Skip to main content
Chemistry LibreTexts

Extra Credit 13

  • Page ID
    82769
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Question 17.2.2

    Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions.

    1. Mg(s)│Mg2+(aq)║Cu2+(aq)│Cu(s)
    2. Ni(s)│Ni2+(aq)║Ag+(aq)│Ag(s)

    Answer 17.2.2

    Electrochemical reactions can be represented by electrochemical cell notation. Cell notation are shorthand description of voltaic or galvanic (spontaneous) cells. The reaction conditions such as pressure, temperature, concentration, etc., the anode, cathode and the electrode components are all included in this shorthand notation. It is also form this shorthand notation we can calculate standard cell potentials for both oxidation and reduction reactions.

    We are asked to identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent. The anode half-cell comes first in the cell notation (the species involved in the oxidation half-reaction, to the left of the double vertical lines), followed by the cathode (species involved in the reduction half-reaction, to the right of the double vertical lines). With the given half reactions, the reactants are described first, followed by the products. So as one reads the cell notation, their eyes move in the direction of electron flow. Its also important to note that spectator ions are not included in the cell notation. The double vertical line represents the salt bridge connecting the two different solutions. Each single vertical line represents a phase boundary, separating species of different phases. The phase of each chemical species is shown in paranthesis, and if the electrolytes are not in standard conditions then concentrations/pressure are included in parantheses along with the phase notation. If no concentration/pressure is noted then the reaction is assumed to be taking place under standard conditions (1.00 M/ 1.00 atm and 298K). The oxidizing agent is the species that accepts electrons. Thus, the oxidizing agent is also the species that is reduced. The reducing agent is the specie that donates electrons. Thus, it is also the specie that is oxidized.

    1. From the cell notation, we can tell that Mg(s) and Mg2+(aq) are involved in the oxidation half-reaction since they are written to the left of the double vertical lines meaning

    \[ \text{ Mg}(\mathit{ s}) \rightarrow \text{ Mg}^\text{ 2+}(\mathit{ aq})+\text{ 2 e}^- \]

    and Cu(s) and Cu2+(aq) are involved in the reduction half-reaction since they are written to the right of the double vertical lines meaning

    \[ \text{ Cu}^\text{ 2+}(\mathit{ aq})+\text{ 2 e}^- \rightarrow \text{ Cu}(\mathit{ s}) \]

    Mg2+(aq) is the specie that donates electrons so it is the reducing agent and the specie that is oxidized. Cu2+(aq) is the specie that accepts electrons so it is the oxidizing agent and the species that is reduced.

    1. From the cell notation, we can tell that Ni(s) and Ni2+(aq) are involved in the oxidation half-reaction since they are written to the left of the double vertical lines meaning

    \[ \text{ Ni}(\mathit{ s}) \rightarrow \text{ Ni}^\text{ 2+}(\mathit{ aq})+\text{ 2 e}^- \]

    and Ag(s) and Ag+(aq) are involved in the reduction half-reaction since they are written to the right of the double vertical lines meaning

    \[ \text{ Ag}^\text{ +}(\mathit{ aq})+\text{ e}^- \rightarrow \text{ Ag}(\mathit{ s}) \]

    Ni2+(aq) is the specie that donates electrons so it is the reducing agent and the specie that is oxidized. Ag+(aq) is the specie that accepts electrons so it is the oxidizing agent and the species that is reduced.

    Question 19.1.11

    Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?

    Answer 19.1.11

    To answer this question, we must first identify the net ionic equation from the given half-reactions. We can write the oxidation and reduction half-reactions:

    \[ \text{ oxidation:} \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+} \]

    \[ \text{ reduction: }\ce{Cr2O7^2-} \rightarrow \text{ Cr}^\text{ 3+} \]

    We can quickly balance the oxidation half-reaction by adding the appropriate number of electrons to get

    \[ \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}+\text{ e}^- \]

    The first step in balancing the reduction half-reaction is to balance elements in the equation other than O and H. In doing so, we get

    \[\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+} \]

    The second step would be to add enough water molecules to balance the oxygen.

    \[ \ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \]

    Next, we add the correct amount of H+ to balance the hydrogen atoms.

    \[ \ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \]

    Finally, we add enough electrons to balance charge.

    \[ \ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \]

    The electrons involved in both half-reactions must be equal in order for us to combine the two to get the net ionic equation. This can be done by multiplying each equation by the appropriate coefficient. Scaling the oxidation half-reaction by 6, we get

    \[ 6 \text{ Fe}^\text{ 2+} \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^- \]

    Now we can combine both half-reactions to get

    \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\]

    The electrons cancel out, so you get:

    \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 6 \text{ Fe}^\text{ 3+}+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\]

    From this we can see that the mole ratio of Cr2O72- to Fe2+ is 1:6. Given that 19.17 mL (or 0.01917 L) of 0.01 M Na2Cr2O7 was needed for titration we know that

    \[ 0.01917\text{ L} \times 0.01 \text{ M} = 1.917 \times 10^{-4} \text{ mol} \]

    of Na2Cr2O7 reacted. Also, since any number of moles of Na2Cr2O7 produces the same number of moles of Cr2O72- in solution

    \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Na2Cr2O7} =1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\]

    We can use the mole ratio of Cr2O72- to Fe2+ to determine how many moles of iron (ii) was in the solution. The number of moles of iron (ii) is the same as the number of moles of pure iron in the sample since all of the iron was converted into iron (ii).

    \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\times \frac{6\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}}{1\text{ mol}\text{ of }\ce{Cr2O7^2-}} = 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} \]

    \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} = 0.0011502\text{ mol}\text{ of } \text{ Fe} \]

    Now we can find the number of grams of iron that were present in the 2.5 g iron ore sample.

    \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}\times\frac{55.847\text { g}}{1\text{ mol}} = 0.0642352194\text{ g}\text{ of }\text{ Fe} \]

    Finally, we can answer the question and find what percentage of the ore sample was iron.

    \[ \frac{0.0642352194\text{ g}}{2.5\text{ g}} \times 100 \approx 2.57\text{%} \]

    So 2.57% of the ore sample was iron.

    Question 19.3.3

    Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3.

    Answer 19.3.3

    The oxidation state of the metal can be found by identifying the charge of one of each molecule in the coordinate compound, multiplying each molecule's charge by the respective number of molecules present, and adding the products. This final sum represents the charge of the overall coordination compound. You can then solve for the oxidation state of the metal algebraically. In this case, one chloride anion Cl- has a charge of -1. So three chloride anions have a total charge of -3. One ammine ligand NH3 has no charge so six ammine ligands have a total charge of zero. Finally, we are trying to solve for the oxidation state of a cobalt ion. Now we can write the equation that adds the total charges of each molecule or ion and is equal to the total charge of the overall coordinate compound.

    \[ (\text{oxidation}\text{ state}\text{ of}\text{ Co})+(-3)+0=0 \]

    \[ \text{oxidation}\text{ state}\text{ of}\text{ Co}=+3 \]

    So the oxidation state of Co is +3. Now we need to identify the number of d-electrons in the Co3+ ion. The electron configuration for cobalt that has no charge is

    \[ [\text{Ar}]4\text{s}^23\text{d}^7 \]

    However, a Co3+ ion has 3 less electrons than its neutral counterpart and has an electron configuration of

    \[ [\text{Ar}]3\text{d}^6 \]

    For transition metals, the \( \text{s} \) electrons are lost first. So cobalt loses its two \( 4\text{s} \) electrons first and then loses a single \( 3\text{d} \) electron meaning Co3+ ion has 6 \( \text{d} \) electrons. To predict the number of unpaired electrons, we must first determine if the complex is high spin or low spin. Whether the complex is high spin or low spin is determined by the ligand in the coordinate complex. Specifically, the ligand must be identified as either a weak-field ligand or a strong-field ligand based on the spectrochemical series. Weak-field ligands induce high spin while strong-field ligands induce low spin. We can then construct the energy diagram or crystal field diagram of the designated spin that has the proper electron placings. The geometric shape of the compound must also be identified to construct the correct diagram. Finally, from this crystal field diagram we can determine the number of unpaired electrons. The number of electrons in the diagram is equal to the number of \( \text{d} \) electrons of the metal. The ligand in this case is NH3, which is a strong field ligand according to the spectrochemical series. This means that the complex is low spin. Additionally, six monodentate ligands means the ligand field is octahedral. The number of electrons that will be in the diagram is 6 since the metal ion Co3+ has 6 \( \text{d} \) electrons. Now the proper crystal field diagram can be constructed.

    fb01ee6b509c632695d226337095e4df (2).jpg

    From the crystal field diagram, we can tell that the complex has no unpaired electrons.

    Question 12.4.3

    Use the data provided in a graphical method to determine the order and rate constant of the following reaction: \[ 2\mathit{P}\rightarrow\mathit{Q}+\mathit{W} \]

    Time (s) 9.0 13.0 18.0 22.0 25.0
    [P] (M) 1.077 × 10−3 1.068 × 10−3 1.055 × 10−3 1.046 × 10−3 1.039 × 10−3

    Answer 12.4.3

    Zero order reactions are when the rate of a reaction is independent of the reactant concentrations, hence the rate of the reaction is equal to the constant k. First order reactions are reactions that proceed at a rate that linearly depends on one of the concentrations of the reactants. Differential rate laws describes what is occurring at the molecular level during a reaction whereas the integrated rate law determines reaction order and rate constants from experimental measurements ( which we have in the table above).

    Initial rates can be measured in a series of experiments at different initial concentrations of reactant A with other concentrations kept constant. However this is not always a reliable method because the measurement of the initial requires accurate determination of small changes in concentration, plus if the rate depends on substances not present at the beginning of the reaction, like intermediates or products, then the rate equation cannot completely be determined. This is why it is determined with the integral method over a long period of time.

    The order of a reaction refers to the sum of the concentrations in a reaction. The rate of a reaction depends on the order of a reaction. This is the sum of the exponents in a rate law equation. This is also an experimental quantity ( which once again is why we use the table above).

    Given data of concentration versus time, plots of [A] vs. time, natural logarithm (ln) of [A] vs. time, and 1/[A] vs. time can be constructed and used to determine the order of the reaction by identifying which graph yields a straight line. Each graph is representative of the integrated rate law of either zero-order reactions, first-order reactions, or second-order reactions, respectively. The rate constant k is represented by the slope of the graph. So the data that best yields this slope (in other words, most closely resembles a straight line) fits into the integrated rate law for that certain order reaction. Writing the appropriate rate law for the reaction and substituting in known values allows us to calculate the rate constant. The graphs with their respective data values would be

    Time (s) 9.0 13.0 18.0 22.0 25.0
    [P] (M) 1.077 × 10−3 1.068 × 10−3 1.055 × 10−3 1.046 × 10−3 1.039 × 10−3

    Screenshot (6).png

    Time (s) 9.0 13.0 18.0 22.0 25.0
    ln [P] (M) -6.83358 -6.84197 -6.85421 -6.86278 -6.8695

    Screenshot (9).png

    Time (s) 9.0 13.0 18.0 22.0 25.0
    1/[P] (M) 928.5051 936.3296 947.8673 956.0229 962.4639

    Screenshot (10).png

    We can see that all three graphs yield a straight line, so we cannot determine the order and rate constant of the reaction. More data is necessary to be able to identify the order and rate constant.

    Question 12.7.6

    For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:

    (a)

    CNX_Chem_12_07_Exercise6a_img.jpg

    (b)

    CNX_Chem_12_07_Exercise6b_img.jpg

    Answer 12.7.6

    In a reaction diagram, the activation energy of the reaction Ea is the difference between the energy of the transition state and the energy of the reactants. In simpler terms, it is the difference between the high point of the graph and the low point on the left side of the graph.

    1. The energy of the transition state is about 35 kJ while the energy of the reactants is about 10 kJ. The difference between the two would be

    \[ 35\text{ kJ}-10\text{ kJ}=25\text{ kJ} \]

    So the activation energy Ea of the reaction is estimated to be 25 kJ.

    1. The energy of the transition state is about 20 kJ while the energy of the reactants is about 10 kJ. The difference between the two would be

    \[ 20\text{ kJ}-10\text{ kJ}=10\text{ kJ} \]

    So the activation energy Ea of the reaction is estimated to be 10 kJ.

    Question 21.5.1

    Write the balanced nuclear equation for the production of the following transuranium elements:

    1. berkelium-244, made by the reaction of Am-241 and He-4
    2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons
    3. lawrencium-257, made by the reaction of Cf-250 and B-11
    4. dubnium-260, made by the reaction of Cf-249 and N-15

    Answer 21.5.1

    When writing balanced nuclear equations, the total mass number and total atomic number of the reactants must be equal to those of the products.

    1. From the given information we can write the nuclear equation

    \[ ^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk} \]

    On the reactants side we see that the total mass number is

    \[ 241 + 4 = 245 \]

    and the total atomic number is

    \[ 95 + 2 = 97 \]

    On the products side we see that the total mass number is 244 and the total atomic number is 97. This shows that the total mass number is unbalanced and the total atomic number is balanced. Simple math shows that the total mass number on the products side needs to increase by one while the total atomic number stays the same. This can be done by adding one neutron. Thus, the balanced nuclear equation is

    \[ ^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk}+^1_0\text{n} \]

    1. From the given information we can write the nuclear equation

    \[ ^{239}_{94}\text{Pu}+\text{x }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm} \]

    On the reactants side we see that the total mass number is the sum of \( 239+(1)\text{x}\). on the products side we see that the total mass number is 254. since the total mass number of the reactants must equal that of the products we can write

    \[ 239+\text{x}=254 \]

    and solving for x gives the total number of neutrons needed to balance the total mass number. 15 neutrons are needed so now we can adjust the equation

    \[ ^{239}_{94}\text{Pu}+15\text{ }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm} \]

    Now that the total mass number of the equation is balanced, we can move onto balancing the total atomic number of the equation. On the reactants side we see that the total atomic number is 94 while the total atomic number on the products side is 100. This means that the total atomic number on the products side needs to be reduced by 6 which can be done by adding 6 electrons to the products side. While electrons do not actually have an atomic number they do have a charge. This charge can be treated as an atomic number in nuclear equations since atomic number of other particles or elements are the number of charged protons the particle or element has. Therefore, the balanced nuclear equation is

    \[ ^{239}_{94}\text{Pu}+15\text{ }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm}+\ce{6 ^0_{-1}e} \]

    1. From the given information we can write the nuclear equation

    \[ ^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr} \]

    On the reactants side the total mass number is

    \[250+11=261 \]

    and the total atomic number is

    \[98+5=103 \]

    On the products side the total mass number is 257 and the total atomic number is 103. This means that the total mass number on the products side needs to be increased by 4 while the total atomic number stays the same. This can be done by adding 4 neturons to the products side. The balanced nuclear equation is

    \[ ^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr}+4\text{ }^1_0\text{n} \]

    1. From the given information we can write the nuclear equation

    \[ ^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Db} \]

    On the reactants side the total mass number is

    \[249+15=264 \]

    and the total atomic number is

    \[ 98+7=105 \]

    On the products side the total mass number is 260 and the total atomic number is 105. This means that the total mass number on the products side needs to be increased by 4 while the total atomic number stays the same. This can be done by adding 4 neturons to the products side. The balanced nuclear equation is

    \[ ^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Lr}+4\text{ }^1_0\text{n} \]

    Question 20.3.15

    Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.

    1. 2Cl(aq) + Br2(l) → Cl2(g) + 2Br(aq)
    2. N2O4(g) + H2O → HNO2(aq) + H+(aq) + NO3(aq)
    3. 2H2O(l) + 2Cl(aq) → H2(g) + Cl2(g) + 2OH(aq)
    4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

    Answer 20.3.15

    1. From the overall reaction given we can write the incomplete oxidation and reduction half-reactions:

    \[ \text{ oxidation: } 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) \]

    \[ \text{ reduction: }\text{Br}_2 (\mathit{l}) \rightarrow 2\text{ Br}^- (\mathit{aq}) \]

    Both equations can be quickly balanced by adding the correct amount of electrons to the appropriate side of the half-reaction.

    \[ \text{ oxidation: } 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g})+2\text{ e}^- \]

    \[ \text{ reduction: }\text{Br}_2 (\mathit{l}) +2\text{ e}^-\rightarrow 2\text{ Br}^- (\mathit{aq}) \]

    The spontaneity of the overall reaction can be determined by calculating the standard cell potential using the equation

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    and the standard reduction potentials for the half-reactions involved. If \(\text{E}^°_\text{cell} \) is positive then the reaction is spontaneous. If \(\text{E}^°_\text{cell} \) is negative then the reaction is not spontaneous. The reaction accurring at the anode and the reaction occurring at the cathode must be determined to properly write the equation. The oxidation half-reaction normally occurs at the anode and the reduction half-reaction normally occurs at the cathode. With this information and the table of standard reduction potentials the reactions occurring at the cathode and anode with their respective standard reduction potentials is:

    • anode: \[ 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g})+2\text{ e}^- \;\;\; \text{E}^° = 1.36\text{ V} \]
    • cathode: \[ \text{Br}_2 (\mathit{l}) +2\text{ e}^-\rightarrow 2\text{ Br}^- (\mathit{aq}) \;\;\; \text{E}^° = 1.07\text{ V}\]

    Now we can find the standard cell potential.

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    \[ \text{E}^°_\text{cell} = 1.07\text{ V} - 1.36\text{ V} \]

    \[ \text{E}^°_\text{cell} = -0.29\text{ V} \]

    Since \(\text{E}^°_\text{cell} \) is negative the reaction is not spontaneous in the forward direction. However, the reaction is spontaneous in the reverse direction. When writing cell diagrams, the species involved in the reaction occurring at the anode are written in the left half of the diagram while the species involved in the reaction occurring at the cathode are written in the right half of the diagram. Species of different phases are separated by a single vertical line. The left half or right half of the cell diagram, which includes the anode, cathode, and respective half-reactions, are called half-cells. Each half-cell requires a conducting metal that allows electrons to flow from one cell to the other. If no conducting metal is already present in the half-reaction, then an inert electrode can be added to the side of the cell diagram that needs it. Pt (s) and C (s) can act as inert electrodes. Sometimes a "salt bridge" is needed to allow charged ions to flow from one half-cell to the other and thus prevent the build-up of charge in one half-cell. This keeps the overall cell electrically active. A salt bridge is represented by a double vertical line separating the two half-cells. Electrodes are written at the ends of the cell diagram and the species in solution connected by the salt bridge are written before and after the double vertical line. Now we can write the cell diagram for a glavanic cell in which this reaction is spontaneous.

    \[ \text{Pt }(\mathit{s}) | \text{Br}_2(\mathit{ l}) | \text{Br}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq}) | \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s}) \]

    1. From the overall reaction given we can write the incomplete oxidation and reduction half-reactions:

    \[ \text{ oxidation: }\ce{N2O4} (\mathit{g}) \rightarrow \text{ NO}_3^- (\mathit{aq}) \]

    \[ \text{ reduction: } \ce{N2O4} (\mathit{g}) \rightarrow \text{HNO}_2 (\mathit{aq}) \]

    We will start with the oxidation half-reaction and balance the nitrogen atoms by scaling the appropriate side.

    \[ \ce{N2O4} (\mathit{g}) \rightarrow 2\text{ NO}_3^- (\mathit{aq}) \]

    Next, balance the number of oxygen atoms by adding enough H2O molecules to the correct side.

    \[ \ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) \]

    Now we balance the number of hydrogen molecules by adding enough H+.

    \[ \ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+\]

    Finally, add electrons to balance charge.

    \[ \ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+ + 2\text{ e}^- \]

    Now move onto the reduction half-reaction. First, we balance elements in the equation other than O and H. In this case, balance nitrogen atoms.

    \[ \ce{N2O4} (\mathit{g}) \rightarrow 2\text{ HNO}_2 (\mathit{aq}) \]

    Oxygen is already balanced so move onto balancing hydrogen.

    \[ \ce{N2O4} (\mathit{g}) + 2\text{ H}^+\rightarrow 2\text{ HNO}_2 (\mathit{aq}) \]

    Finally, add electrons to balance charge and obtain the balanced reduction half-reaction.

    \[ \ce{N2O4} (\mathit{g}) + 2\text{ H}^+ + 2\text{ e}^-\rightarrow 2\text{ HNO}_2 (\mathit{aq}) \]

    The following diagram can be used to find the standard reduction potentials of the half-reactions in question. Notice in our oxidation half-reaction, N2O4 goes to NO3-. From the diagram we see that the potential for this half-reaction would be -0.79 V since the reverse direction would mean a negative value rather than a positive value. However, this would be the standard oxidation potential which is not ideal for use in the equation

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    So to get the standard reduction potential for the oxidation half-reaction, we need to go from NO3- to N2O4 instead of going from N2O4 to NO3-. So the standard reduction potential for our oxidation half-reaction would be 0.79 V. The reactions occurring at the cathode and anode with their respective standard reduction potentials is:

    Nitrogen_electrode_potentials.svg.png

    • anode: \[ \ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+ + 2\text{ e}^- \;\;\; \text{E}^° = 0.79\text{ V} \]
    • cathode: \[ \ce{N2O4} (\mathit{g}) + 2\text{ H}^+ + 2\text{ e}^-\rightarrow 2\text{ HNO}_2 (\mathit{aq}) \;\;\; \text{E}^° = 1.07\text{ V} \]

    Now we can calculate the standard cell potential of the overall reaction and determine spontaneity.

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    \[ \text{E}^°_\text{cell} = 1.07\text{ V} - 0.79\text{ V} \]

    \[ \text{E}^°_\text{cell} = 0.28\text{ V} \]

    Since \(\text{E}^°_\text{cell} \) is positive the reaction is spontaneous in the forward direction. Now we go on to write the cell diagram for a galvanic cell of this reaction. The overall reaction does not have any conducting metals so both half-cells need either a Pt (s) or C (s) electrode to transfer electrons from one half-cell to the other. A salt bridge is also needed to allow charged ions to flow between the solutions of the half-cells and complete the circuit. Remember to put the electrodes at the ends of the cell diagram and write species in solution connected by a salt bridge before and after the double vertical line.

    \[ \text{Pt }(\mathit{s}) | \ce{N2O4}(\mathit{g}) | \text{ NO}_3^- (\mathit{aq}) || \text{ HNO}_2 (\mathit{aq}) | \ce{N2O4}(\mathit{g}) | \text{Pt} (\mathit{s}) \]

    1. From the overall reaction given we can write the incomplete oxidation and reduction half-reactions:

    \[ \text{ oxidation: }2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) \]

    \[ \text{ reduction: } \ce{2H2O} (\mathit{ l}) \rightarrow \ce{H2} (\mathit{g}) \]

    The oxidation half-reaction can be quickly balanced by adding electrons to balance charge.

    \[ 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) + 2\text{ e}^-\]

    To balance the reduction half-reaction, we start by adding water molecules to balance oxygen atoms.

    \[ \ce{2H2O} (\mathit{ l}) \rightarrow \ce{H2} (\mathit{g}) + \ce{2H2O} \]

    Now balance hydrogen atoms by adding H+.

    \[ \ce{2H2O} (\mathit{ l}) + 2\text{ H}^+\rightarrow \ce{H2} (\mathit{g}) + \ce{2H2O} \]

    We can see that the overall reaction contains OH- molecules so we need to perform the step that is unique to balancing redox reactions in basic solution. Adding the same number of OH- molecules to both sides as H+ molecules were added we get:

    \[ \ce{2H2O} (\mathit{ l}) + 2\text{ H}^+ + 2\text{ OH}^-\rightarrow \ce{H2} (\mathit{g}) + \ce{2H2O} + 2\text{ OH}^- \]

    OH- and H+ molecules combine to form H2O molecules.

    \[ \ce{2H2O} (\mathit{ l}) + \ce{2H2O}\rightarrow \ce{H2} (\mathit{g}) + \ce{2H2O} + 2\text{ OH}^- \]

    Cancel out like terms that appear on both sides.

    \[ \ce{2H2O} (\mathit{ l}) \rightarrow \ce{H2} (\mathit{g}) + 2\text{ OH}^- \]

    Finally, add electrons to balance charge to obtain the balanced reduction half-reaction.

    \[ \ce{2H2O} (\mathit{ l}) + 2\text{ e}^-\rightarrow \ce{H2} (\mathit{g}) + 2\text{ OH}^- \]

    Now that we have the balanced oxidation and reduction half-reactions, we can identify which occurs at the anode and which occurs at the cathode. Then determine the standard reduction potentials of both and finally calculate the standard cell potential.

    • anode: \[ 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) + 2\text{ e}^- \;\;\; \text{E}^° = 1.36\text{ V} \]
    • cathode: \[ \ce{2H2O} (\mathit{ l}) + 2\text{ e}^-\rightarrow \ce{H2} (\mathit{g}) + 2\text{ OH}^- \;\;\; \text{E}^° = -0.83\text{ V}\]

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    \[ \text{E}^°_\text{cell} = -0.83\text{ V} - 1.36\text{ V} \]

    \[ \text{E}^°_\text{cell} = -2.19\text{ V} \]

    Since \(\text{E}^°_\text{cell} \) is negative the reaction is not spontaneous in the forward direction. However, the reaction is spontaneous in the reverse direction. The overall reaction does not have any conducting metals so both half-cells need either a Pt (s) or C (s) electrode to transfer electrons from one half-cell to the other. A salt bridge is also needed to allow charged ions to flow between the solutions of the half-cells and complete the circuit. Remember to put the electrodes at the ends of the cell diagram and write species in solution connected by a salt bridge before and after the double vertical line.

    \[ \text{Pt }(\mathit{s}) | \text{H}_2(\mathit{ g}) | \text{OH}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq})| \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s}) \]

    1. Notice that this reaction is a combustion reaction. Combustion reactions are known to have negative ΔG values which means they are spontaneous. Therefore, the reaction in question is spontaneous. However, we can also determine spontaneity by calculating \(\text{E}^°_\text{cell} \). We won't go over the exact numbers for this process, but we'll detail the steps on how to calculate the desired \(\text{E}^°_\text{cell} \) value. From the overall reaction given we can write the incomplete oxidation and reduction half-reactions:

    \[ \text{ oxidation: }\ce{C3H8 }(\mathit{g}) \rightarrow 3\text{ CO}_2 (\mathit{g}) \]

    \[ \text{ reduction: } \ce{5O2} (\mathit{ g}) \rightarrow \ce{4H2O} (\mathit{g}) \]

    In the oxidation half-reaction, elements other than O and H are already balanced so now we add H2O molecules to balance oxygen atoms.

    \[ \ce{C3H8 }(\mathit{g}) + \ce{6H2O}(\mathit{l})\rightarrow 3\text{ CO}_2 (\mathit{g}) \]

    Next is balancing hydrogen atoms.

    \[ \ce{C3H8 }(\mathit{g}) + \ce{6H2O}(\mathit{l}) \rightarrow 3\text{ CO}_2 (\mathit{g}) + 20\text{ H}^+\]

    Finally, add electrons to balance charge and get the balanced oxidation half-reaction.

    \[ \ce{C3H8 }(\mathit{g}) + \ce{6H2O} (\mathit{l}) \rightarrow 3\text{ CO}_2 (\mathit{g}) + 20\text{ H}^+ + 20\text{ e}^- \]

    For the reduction half-reaction we start by balancing oxygen atoms.

    \[ \ce{5O2} (\mathit{ g}) \rightarrow \ce{4H2O} (\mathit{g}) + \ce{6H2O} \]

    \[ \ce{5O2} (\mathit{ g}) \rightarrow \ce{10H2O} (\mathit{g}) \]

    Now balance hydrogen atoms.

    \[ \ce{5O2} (\mathit{ g}) + 20\text{ H}^+\rightarrow \ce{10H2O} (\mathit{g}) \]

    Finally, add electrons to balance charge and simplify to obtain the balanced reduction half-reaction.

    \[ \ce{5O2} (\mathit{ g}) + 20\text{ H}^+ + 20\text{ e}^-\rightarrow \ce{10H2O} (\mathit{g}) \]

    \[ \ce{O2} (\mathit{ g}) + 4\text{ H}^+ + 4\text{ e}^-\rightarrow \ce{2H2O} (\mathit{g}) \]

    Now that we have the balanced oxidation and reduction half-reactions, we can identify which occurs at the anode and which occurs at the cathode. Then determine the standard reduction potentials of both and finally calculate the standard cell potential.

    • anode: \[ \ce{C3H8 }(\mathit{g}) + \ce{6H2O}(\mathit{l}) \rightarrow 3\text{ CO}_2 (\mathit{g}) + 20\text{ H}^+ + 20\text{ e}^- \;\;\; \text{E}^° = ??????\text{ V}\]
    • cathode: \[ \ce{O2} (\mathit{ g}) + 4\text{ H}^+ + 4\text{ e}^-\rightarrow \ce{2H2O} (\mathit{g}) \;\;\; \text{E}^° = 1.229\text{ V}\]

    Notice that the standard reduction potential for the anode reaction cannot be found in a table of standard reduction potential values. To calculate the standard reduction potential for the reaction, the first step is to calculate ΔG. To do so first look up the ΔGf° values of the products and add them. Then do the same for the reactants. Next, subtract the total ΔGf° of the reactants from that of the products. You should get a negative value. The difference can then be used in the following equation to solve for the standard reduction potential of the half-reaction in question.

    \[ \text{ΔG} = -\text{nF}\text{E}^°_\text{cell} \]

    where "n" is the number of moles of electrons involved in the half-reaction and "F" is Faraday's constant. While the equation is meant to involve the cell potential for the overall reaction it can be used to involve the cell potential for a half-reaction. Notice how this is the standard oxidation potential for the anode reaction. To be able to use it in the following equation we must change the standard oxidation potential into its standard reduction potential counterpart which can be done by applying a negative sign to the value. Now use the equation

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    to determine the specific \(\text{E}^°_\text{cell} \) value for the overall reaction and conclude whether the reaction is spontaneous or not. Now we go on to write the cell diagram for a galvanic cell of this reaction. The overall reaction does not have any conducting metals so Pt (s) or C (s) electrodes are needed to transfer electrons between half-cells. The salt bridge in this case would be a semi-permeable member that allows charged spectator ions to flow between the cells, thus closing the circuit.

    \[ \text{Pt }(\mathit{s}) | \ce{C3H8 }(\mathit{g}) | \text{ CO}_2 (\mathit{g}) || \ce{O2} (\mathit{ g}) | \ce{H2O} (\mathit{g}) | \text{Pt} (\mathit{s}) \]

    Question 20.5.28

    Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CNas a complexing agent for Mn3+/Mn2+?

    Mn3+(aq) + e → Mn2+(aq) E° = 1.51 V

    Mn(CN)63−(aq) + e → Mn(CN)64− E° = −0.24 V

    Answer 20.5.28

    To better understand how to solve this problem we can paint a bigger picture of the problem. We can assume the process is spontaneous since the net stabilization of a species means the new species is more stable than the old specie. Thus, the new specie is lower in free energy than the old specie and when the change in free energy ΔG is negative then the reaction is spontaneous. Now, given the two half-reactions and the fact that the described process is spontaneous, we can imagine a galvanic cell for the entire process. Since the process is spontaneous we know that \(\text{E}^°_\text{cell} \) must be positive. For this to be true we use the equation

    \[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

    and substitute the given standard reduction potentials into the appropriate places.

    \[ \text{E}^°_\text{cell} = 1.51\text{ V} - (-0.24)\text{ V} \]

    \[ \text{E}^°_\text{cell} = 1.75 \text{ V} \]

    So now we know which reaction occurs at the anode and which reaction occurs at the cathode.

    \[ \text{cathode: }\text{Mn}^{3\text{+}}(\mathit{ aq})+\text{ e}^- \rightarrow \text{ Mn}^{2\text{+}}(\mathit{ aq}) \]

    \[ \text{anode: } \text{ Mn}(\text{CN})_6^{4\text{-}} \rightarrow \text{Mn}(\text{CN})_6^{3\text{-}}(\mathit{ aq})+\text{ e}^- \]

    While this bit of information is unnecessary to solving the problem, it does help in understanding what is going on. We can find the net thermodynamic stabilization energy since it is synonymous with the change in free energy ΔG of the system by the equation

    \[ \text{ΔG} = -\text{nF}\text{E}^°_\text{cell} \]

    The variable "n" is the number of moles electrons transferred in the overall reaction. From the given half-reactions we can see that one electron is involved in the overall reaction (n=1). We determined \(\text{E}^°_\text{cell} \) earlier, and Faraday's constant "F" is 96486 J/(V \( \times \) mol of e-). So now we can calculate ΔG

    \[ \text{ΔG} = -(1\text{ mol}\text{ of}\text{ e}^-)\times\frac{96486\text{ J}}{\text{V}\times\text{mol}\text{ of}\text{ e}^-}\times1.75 \text{ V} \]

    \[ \text{ΔG} = -168850.5\text{ J} \approx -1.69\times10^5\text{ J} \]

    The net thermodynamic stabilization energy is about -1.69 \( \times \) 105 J or -169 kJ.


    Extra Credit 13 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?