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7.8: Electrochemistry (Exercises)

  • Page ID
    120520
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    Q7.1

    For an electrochemical cell driven by the reaction:

    \[\ce{Cl2 + 2Tl → 2Cl- + 2Tl+}\]

    Determine the electromotive force of the cell under standard conditions. Standard reduction potentials for the half cells are given below

    \[\ce{Cl2 (g) + 2e- → 2Cl- }\]

    with Eo= +1.36 V

    \[\ce{Tl+ + e- → Tl}\]

    with Eo= -0.336 V

    S7.1

    The change in gibbs free energy is an extrinsic quality that can be used to determine the EMF of the cell based on the SRPs of the half-cells. Because the number of electrons exchanged in each reaction are known, we can cancel out the Faraday constants in the equation to relate Gibbs' free energy directly to the EMF of the cell.

    ΔGTot = ΔG1 + ΔG2 → EototntotF = Eo1n1F + Eo2n2F → Eotot = (Eo1n1 + Eo2n2)/ntot = (1.36V*(2) -0.336V(1))/2 = 2.38 V

    Q7.8

    For the following electrochemical cells, the value of K is given. Determine the change in Gibbs free energy for the reaction in the cell described, and determine the standard potential of the cell.

    Cu2+ + Sn -> Sn2+ + Cu; K = e38.8

    2Li + Hg2+ -> 2Li+ + Hg; K = e381

    Au3+ + 3Zn -> 3Zn+ + Au; K = e264

    Q7.9

    Refer to the reaction below for the following question.

    \[Zn + Sn^{4+} \rightleftharpoons Zn^{2+} +Sn^{2+}\]

    Calculate the for the specified reaction at 298 K.

    S7.9

    First, identify the anode and the cathode:

    Anode: CodeCogsEqn-63.gif

    Cathode: CodeCogsEqn-64.gif

    With this information established, you can then calculate the using the half-reactions.

    CodeCogsEqn-65.gif

    Q7.9

    \[Cu^{2+} + 2Tl \rightarrow Cu + 2Tl^{2+}\]

    If the equilibrium constant is \(8.66 \times 10^{22}\), what is the electric potential of the cell?

    Q7.10a

    In an experiment using standard conditions, you find the cell emf to be 0.0428 V for the following cell:

    \[ Ag_{(s)}|Ag^{+}||Ag^{+}|Ag_{(s)}\]

    If the concentration of Ag+ is 4.5 M at the cathode, calculate the concentration of Ag+ at the anode.

    S7.10a

    Firstly, note the half reaction.

    \[ Ag^{+} + e^{-} \rightarrow Ag\]

    Then solve using the Nernst equation, noting that E is 0 for concentration cells (same electrodes with different concentrations).

    \[ \begin{align}
    E &= E^o - \dfrac{RT}{vF}\ln \dfrac{[Anode]}{[Cathode]} \\
    0.0428&=0 - \frac{(8.3145 \ J \ mol^{-1})(298^{\circ}K)}{(1)(96,485 J)} \ \cdot \ \ln\frac{X}{4.5M} \\
    0.0428&=-0.0257 \ \cdot \ \ln\frac{X}{4.5M} \\
    -1.665&= \ln(x) - \ln(4.5) \\
    \ln(4.5)+ \left (-1.665 \right) &= \ln (x) \\
    e^{-0.161} &= e^{\ln(x)} \\
    X&=0.85 M \rightarrow \underline{ [Ag^{+}]=0.85M }
    \end{align} \]

    Click here for more information on concentration cells.

    Q7.10b

    A concentration cell contains \(I\) and \(I_2\). What is the pressure of \(I\) if the cell emf is -.05 V and the pressure of \(I_2\)is 3.75 bar?

    S7.10b

    \[2I_{(g)} \rightleftharpoons I_{2(g)} \]

    \[E=\dfrac{-.0257V}{2}ln\dfrac{3.75bar}{x}\rightarrow x=13.0bar\]

    Q7.14

    Cu(s)|Cu2+ (0.5M) || Cu2+(aq)|Cu(s)

    Calculate the electric potential difference for the above concentration cell

    Q7.14

    CodeCogsEqn-25.gif

    Using the above concentration cell, determine the emf. Assume the temperature is at 25°C.

    S7.14

    CodeCogsEqn-24.gif

    CodeCogsEqn-27.gif

    The overall reaction is:

    CodeCogsEqn-28.gif

    The emf of the cell is dependent upon the concentrations of Co2+ at the anode and at the cathode.

    CodeCogsEqn-29.gif

    Q7.16

    Referring to the reaction:

    CodeCogsEqn-31.gif

    and taking into account that at 298 K, the E°' is -0.219 V. Use this information to determine the E' at pH = 3. Note: Both FAD and FADH2 have equal moles of concentration.

    S7.16

    CodeCogsEqn-38.gif

    CodeCogsEqn-36.gif

    CodeCogsEqn-39.gif

    Q7.15

    Calculate E for the following reaction:

    [(Zn(s)+Cu^{2+}(aq) \rightleftharpoons Cu(s) + Zn^{2+}(aq)\]

    S7.15

    \(Zn(s) \rightarrow 2e^- + Zn^{2+}(aq)\) -Anode [Zn2+]=.25M

    \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\) -Cathode [Cu2+]=.35M

    \[E^\circ = E_{cathode}-E_{anode}=.342V-(-.762V)=1.104 V\]

    \[E=E^\circ -\dfrac{.0257V}{2}ln\dfrac{Cu^{2+}}{Zn^{2+}}=1.10V -\dfrac{0.0257V}{2}ln\dfrac{.35M}{.25M}= 1.08V\]

    Q7.14

    Co(s)|Co2+ (0.05M) ||Co2+(aq) (0.1M)|Co(s)

    Calculate the emf for the above concentration cell.

    Q7.21

    To breakdown consumed alcohol, our body converted ethanol into an acetyldehyde by forming a redox reaction with \(NAD^+\).

    \[\text{ethanol} + NAD^+ → \text{acetaldehyde} + NADH + H^+\]

    Determine ΔrG°' of reaction at 298 K and pH = 7 (Hint: Set up half reactions and find E°' of the whole reaction.)

    S7.21

    \[ethanol \rightarrow acetaldehyde + 2H^{+} + 2e^{-} \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = +0.197 V\]

    \[NAD^{+} + H^{+} + 2e^{-} \rightarrow NADH \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = -0.320 V \]

    \[ Overall: \, ethanol + NAD^{+} \rightarrow acetaldehyde + NADH + H^{+} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = -0.123\]

    \[ \Delta rG^{\circ}{'} = -vF\Delta E^{\circ}{'} = -(2)(96500 \frac{C}{mol})(-0.123V) = 23739 J = {\color{red}23.7 kJ/mol}\]

    Q7.21

    The hydrolysis of ATP with the following concentrations:

    ATP4- + H2O  ADP3- + H+ + HPO4²-
    10mM 0.5mM pH 6 5mM

    Calculate the change in biochemical standard Gibbs energy of reaction.

    Q7.22

    Calculate the equilibrium constant and ∆rGo' for hydrolysis of ATP at 25.0oC:

    \[\text{Glucose} + \text{ ATP} \rightarrow \text{ADP} + \text{ glucose-6-phosphate}\]

    • [Glucose]initial = 0.05M
    • [Glucose]equilibrium = 0.01M

    S7.22

    KC = (0.04M/1M)/(0.01M/1M)=4.0

    delta GO = -8.3145J/Kmole *(25+273.15)K*ln 4.0=-3.44 *103kJ/mol

    Q7.23

    Calculate the Δr/ of a certain biochemical reaction that involves the exchange of 2 electrons and where Eº/ = 3.25 V

    S7.23

    \[\Delta _{r}G^{\circ}{}'= -vFE^{\circ}{}'\]

    \[\Delta _{r}G^{\circ}{}'=-(2)(96500 C\ mol^{-1})(3.25 V)\]

    \[\Delta _{r}G^{\circ}{}'= -627 kJ\ mol^{-1}\]

    Q7.25

    In intestinal epithelial cells, transport of lactose across a membrane is coupled with the transport of 2 H+ ions from a high to low concentration.

    \[\Delta _{r}G^{\circ'} = 16.2 kJ\, \, \, \, \, \, \, \, \, \psi=-0.15V\]

    How much will the pH change across the membrane at 298 K and 1 atm be if one lactose molecule is transported?

    Hint: What is the change in Gibbs Free Energy for the movement of 2H+ across this membrane? Write this as an expression as a function of change in pH.

    S7.25

    When we consider the chemical potential of a single H+

    \[\mu _{H^{+}}=\mu^{\circ} _{H^{+}}+RTln[H^{+}]+Fz\psi\]

    • z = the ion charge
    • F = Faraday constant
    • \psi = electrical potential

    \[pH=-log[H^{+}]\]

    \[ln[H^{+}]=-2.3log[H^{+}]=-2.3pH\]

    \[\mu _{H^{+}}=\mu^{\circ} _{H^{+}}-2.3RTpH+F\psi\]

    Change in Gibbs free energy is defined as the initial and final chemical potentials, or:

    \[\Delta _{r}G = \mu _{H^{+}(low)}-\mu _{H^{+}(high)}\]

    \[=-2.3RT\Delta pH+F\Delta \psi\]

    As a function of change in pH, the equation can be written as

    \[\Delta pH=\frac{\Delta_{r}G-F\Delta \psi }{-2.3RT}\]

    Since this equation describes chemical potential across a membrane for a H+ ions, the sign for change in Gibb's free energy should be opposite that of the transport lactose, or -16.2 KJ. Setting change in electric potential as -0.15 V and converting 12.7 KJ to J

    \[\Delta pH= \frac{-16200J-(96500\frac{C}{mol})(-0.15 V)}{(298K)(-2.3)(8.3145\frac{J}{K\cdot mol})}\]

    \[=0.30\]

    Q7.28

    You are studying muscle cells. You notice inside the muscle there is 255 mM of Li+ and 16 mM of Rb+ and there is 58 Mm of Li+ and 249 mM of Rb+ outside the cell. From this information, decide if the membrane is more permeable to Li+ or Rb+.

    S7.28

    CodeCogsEqn (6).gif

    CodeCogsEqn (26).gif

    =-0.075

    CodeCogsEqn (5).gif

    CodeCogsEqn (27).gif

    =0.0407

    So Rb+ has a greater magnitude so the membrane is more permeable to it

    Q7.29

    Calculate the membrane potential of a cell that has [K+]out = 8 and [K+]in = 200. Assume standard temperature of 298 K.

    Q7.30

    Find E° for:
    Cu²+ + e- -> Cu+
    K+ + e- -> K
    How can E° be used to find Ksp?

    Q7.30a

    Given the values of E° for the following reactions:

    Ag+ + e- → Ag E° = +0.851 V

    AgCl + e- → Ag + Cl- E° = +0.222 V

    Determine the solubility product (Ksp) of AgCl at 23.5°C.

    Q7.30b

    Given the half-reduction reaction:

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

    \[FeCl_3 + e^- \rightarrow Fe + 3Cl^-\]

    Calculate the Ksp of FeCl3 at 298 K

    Q7.31

    Given the following reaction at pH 7.0:

    \[CH_3COOH + CO_{2(g)} + 2H^+ + 2e^- \rightarrow CH_3COCOOH + H_2O \]

    1. Write down the reaction that produces \(H^+\) ion in term of Eo
    2. Calculate the concentration of \(CH_3COCOOH\) at 25oC. Given the emf of the cell is -0.70 V.

    Q7.31

    A certain reaction in solution has provides H+ ions and has an Eº of .48V. If a calomel electrode is placed in solution (as part of a pH meter) and has an Eref of .11 V. Find the pH of the solution.

    S7.31

    \[pH = \dfrac{E - E_{ref}}{0.0591V}\]

    \[pH = \dfrac{.37V}{0.0591V}\]

    \[pH = 6.26\]

    Q7.32a

    If a since and iron block were to be exposed for an extended period of time, which would rust first?

    S7.32a

    Zinc is a more electropositive metal with a reduction potential of -0.762 V while iron has a reduction potential of -0.447 V. Since Zinc is more electropositive, it will be more likely to be oxidized so it will rust first.

    Q7.32b

    Aluminum is used widely to be a better protective metal than lead because it forms a corrosion resistance layer in most of environments. Explain this using Standard Reduction Potentials values?

    Q7.34

    The \(\Delta rS^o\) of the Daniell cell is found to be \(-26.8JK^{-1}mol^{-1}\). Given this value at \(95^oC\) calculate the emf of the cell and the temperature coefficient.

    S7.34

    Step 1: (Hint: \(\Delta rS^o\) is directly related to temperature coefficient.)

    \[\Delta rS^o=vF(\dfrac{ ∂ E^o}{\ ∂})_P\]

    \[\Delta rS^o=vF(\dfrac{∂ E^o}{∂T})_P=\dfrac{\Delta rS^o}{vF}=\dfrac{-26.8JK^{-1}\,mol^{-1}}{2(96500C\,mol^{-1})}=-1.38860\times10^{-4}vK^{-1}\]

    Step 2:

    \[\dfrac{∂ E^o}{∂})_P=\dfrac{E^o_{369.15K}-E^o_{298.15K}}{368.15-298.15K}=\dfrac{E^o_{368.15K-1.104V}}{70K}=1.39\times10^{-4}VK^{-1}\]

    \[E^o_{268.15K}=1.094V\]

    Q7.36

    Calculate the equilibrium constant for the following reaction:

    \[FeCl_2\rightleftharpoons Fe^{2+} + 2Cl^-\]

    (Hint: look up the reduction potential for the half reactions)

    S7.36

    Half reactions are\[\begin{align*}
    Fe\rightarrow &Fe^{2+}+2e^-\\
    &\varepsilon ^\circ=0.44V (this-is-oxidation)\\
    Cl_2+ 2e^-\rightarrow &2Cl^- \\
    &\varepsilon ^\circ = 1.36V (this-is-reduction)\end{align*}\]

    \(\varepsilon ^\circ=\varepsilon ^\circ_{cathode}-\varepsilon ^\circ_{anode}=1.36V-0.44V=0.92V\)

    \(K=e^\dfrac{vF\varepsilon ^\circ}{RT}=e^\dfrac{(96500C/mol)(0.92V)}{(8.314J/K*mol)(298.15K)}=3.58*10^{15}\)

    S7.36

    \(\begin{align*}Anode:2K\rightarrow &2K^+ +2e^-\\
    &\varepsilon ^\circ =5.84V\\
    Cathode:2H^+2e^-\rightarrow &H_2\\
    &\varepsilon ^\circ=0V\end{align*}\)

    \(\varepsilon ^\circ=\varepsilon ^\circ_{cathode}-\varepsilon ^\circ_{anode}=-5.84V\\
    \Delta _r\bar{G}^\circ=-vF\varepsilon ^\circ=2\Delta _f\bar{G}^\circ[K^+]+\Delta _f\bar{G}^\circ[H_2]-2\Delta _f\bar{G}^\circ[K]-2\Delta _f\bar{G}^\circ[H^+]\\
    \begin{align*}\Delta _f\bar{G}^\circ[K^+]&=-\dfrac{1}{2}vF\varepsilon ^\circ-\dfrac{1}{2}\Delta _f\bar{G}^\circ[H_2]+\Delta _f\bar{G}^\circ[K]+\Delta _f\bar{G}^\circ[H^+]\\
    &=-\dfrac{1}{2}(2)(96500C/mol)(-5.84V)-\dfrac{1}{2}(0kJ/mol)+(0kJ/mol)+(0kJ/mol)=5.6*10^5J/mol\end{align*}\)

    Q7.36

    Do you expect a high Kw value at 276 K or 373 K? Explain.

    S7.36

    We expect a higher Kw value for 373 K.

    Q7.38a

    What is the the reaction for a solution with Au3++3e- →Au and Ag++e- →Ag

    S7.38a

    The first thing we do is look at the standard reduction potentials table. We see that the reduction potential is higher for Au3+ so that will be the one that gets reduced. This means we reverse the reaction for Ag+.

    \[Au^{3+}+3e^- →Au\]

    3x(Ag→Ag++e-) we need the charges to add up to zero so we multiply it by 3 so there will be 3 electrons here as well

    We end up with:

    \[Au^{3+}+3Ag → Au + 3Ag^+\]

    Q7.38b

    The magnitudes of the standard electrode potentials of two metals, A and B, are

    A2+ + 2e- → A |E°| = 0.63 V

    B2+ + 2e- →B |E°| = 1.24 V

    Where the | | notation denotes that only the magnitude (but not the sign) of the E° value is shown. When the half-cells of A and B are connected, electrons flow from A to B. When A is connected to a SHE, electrons flow from A to SHE. (a) What are the signs for each of the E°? (b) What is the standard emf of a cell made up of A and B?

    Q7.38c

    The electrochemical properties of a pair of newly discovered metals, Q and G, are being studied in a lab. Their associated reduction reactions are shown below. A lab technician collecting the data makes a mistake, and does not record the sign of the standard potential for the half-cells in their notes. However, he remembers that when he connected the cell containing Q to the standard hydrogen electrode, the Q electrode was oxidized, and when he connected G to the standard hydrogen electrode, it was reduced. What are the signs of the standard potentials for the half-cells? When the two half-cells are connected, what is the electromotive force of the complete Q/G cell?

    \[Q+ + e- -> Q= 0.38 V \]

    \[G+ + e- -> G = 0.72 V \]

    Q7.39

    Consider a galvanic (voltaic) cell with silver wire suspended in 0.5M Ag+ solution and a cobalt wire in 1.0M Co2+ solution. Write out the half cell reactions and overall reaction to find the cell's equilibrium constant and observed emf.

    Hint: Look at examples of setting up half reactions

    S7.39

    Half cell reactions:

    \[Cathode: \, \, \, \, \, \, \, \, \,2[Ag^{+}(aq) + e^{-} \rightarrow Ag(s)] \]

    \[Anode: \, \, \, \, \, \, \, \, \,Co(s) \rightarrow Co^{2+}(aq) + 2e^{-} \]

    \[Overall: \, \, \, \, \, \, \, \,2Ag^{+}(aq) + Co(s) \rightarrow 2Ag(s) + Co^{2+}(aq) \]

    \[E^{\circ} = 0.800 \,V - (-.277 \,V) = +1.077 \,V \]

    \[ E^{\circ} = \frac{RT}{vF}lnK \]

    \[+1.077 \,V = \frac{8.314 \frac{J}{K \cdot mol}(298K)}{(2)(26485) \frac{C}{mol}}lnK\]

    \[{\color{red} K = 2.51 \times 10^{36}} \]

    \[E = E^{\circ} - \frac{RT}{v} ln \frac{[Co^{2+}]}{[Ag^{+}]} = +1.077V - \frac{0.0257V}{2}ln\frac{1.0M}{0.5M}\]

    \[{\color{red} E = +1.068 \,V }\]

    Q7.40a

    Answer the following questions about aluminum:

    1. Given that the standard electrical potential of Al3+ is -1.662, why does aluminum tarnish in air?
    2. Would the the following reaction be spontaneous? \[ 2 Al + 3 Br_2 \rightarrow Al_2Br_6\]
    3. Predict the reaction between aluminum and chloride gas.

    Hint: look up values for Eand appropriate half reactions.

    S7.40a

    (a) Silver tarnishes in air because the reduction potential for oxygen is high enough that it oxidizes aluminum.

    \[ \begin{align} & E_{cell}^{\circ}=E_{cathode}^{\circ}-E_{anode}^{\circ} \\ & E_{cell}^{\circ}=E_{oxygen}^{\circ}-E_{aluminum}^{\circ} \\ & E_{cell}^{\circ}=1.23 \ V \ - \left ( -1.662 \ V \right ) > 0 \end{align}\]

    Since E > 0, then rG is less than zero (since rG = -vFE). Thus the reaction is spontaneous and aluminum will be oxidized (tarnish) in contact with air.


    (b) Find the appropriate half reactions to fulfill the equation.

    \[ \begin{align} 2 Al + 3 Br_2 & \rightarrow Al_2Br_6\ \left ( 1 \right ) \\ Al^{3^{+}} + 3 e^{-} & \rightarrow Al, \ where \ E_{Al}^{\circ}=-1.662 \\ Br_2 + 2 e^{-}& \rightarrow 2Br^{-}, \ where \ E_{Br}^{\circ}=+1.087 \\ \end{align}\]

    Multiply both sides by the appropriate integer to fulfill equation (1).

    \[ \begin{align} -2 \ times \ Al^{3^{+}} + 3 e^{-} & \rightarrow Al, \ where \ E_{Al}^{\circ}=-1.662 \\ 3 \ times \ Br_2 + 2 e^{-}& \rightarrow 2Br^{-}, \ where \ E_{Br}^{\circ}=+1.087 \end{align}\]

    Switching around the half reactions you should achieve the following, which matches up to the equation (1).

    \[ 2Al + 3Br_2 + 6e^{-} \rightarrow 2Al^{3^{+}} + 6Br^{-} + 6e^{-}\]

    Use the equation to find the E. Keep in mind cathodes are reduced in the half reaction (you want the electrons to be on the left hand side of the half reaction after balancing) and anodes are oxidized in the half reaction (electrons on the right). Also note how standard reduction potential is an intrinsic property so there is no need to multiply by molar coefficients represented in the balanced equation.

    \[ \begin{align} E_{cell}^{\circ}&=E_{cathode}^{\circ}-E_{anode}^{\circ} \\ E_{cell}^{\circ}&=E_{bromide}^{\circ}- E_{aluminum}^{\circ} \\ E_{cell}^{\circ}&= \left ( 1.087 \ V \right ) - \left ( -1.662 \ V \right ) > 0 \\ E_{cell}^{\circ}&=2.749 \ V \\ \end{align}\]

    Since E > 0, then rG is less than zero (since rG = -vFE). Thus the reaction is spontaneous.


    (c)2Al + 3Cl2 → 2AlCl3 since Al3+ is aluminum's most stable form and Cl-. Then it's a matter of balancing the equation.

    Check here for additional information on Redox Chemistry.

    Q7.40b

    Using the electrochemical principle to explain how the reduction potential for oxygen is not adequate to oxidize the gold metal.

    Q7.40c

    Answer the following questions for the half-reaction below and E°[Pt|O2,H2O] = 1.229 V.

    Fe2+(aq) + e- → Fe(s) E° = -0.447 V

    (a) Predict whether or not iron will rust in air.

    (b) Will the following half-reaction have a positive E°? (E°[Pt|Fe3+,Fe2+] = 0.771 V)

    Fe(s) → Fe3+(aq) + 3e-

    (c) Predict the reaction between iron and bromine gas (E°[Br2] = 1.087 V)

    Q7.40d

    The standard reduction potential for Platinum, Pt, is as follows:

    \[Pt^{2+} + 2e^{-} \rightarrow Pt_{(s)}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.2 V\]

    1. Platinum is often used as a metal in fine jewelry because it does not tarnish noticeably. Why is this the case?
    2. Based on the above standard reduction potential, will the following reaction be spontaneous,

    \[Pt^{2+}+Cl^{2-}\rightarrow PtCl_{4}^{2-}\]

    given that

    \[PtCl_{4}^{2-}+ 2e^{-}\rightarrow Pt_{(s)}+4Cl^{-}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +0.73 V\]

    c.) Gold is also used as a metal in fine jewelry. Based on your answer to part (a.) and the standard reduction potential for gold Au3+, Eo= 1.69, does gold tarnish?

    S7.40d

    a.) Platinum does not tarnish, or oxidize, very quickly in air since its reduction potential is only slightly lower than that of oxygen. Thus, any oxidation which does occur does so very slowly.

    \[O_{2}+4H^{+}+4e^{-}\rightarrow H_{2}O\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.229 V\]

    b.) If the reaction has a positive standard reduction potential, then it is spontaneous. The reaction can be broken up into two half reactions whose values are provided.

    \[Pt_{(s)}+4Cl^{-}\rightarrow PtCl_{4}^{2-}+ 2e^{-}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= -0.73 V\]

    \[\, \, \, \, \, \, Pt^{2+} + 2e^{-} \rightarrow Pt_{(s)}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.2 V \]

    \[Pt^{2+}+Cl^{2-}\rightarrow PtCl_{4}^{2-}\]

    so the standard reduction potential is:

    \[E^{o}=-0.73V+1.2V=+0.47V\]

    Thus, the reaction is spontaneous.

    c.) No. Since the standard reduction potential of gold is greater than that of oxygen, oxygen cannot spontaneously react with gold and oxidize it.

    Q7.41

    In figure 7.1 we observe a voltaic cell. Here the anode is oxidized (loss of electrons) and the cathode is reduced (gains electrons). The Zinc metal loses electron and Zn (s) becomes Zn 2+ ions. Explain how the Cu2+ ions become solid Copper.

    Q7.42a

    In the following reaction at 298K,

    \[Ca_{(s)}+2H^+ \rightarrow \leftarrow Ca_{(aq)}+H_{2(g)}\]

    Calculate the pressure of \(H_2\) in bar to maintain equilibrium. The concentration of \([Ca^{2+}]=0.026M and the pH of the solution is buffered at 1.60 pH.

    S7.42a

    Step1:

    \[Anode: Ca^{+2}\rightarrow Ca^{+2}+2e^-\]

    \[Cathode: 2H^++2e^-\rightarrow H_2\]

    \[E^o=0V-(-0.34)=0.34\]

    Step 2: EMF calculation

    \[E=E^o-\dfrac{0.0257V}{v}ln\dfrac{[Ca^{+2}]P_{H_{2}}}{[H^+]^{2}}\]

    \(From\;the\;pH, [H^+]=10^{-1.60}=2.51\times10^{-2}M\)

    At E=0,

    \[0=0.34-\dfrac{0.0257V}{2}ln\dfrac{(0.026)P_{H{2}}}{(2.51\times10^{-2})^2}\]

    \[P_{H_2}=7.5\times10^{9}bar\]

    Q7.42b

    For the reaction at 300K:

    \[Mg(s) +2H^+(aq) \rightleftharpoons Mg^{2+} (aq) + H_2(g)\]

    Given pressure following the reaction is 296.07 atm, and molarity of acid used is 0.025M. Compute the molarity of the Mg2+.

    Q7.44a

    For the ion \(Sn^{2+}\), use the electric potential table to determine the value of \(\Delta f\bar G^o\).

    S7.44a

    \[Sn^{2+}_{(aq)}+2e^{-}\rightarrow Sn_{(s)}\]

    \[E^o=-0.137\]

    \(Anode: Sn_{(s)}\rightarrow Sn^{2+}+2e^-\)

    \(Cathode: 2H^{+}_{(aq)}+2e^{-}\rightarrow H_{2(g)}\)

    Overall Equation,

    \[Sn_{(s)}+2H^+_{(aq)}\rightarrow Sn^{2+}_{(aq)}+H_{2}\;\;\;E^o=0.137V\]

    \[\Delta rG^o=-vFE^o=\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]+\Delta _f\bar{G}^o[H_{2(g)}]-\Delta _f\bar{G}^o[Sn_{(s)}]-2\Delta _f\bar{G}^o[H^+_{(aq)}]\]

    \[-(2)(96500C\,mol^{-1})(0.137V)=\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]+0kj\,mol^{-1}-0kj\,mol^{-1}-2(0kj\,mol^{-1})\]

    \[\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]=-2.6\times10^4J\,mol^{-1}\]

    Q7.44b

    Calculate the \(\Delta _f\bar{G}^\circ\) for \(K^+\) in the following reaction: \[2K+2H^+\rightarrow 2K^+ +H_2\]

    Q7.46a

    Determine the standard emf Eo for the following reaction (at 298 K and pH 0)

    I-(aq) + H+(aq) + Cr2O72- (aq) → I2(s) + Cr3+(aq) + H2O(l)

    Hint: Use standard change in free energy of formation and use the relationship between Go and Eo

    S7.46a

    \[oxidation: 6I^{-}(aq) \rightarrow 3I_{2} + 6e^{-}\]

    \[reduction: 6e^{-} + 14H^{+}(aq) + Cr_{2}O{7}^{-2}(aq) \rightarrow 2Cr^{3+}(aq) +7H_{2}O (l)\]

    \[Overall: 6I^{-}(aq) + 14H^{+}(aq) + Cr_{2}O{7}^{-2}(aq) \rightarrow 2Cr^{3+}(aq) +7H_{2}O +3I_{2}(s)\]

    \[\Delta _{r}G^{\circ} = 2\Delta _{f}G^{\circ}(Cr^{3+})+3\Delta _{f}G^{\circ}(I_{2})+7\Delta _{f}G^{\circ}({H_{2}O}) - 6\Delta _{f}G^{\circ}(I^{-})-14\Delta _{f}G^{\circ}(H^{+})-\Delta _{f}G^{\circ}(Cr_{2}O_{7}\,^{2-})\]

    \[\Delta _{r}G^{\circ} = [2(-214.2) + 3(0) + 7(-237.2) -6(51.57)-14(0)-(-1301.1)]\frac{kJ}{mol} = -1097.12 \frac{kJ}{mol}\]

    \[E^{\circ} = -\frac{ \Delta _{r}G^{\circ}}{vF} = -\frac{-1097.12 \frac{kJ}{mol} \times \frac{1000J}{1kJ}}{(6)(96500 \frac{C}{mol})} = {\color{red} +1.89V}\]

    Q7.46b

    Calculate the \(\varepsilon ^\circ\) for the combustion of ethanol given that \(\Delta _f\bar{G}^\circ_{ethanol} =-174.8kJ/mol\).

    S7.46b

    \(Overall:C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O\\
    Anode: C_2H_5OH +3H_2O \rightarrow 2CO_2+12H^+ +12e^-\\
    Cathode: 3O_2+12H^+ +12e^- \rightarrow 6H_2O\)

    \(\begin{align*} \Delta _rG^\circ&=2\Delta _f\bar{G}^\circ[CO_2]+3\Delta _f\bar{G}^\circ[H_2O]-\Delta _f\bar{G}^\circ[C_2H_5OH]-3\Delta _f\bar{G}^\circ[O_2]\\&=2(-394.4kJ/mol)+3(-237.14kJ/mol)-(-174.8kJ/mol)-3(0kJ/mol)\\&=-1325.42kJ/mol\end{align*}\)

    \(\varepsilon ^\circ=-\dfrac{\Delta _rG^\circ}{vF}=-\dfrac{-1325.42*10^3J/mol}{(12)(96500C/mol)}=1.145V\)

    Q7.47a

    Explain the biological significance of cytochromes by comparing the following half reaction in cytochrome c, a, or f vs. an electrode:

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

    S7.47a

    For cytochrome c, E∘' = +0.254

    For Pt|Fe3+ , Fe2+, E = +0.771

    Since cytochrome c has a smaller reduction potential, it makes it easier for for Iron to exist in both the oxidized form Fe3+ or the reduced form Fe2+. This allows cytochromes to transfer electrons easily between other cytochromes in the electron transport chain. keep in mind that different cytochromes have slightly different reduction potentials.

    See here more details about Electron Transfer Proteins.

    Q7.47b

    Practice

    The standard reduction potential for the reduction of copper in aqueous solution is as follows:

    \[Cu^{2+} + e^{-} \rightarrow Cu^{+}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +0.153 V\]

    In hemocyanin, the primary oxygen carrier in invertebrate blood, however, the standard reduction potential is:

    \[Cu^{2+} + e^{-} \rightarrow Cu^{+}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ'}= +0.540 V\]

    Why do these reduction potentials differ?

    S7.47b

    The standard reduction potential is used to measure the readiness with which an atom may be reduced at standard conditions (25 oC, 1 atm, reactants at 1M). In biological half-reactions, standard reduction potentials differ because the process cannot occur at standard concentrations. Eo' occurs at pH 7, as opposed to pH 0, the reference point for electrode half reactions. To obtain a more intuitive understanding, we may use the Nernst equation:

    \[E^{o'}=E^{o}-\frac{RT}{\gamma F}\ln \frac{[Cu^{+}]}{[Cu^{2+}]}\]

    At a higher pH, more electrons are found in solution, due to the formation of Hydrogen ions. By le chatelier's principle, this pushes the above half reaction forward so that the concentration of Cu2+ is greater than that of Cu+. This makes the value produced by the natural log negative. Thus. Eo' is greater than Eo at higher pH.


    7.8: Electrochemistry (Exercises) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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