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Extra Credit 23

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    Edited by Ian Good

    17.3.2 For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

    1. \[ \ce{Mn}(\mathit{ s})+ \ce{Ni^2+} (\mathit{ aq}) \rightarrow \ce{Mn^2+}(\mathit{ aq})+ \ce{Ni} (\mathit{ s})\]
    2. \[ \ce{3Cu^2+}(\mathit{ aq})+ \ce{2Al} (\mathit{ s}) \rightarrow \ce{2Al^3+}(\mathit{ aq})+ \ce{2Cu} (\mathit{ s})\]
    3. \[ \ce{Na}(\mathit{ s})+ \ce{LiNO3} (\mathit{ aq}) \rightarrow \ce{NaNO3}(\mathit{ aq})+ \ce{Li} (\mathit{ s})\]
    4. \[ \ce{Ca(NO3)2}(\mathit{ aq})+ \ce{Ba} (\mathit{ s}) \rightarrow \ce{Ba(NO3)2}(\mathit{ aq})+ \ce{Ca} (\mathit{ s})\]

    Solutions:

    First, let us establish the knowledge required to correctly solve these problems.

    We know that the equation for standard cell potential is as follows: \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    We also know that the oxidation reaction occurs at the anode and the reduction reaction occurs at the cathode.

    Concerning the spontaneity of a reaction, a positive cell potential corresponds to a negative change in the Gibbs free energy which corresponds to a spontaneous reaction. On the other hand, a negative cell potential corresponds to a positive change in the Gibbs free energy and a non-spontaneous reaction.

    1. Looking at the table of standard electrode potentials, the standard electrode potential for \[ \ce{Mn^2+}(\mathit{ aq})+ \text{ 2e}^- \rightarrow \ce{Mn}(\mathit{ s}) \] is -1.185V

    the standard electrode potential for \[ \ce{Ni^2+}(\mathit{ aq})+ \text{ 2e}^- \rightarrow \ce{Ni}(\mathit{ s}) \] is -0.257V.

    Following the formula to determine the standard cell potential \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \] since manganese is oxidized it occurs at the anode and nickel occurs at the cathode since it is reduced, you can plug in the respective standard electrode potentials into the formula to find the standard cell potentials.

    \[ \text{E}^°_\text{cell} = -0.257\text{ V} - (-1.185)\text{ V} \]\[ \text{E}^°_\text{cell} = 0.928\text{ V} \]

    Since the standard cell potential is positive, it is spontaneous at standard conditions.

    Phase II : number one is correct

    2. Looking at the table of standard electrode potentials, the standard electrode potential for \[ \ce{Cu^2+}(\mathit{ aq})+ \text{ 2e}^- \rightarrow \ce{Cu}(\mathit{ s}) \] is 0.34V

    the standard electrode potential for \[ \ce{Al^3+}(\mathit{ aq})+ \text{ 3e}^- \rightarrow \ce{Al}(\mathit{ s}) \] is -1.662V.

    Following the formula to determine the standard cell potential \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \] since aluminum is oxidized it occurs at the anode and copper occurs at the cathode since it is reduced, you can plug in the respective standard electrode potentials into the formula to find the standard cell potentials.

    \[ \text{E}^°_\text{cell} = 0.34\text{ V} - (-1.662)\text{ V} \]\[ \text{E}^°_\text{cell} = 2.002\text{ V} \]

    Since the standard cell potential is positive, it is spontaneous at standard conditions.

    Phase II: number 2 is correct.

    3. Looking at the table of standard electrode potentials, the standard electrode potential for \[ \ce{Na^+}(\mathit{ aq})+ \text{ e}^- \rightarrow \ce{Na}(\mathit{ s}) \] is -2.71V

    the standard electrode potential for \[ \ce{Li^+}(\mathit{ aq})+ \text{ e}^- \rightarrow \ce{Li}(\mathit{ s}) \] is -3.04V.

    Following the formula to determine the standard cell potential \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \] since sodium is oxidized it occurs at the anode and since lithium is reduced it occurs at the cathode, you can plug in the respective standard electrode potentials into the formula to find the standard cell potentials.

    \[ \text{E}^°_\text{cell} = -3.04\text{ V} - (-2.71)\text{ V} \]\[ \text{E}^°_\text{cell} = -0.33\text{ V} \]

    Since the standard cell potential is negative, it is not spontaneous at standard conditions.

    Phase II: number 3 is correct.

    4. Looking at the table of standard electrode potentials, the standard electrode potential for \[ \ce{Ca^2+}(\mathit{ aq})+ \text{ 2e}^- \rightarrow \ce{Ca}(\mathit{ s}) \] is -2.868V

    the standard electrode potential for \[ \ce{Ba^2+}(\mathit{ aq})+ \text{ 2e}^- \rightarrow \ce{Ba}(\mathit{ s}) \] is -2.912V.

    Following the formula to determine the standard cell potential \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \] since barium is oxidized it occurs at the anode and since calcium is reduced it occurs at the cathode, you can plug in the respective standard electrode potentials into the formula to find the standard cell potentials.

    \[ \text{E}^°_\text{cell} = -2.868\text{ V} - (-2.912)\text{ V} \]\[ \text{E}^°_\text{cell} = 0.044\text{ V} \]

    Since the standard cell potential is positive, it is spontaneous at standard conditions.

    Phase II: number 4 is correct

    19.1.21 Predict the products of each of the following reactions and then balance the chemical equations.

    1. Fe is heated in an atmosphere of steam.
    2. NaOH is added to a solution of Fe(NO3)3.
    3. FeSO4 is added to an acidic solution of KMnO4.
    4. Fe is added to a dilute solution of H2SO4.
    5. A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
    6. FeCO3 is added to a solution of HClO4.
    7. Fe is heated in air.

    Solutions:

    First, we recognize that Fe, Iron, is a transition metal the typically exists at a cation. To correctly predict the products of these reactions, one must be sure to note the phases of the reactants if possible. One should also determine the oxidation numbers of the atoms in molecules correctly.

    1. \[ \ce{3Fe}(\mathit{ s})+ \ce{4H2O} (\mathit{ g}) \rightarrow \ce{Fe3O4}(\mathit{ s})+ \ce{4H2} (\mathit{ g})\]

    Phase II: the equation is correct; this equation is single replacement,

    $${3Fe}(\mathit{ s})+ \ce{4H2O} (\mathit{stream}) \rightarrow \ce{Fe3O4}(\mathit{ s})+ \ce{4H2} (\mathit{ g})$$

    2. \[ \ce{Fe(NO3)3}(\mathit{ aq})+ \ce{3NaOH} (\mathit{ aq}) \rightarrow \ce{Fe(OH)3}(\mathit{ s})+ \ce{3NaNO3} (\mathit{ aq})\]

    Phase II: the equation is correct; this equation is double replacement,

    3. \[ \ce{8H2SO4}(\mathit{ aq})+ \ce{10FeSO4} (\mathit{ aq})+ \ce{2KMnO4} (\mathit{ aq}) \rightarrow \ce{2MnSO4}(\mathit{ aq})+ \ce{5Fe2(SO4)3} (\mathit{ aq})+ \ce{K2SO4} (\mathit{ aq})+ \ce{4H2O} (\mathit{ l})\]

    Phase II: the equation is incorrect; there should be 8 water at the product side .The equation should be

    \[ \ce{8H2SO4}(\mathit{ aq})+ \ce{10FeSO4} (\mathit{ aq})+ \ce{2KMnO4} (\mathit{ aq}) \rightarrow \ce{2MnSO4}(\mathit{ aq})+ \ce{5Fe2(SO4)3} (\mathit{ aq})+ \ce{K2SO4} (\mathit{ aq})+ \ce{8H2O} (\mathit{ l})\]

    4. \[ \ce{Fe}(\mathit{ s})+ \ce{H2SO4} (\mathit{ aq}) \rightarrow \ce{FeSO4}(\mathit{ aq})+ \ce{H2} (\mathit{ g})\]

    Phase II: the equation is correct

    5. \[ \ce{2HNO3}(\mathit{ aq})+ \ce{2Fe(NO3)2} (\mathit{ aq}) \rightarrow \ce{2Fe(NO3)3}(\mathit{ aq})+ \ce{H2} (\mathit{ g})\]

    Phase II: the equation is incorrect

    $$3Fe(NO3)_2 (aq)+ 4HNO3(aq) \rightarrow 3Fe(NO3)3 (aq)+ NO(g) + 2H2O(l)$$

    6. \[ \ce{FeCO3}(\mathit{ aq})+ \ce{2HClO4} (\mathit{ aq}) \rightarrow \ce{2Fe(ClO4)2}(\mathit{ aq})+ \ce{H2CO3} (\mathit{ aq})\]

    PhaseII: the equation is incorrect. For the reactant, the FeCO3 is insiluble; for the product, the production should include carbondioxide and water.

    $$FeCO3(s) + 2 HClO_4(aq) \rightarrow Fe(ClO_4)_2(aq) + CO_2(g) + H_2O(l)$$

    7. \[ \ce{4Fe}(\mathit{ s})+ \ce{3O2} (\mathit{ g}) \rightarrow \ce{2Fe2O3}(\mathit{ s})\]

    Phase II: the equation is correct

    19.3.13 [CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?

    Solution: [Cu(H2O)6]2+ absorbs the higher energy photons because since the complex is blue, it absorbs orange light and orange light has a shorter wavelength than red which is what [CuCl4]2− is absorbing. Shorter wavelength corresponds to higher energy. Because [Cu(H2O)6]2+ absorbs a higher energy photon, we know that it has a larger crystal field splitting energy than [CuCl4]2−. Because we know the crystal field splitting of [CuCl4]2− is smaller than the crystal field splitting of [Cu(H2O)6]2+ we know that [CuCl4]2− is tetrahedral, not square planar, because the crystal field splitting of tetrahedral complexes are smaller than that of octahedral complexes, but the crystal field splitting of square planar complexes are larger than the crystal field splitting energy of octahedral complexes. [CuCl4]2− is predicted to have the larger (THIS IS INCORRECT! It should read: "SMALLER") crystal field splitting because [Cu(H2O)6]2+ is octahedral and the relationship between the crystal field splitting of tetrahedral and octahedral complexes is Δt= 0.44Δo. Therefore, the crystal field splitting of an octahedral complex is 0.4 times greater than the crystal field splitting energy of a tetrahedral complex. THIS IS INCORRECT! It should read: "The crystal field splitting energy of a tetrahedral complex is only 44% of the total crystal field splitting energy of octahedral complexes".

    Phase II: the conclusion is correct. In the later explanation part, [CuCl4]2- should have smaller crystal filed splitting energy; and the crystal filed splitting energy of a tetrahedral should be less the that crystal filed splitting energy of an octahedral

    12.4.13 Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?

    Solution: To figure out this problem we have to use the equation \[ \text{ Fraction remaining}= 0.5^{t\over t_{1/2}}\] where t is the time elapsed and t1/2 is the half life of technetium and thallium. Since the half lives are in units of hours we have to plug in 48 hours into the equation, not two days, in order to keep the units constant. After plugging in all the values we know, we can multiply by 100% to get the percent of radioactivity that would remain of each isotope after 2 days.

    For technetium-99,

    \[ \text{ Fraction remaining}= 0.5^{t\over t_{1/2}}\] where t is the time elapsed and t1/2 is the half life of technetium-99

    \[ \text{ Fraction remaining}= 0.5^{48\over 6}\]

    \[ \text{ Fraction remaining}= 0.5^{8}=0.00390625\]

    \[ \text{ Fraction remaining}=0.00390625\times100 \%\]

    \[ \text{ Fraction remaining}=0.390625\%\text{remaining}\]

    For thallium-201,

    \[ \text{ Fraction remaining}= 0.5^{t\over t_{1/2}}\] where t is the time elapsed and t1/2 is the half life of thallium-201

    \[ \text{ Fraction remaining}= 0.5^{48\over 73}\]

    \[ \text{ Fraction remaining}= 0.5^{0.65753}=0.6339608932\]

    \[ \text{ Fraction remaining}=0.6339608932\times100 \%\]

    \[ \text{ Fraction remaining}=63.396\%\text{remaining}\]

    Therefore, after two days, or 48 hours, there is .390625% of the technetium-99 remaining and 63.369% of the thallium-201 remaining.

    PhaseII, the equation and calculation is correct.

    21.2.8 The mass of the atom 23Na is 22.9898 amu.

    a.Calculate its binding energy per atom in millions of electron volts.

    b.Calculate its binding energy per nucleon.

    Solution:

    a. To calculate the binding energy per atom in millions of electron volts, use the equation E=mc2.

    From 23Na, we know that from the periodic table sodium has 11 protons therefore 23-11=12 neutrons.

    Then multiply 11 protons by the mass of each individual proton and do the same with the 12 neutrons.

    \[ \text{ Mass of neutrons}= 12\times1.008665= 12.10398\text{amu} \]

    \[ \text{ Mass of protons}= 11\times1.007825= 11.086075\text{amu} \]

    Then add the two masses of protons and neutrons together to get the mass of the free nucleons.

    \[ \text{ Mass of free nucleons}= 12.10398+11.08677= 23.190055 amu \]

    Subtract the mass of the nucleus which is given in the problem, 22.9898 amu, from the mass of the free nucleons to get Δm. Δm is the mass defect.

    \[ \text{ Δm}= 23.190055 amu -22.9898= 0.200amu \]

    Then convert amu to kg.

    \[ 0.200\text{ amu}\times\frac{1.661\times10^{-27}\text { kg}}{1\text{ amu}} = 3.325x10^{-28}\text{ kg} \]

    Then plug in m into E=mc2 to find E, where c= 3.00x108m/s

    \[ \text{ E}= 3.325x10^{-28}\times{(3.00\times10^8)}^2= 2.99\times10^{-11} \] (J/atom)

    Because the answer above is in J and we want our answer in meV, we have to do the conversion to get to millions of electron volts

    \[ 2.99\times10^{-11}\text{ J}\times\frac{1\text{ meV}}{1.602\times10^{-13}\text { J}}= 186.64\text{ meV} \]

    PhaseII: part(a) is correct, there are some units missing.

    b. To calculate the binding energy per nucleon, take the calculated answer from part a, 186.64 meV and divide it by the number of nucleons which is 23 in this case (12 neutrons+ 11 protons)

    \[186.64\text{ meV}/23=8.11\text{ meV}\text{ per}\text{ nucleon} \]

    PhaseII: for part(b), it's better to express the answer in J/necleon. Therefore, we can take $$E=2.99\times10^{-11} \] J/atom$$ and divided by the number of nucleon which is 23. The energy should be 1.299E-12 J/necleon.

    21.6.3 Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β− emission.

    Write an equation for the decay.

    1. How long will it take for 95.0% of a dose of I-131 to decay?

    Solution:

    1. \[ ^{113}_{53}\text{I}\rightarrow^{0}_{-1}\ce{ß^-}+^{113}_{54}\text{Xe} \]

    2. 37.6 days.

    To find how long it would take for 95% to decay, we have to find how much would remain which is 1-0.95= 0.05. Then we have to find how long it would take in terms of half life for 0.05 or 5% to remain and multiply it by the length of a half life which in this case is 8.70 days.

    0.95= amount decayed, 0.05= amount remaining

    h is the number of half life periods \[(1/2)^h=0.05 \]

    \[h=ln(0.05)/ln(1/2)=4.321928 \]

    \[ 4.321928\times8.70\text{days}= 37.6\text{days} \]

    Therefore, it will take 37.6 days for 95% of I-131 to decay.

    Phase II: there is an alternative way of solving this problem. Because of the nuclear decay equation, this reaction is first order. In order to find out the decay time, we should use the equation

    t1/2=0.693/k

    k=0.0796 s-1

    $$ln[A]_0=-kt ln[a]_0$$

    $$ln[0.05]_0=-(0.0796)t+ln[1]_0$$

    t=37.6 days

    20.4.9 All reference electrodes must conform to certain requirements. List the requirements and explain their significance.

    Solution:

    Ultimately, reference electrodes are used as set standards for other electrodes to be compared to, therefore there are specific requirements all reference electrodes must meet.

    1. Reference electrodes should be made using half-cell components that are stable over a long period of time and stable with varying temperature. This is an important requirement because all other potentials are being compared and measured relative to the reference electrode. If the reference electrode is not stable over time or with changing temperature, it will be hard to achieve accurate measurements of potentials in experiments that one is trying to conduct, especially with experiments that would last over a long period of time. Often, buffers will be used to establish the stability of reference electrodes.

    2. Reference electrodes should also have electrode potentials that are fixed, not varying when measured repeatedly. This is significant because to be a reference electrode, it must be constant or measurements of applied potentials will vary for each trial. If the potential of the reference is not fixed, then it is not much of a reference at all and can't be relied on or compared to. Again, this illustrates the fact that reference electrodes must be stable.

    Phase II, the explanation is good.

    20.8.1 Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?

    Solution:

    You would expect a bent nail to corrode more rapidly than a straight nail because corrosion tends to begin at an area in which the metal is under stress, for example where the nail is bent. Also, when you bend a nail, the sacrificial plating on the outside of the nail may be cracked when bent, therefore speeding up the oxidation of the metal and the corrosion process.

    It is also possible that bending the nail would increase the surface area of the nail creating more surface area on which oxidation can take place.

    Phase II : the answer is correct. There is a point when nail is bent and for the reaction of Iron with Oxygen happens which becomes anode; the equation will be Fe→Fe2++2e-; the reduction will be O2+2H2O+4e-→4OH

    Extra Credit 23 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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