron, this produces a low spin configuration, in which electrons are paired first before all of the shells are filled. Therefore there is only one unpaired electron in this compound.
Q24.11C
Use the table below to answer the following questions:
- Find the order of the reaction A vs. B
- Find the overall order
- Find the rate constant
| EXPT | [A],M | [B],M | Initial Rates, Ms-1 |
|---|---|---|---|
| 1 | 0.175 | 0.138 | 3.75 × 10-4 |
| 2 | 0.175 | 0.185 | 1.25 × 10-3 |
| 3 | 0.365 | 0.138 | 3.75 × 10-4 |
| 4 | 0.365 | 0.185 | 1.25 × 10-3 |
A 24.11C
- To find the order of reaction for B you must first look on the table above and find the 2 equations where A is constant and B changes. In this case, it is Reaction 1 and Reaction 2. We then divide \({\dfrac{Reaction\:2}{Reaction\:1}}\) and set up a proportion where [A][B] ratio = Initial Rates ratio
\(\mathrm{\dfrac{Reaction\: 2}{Reaction\: 1} = \dfrac{1.25 \times 10^{-3}}{ 3.75 \times 10^{-4}} = \dfrac{[0.175]^m [0.185]^n}{[0.175]^m[0.138]^n}}\)
\(4 = 2^n\)
\(n=2\)
Therefore the reaction is second order with respect to \(\ce{B}\). This is because the reaction rate will increase four fold if the concentration of B is doubled.
We follow the same process to find the order of reaction with respect to A. We first find where B is constant and A changes (Reaction 3 and Reaction 1). Then we divide:
\(\mathrm{\dfrac{Reaction\: 3}{Reaction\: 1} = \dfrac{4.41 \times 10^{-4}}{ 2.205 \times 10^{-4}} = \dfrac{[0.406]^m[0.662]^n}{[0.203]^m[0.662]^n}}\)
\(\mathrm{\dfrac{R3}{R1} = 2 = 2^m}\)
\(\mathrm{m = 1}\)
Therefore the reaction is first order with respect to \(\ce{A}\). The reaction rate will double if the concentration of A is doubled.
b. The overall order of reaction is equal to the sum of the individual reaction orders of the system.
\(\mathrm{Reaction\: order = \textrm{reaction order of A} + \textrm{reaction order of B} = 3^{rd}\textrm{ order reaction}}\)
c. To find the rate constant we can plug in any set of data into the rate equation and solve for k using simple algebra.
\(rate= k \times [A]^{m}[B]^{n}\)
\(\mathrm{2.205 \times 10^{-4} = k \times [0.331][0.203]^2}\)
\(\mathrm{k=1.6 \times 10^{-2}}\)
Q24.51C
The rate constant for the reaction \(H_2(g)+F_2(g)→2HF\) has been determined at the following temperatures 650 degrees k, \(k=4.8×10^{−4}M^{−1}s^{−1}; 700 degrees K, \(k=3.0×10^{−2}M^{−1}s^{−1}. Calculate the activation energy for the reaction.
A24.51C
To find the rate constant for this reaction at 2 different temperatures we will use the Arrhenius equation and plug in for what we know and solve.
\(\mathrm{\ln\dfrac{k_1}{k_2} = \dfrac{E_a}{R} \left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)}\)
Where k is the rate constant, \(E_a\) is the activation energy, R is the ideal gas (\(\mathrm{R=8.314\: j/(mol\: k)}\)), and T is temperature in Kelvin.
\(\mathrm{\ln\dfrac{k_1}{k_2} = \dfrac{E_a}{R} \left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right) =\ln\dfrac{4.80 \times 10^{-4}\, M^{-1}s^{-1}}{3.00 \times 10^{-2}\, M^{-1}s^{-1}} = \dfrac{E_a}{R} \left(\dfrac{1}{700 K} -\dfrac{1}{650 K}\right)}\)
\(\mathrm{(E_a)(-1.099 \times 10^{-4}) = -4.135R}\)
\(\mathrm{E_a =\dfrac{(-4.135)(8.314\: J/(mol\: k))}{-1.099 \times 10^{-4}} = 313\: kJ/mol}\)
Q25.27C
A piece of wood is claimed to found in Qin Mausoleum and is offered for sale to an art museum. Radiocarbon dating of the object reveals a disintegration rate of 5.0 dis min-1 g-1. Do you think the object is real? Explain your answer. Given that the \(\mathrm{\lambda = \dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}}\) and the initial rate of decay is about 15 dis/min.
A25.27C
To determine whether the object is really from Qin Mausoleum you must determine how old the wood is by calculating the rate of decay, using the formula:
\(\ln\dfrac{N}{N_0}= -kt\)
Decay reactions follow the same rates of change as first order reactions do.
Where \(k=\lambda\), the rate constant and N and\(N_0\) are the final and intial disintegration rates, respectively. Therefore, we plug in the values we have and solve for t (time).
\(\mathrm{\ln\dfrac{5\, dis/min}{15\, dis/min} = -1.0986\,s=-\lambda t}\)
\(\mathrm{t = \dfrac{1.0976}{1.21\times10^{-4}\,y^{-1}}=9071\,years}\)
The piece of wood is around 9,070 years old. Therefore, I do not believe this object is really from the Quin Mausoleum because this is around 3000B.C.
Q18.8
The following table contains data obtained by measuring the voltage between two metals in an experiment like the one you did with a lemon:
|
Voltaic Cell |
Anode (-) |
Cathode (+) |
Cell Voltage (v) |
|---|---|---|---|
|
Pb/Ni |
Ni |
Pb |
0.10 volts |
|
Pb/Au |
Pb |
Au |
0.80 volts |
|
Pb/Fe |
Fe |
Pb |
0.25 volts |
|
Ni/Au |
Ni |
Au |
0.90 volts |
|
Ni/Fe |
Fe |
Ni |
0.15 volts |
|
Fe/Au |
Fe |
Au |
1.05 volts |
- From the data in the table above, which metal is the strongest reducing agent?
- From the data in the table above, which metal is the weakest reducing agent?
- Using the reduction of lead (Pb) as a reference, construct a half-cell voltage table from the experimental data above.
A18.8
1. The strongest reducing agent is the one who is most likely to initiate reduction by becoming oxidized and losing electrons. Therefore, the strongest reducing agent in the table above is Iron (Fe) because it is always oxidized in every reaction while other metals are being both oxidized and reduced.
2. The weakest reducing agent is the one who rarely initiates reduction because it has a high reduction potential and is rarely oxidized. In the table above, we can see that Gold (Au) us the weakest reducing agent because it never becomes oxidized in any of the reactions listed above.
3. To construct a half-cell voltage table we must define the half-reactions and their corresponding Standard Reduction Potentials (\(E^o_{red}\)).
We use Pb as a reference and therefore it's \(E^o_{red}\) is 0v. After that, we list the rest of the half-reaction of each metal and corresponding electron transfers to make the charges on each side of the reaction be equal.
We can easily find their \(E^o_{red}\) by looking at the table above. The reactions of Ni and Fe in the table are oxidation reactions and, therefore, we must make these values negative so that they are listed as reduction potentials.
Lastly, we list the half reactions in decreasing \(E^o_{red}\) so that the strongest oxidizing agents are at the top and the strongest reducing agents are at the bottom.
| Half-Reactions | \(E^o_{red}\) (volts) |
| \(Au^{3+} + 3e^- \leftrightarrow Au\) | +0.80 v |
| \( Pb^{2+} + 2e^- \leftrightarrow Pb\) | 0.00 v |
| \(Ni^{2+} + 2e^- \leftrightarrow Ni\) | -0.10 v |
| \(Fe^{2+} + 2e^- \leftrightarrow Fe\) | -0.25 v |
Q21.4.5
Azomethane (CH3N2CH3) decomposes at 600 K to C2H6 and N2. The decomposition is first order in azomethane.
a. Calculate t½ from the data in the following table.
b. How long will it take for the decomposition to be 99.9% complete?
|
Time (s) |
PCH3N2CH3(atm) |
| 0 | 8.2 × 10−2 |
| 2000 | 3.99 × 10−2 |
| 4000 | 1.94 × 10−2 |
A21.4.5
a. To find the half life of Azomethane you must first determine the rate constant (k) by plugging in known values from the table to the rate law equation and solving for k.
Rate Law of First Order Reaction: \([A]=[A]_0e^{-kt}\)
\(3.99 x 10^{-2}=(8.2 x 10^{-2})e^{(-k)(2000)}\)
\(0.487 = e^{-2000k}\)
\(\ln(0.487) = -2000k\)
\(-0.72 = -2000k\)
\(k=3.6 x 10^{-4}\)
Know that we know the rate constant, we can plug values into half-life equation and solve for \(t_{1/2}\).
\[t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k} \label{5}\]
\[t_{1/2} = \dfrac{0.693}{3.6 x 10^{-4}}\]
\[t_{1/2} = 1.925 x 10^{3}\]
\[t_{1/2} = 1925s\]
b. To calculate how long it will take for the decomposition to be 99.9% complete we plug "fake values" into the rate law equation and solve for time (t). The fake values are representative of 0.01/100 of the sample being left after t, time, and therefore we say \([A]=0.01\) and \([A_0]=100\). We also know that \(k=3.6 x 10^{-4}\).
\([A]=[A]_0e^{-kt}\)
\([0.01]=[100]_0e^{(-t)(3.6 x 10^{-4})}\)
\(0.0001 = e^{-3.6 x 10^{-4}t}\)
\(\ln(0.0001) = -3.6 x 10^{-4}t\)
\(-9.210 = -3.6 x 10^{-4}t\)
\(t=25584.3s \approx 7.11 hour\)