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19: Electron Transfer Reactions

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    19.2 Oxidation-Reduction Reactions

    Balancing Redox Reactions in Acidic Solutions

    Exercise \(\PageIndex{2.1}\)

    Balance the reaction in an acidic solution:

    \(Cu(s)+NO_{3}^{-}(aq)\rightarrow Cu^{2+}(aq)+NO_{2}(g)\)

    Answer

    \[Cu(s)+NO_{3}^{-}(aq)\rightarrow Cu^{2+}(aq)+NO_{2}(g)\]

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

    \[NO_{3}^{-}(aq)\rightarrow NO_{2}(g)\]

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

    \[NO_{3}^{-}(aq)+4H^{+}(aq)+2e^{-}\rightarrow NO_{2}(g)+2H_{2}O(l)+Cu^{2+}(aq)\]

    Exercise \(\PageIndex{2.2}\)

    Balance the reaction in an acidic solution:

    \(Mn^{2+}(aq)+BiO_{3}^{-}(aq)\rightarrow Bi^{3+}(aq)+MnO_{4}^{-}(aq)\)

    Answer

    \[Mn^{2+}(aq)+BiO_{3}^{-}(aq)\rightarrow Bi^{3+}(aq)+MnO_{4}^{-}(aq)\]

    \[Mn^{2+}(aq)+4H_{2}O(l)\rightarrow MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\]

    \[BiO_{3}^{-}(aq)+6H^{+}(aq)+2e^{-}\rightarrow Bi^{3+}(aq)+3H_{2}O(l)\]

    \[2Mn^{2+}(aq)+5BiO_{3}^{-}(aq)+14H^{+}(aq)\rightarrow 5Bi^{3+}(aq)+7H_{2}O(l)+2MnO_{4}^{-}(aq)\]

    Exercise \(\PageIndex{2.3}\)

    Balance the reaction in an acidic solution:

    \(Cr_{2}O_{7}^{2-}+(aq)+Cl^{-}(aq)\rightarrow Cr^{3+}(aq)+ClO_{3}^{-}(aq)\)

    Answer

    \[Cr_{2}O_{7}^{2-}+(aq)+Cl^{-}(aq)\rightarrow Cr^{3+}(aq)+ClO_{3}^{-}(aq)\]

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

    \[Cl^{-}(aq)+3H_{2}O(l)\rightarrow ClO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\]

    \[Cr_{2}O_{7}^{2-}+(aq)+14H^{+}(aq)+Cl^{-}(aq)\rightarrow 2Cr^{3+}(aq)+4H_{2}O(l)+ClO_{3}^{-}(aq)\]

    Exercise \(\PageIndex{2.4}\)

    Balance the reaction in an acidic solution:

    \(IO_{3}^{-}(aq)+HNO_{2}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\)

    Answer

    \[IO_{3}^{-}(aq)+HNO_{2}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\]

    \[IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\rightarrow I^{-}(aq)+3H_{2}O(l)\]

    \[HNO_{2}(aq)+H_{2}O(l)\rightarrow NO_{3}^{-}(aq)+3H^{+}(aq)+2e^{-}\]

    \[IO_{3}^{-}(aq)+3HNO_{2}(aq)\rightarrow I^{-}(aq)+3NO_{3}^{-}(aq)+3H^{+}(aq)\]

    Exercise \(\PageIndex{2.5}\)

    Balance the reaction in an acidic solution:

    \(Fe_{2}S(s)+Ce^{4+}(aq)\rightarrow Fe^{2+}(aq)+SO_{4}^{2-}(aq)+Ce^{3+}(aq)\)

    Answer

    \[Fe_{2}S(s)+Ce^{4+}(aq)\rightarrow Fe^{2+}(aq)+SO_{4}^{2-}(aq)+Ce^{3+}(aq)\]

    \[Fe_{2}S(s)+4H_{2}O(l)\rightarrow 2Fe^{2+}(aq)+SO_{4}^{2-}(aq)+8H^{+}(aq)+10e^{-}\]

    \[Ce^{4+}(aq)+e^{-}\rightarrow Ce^{3+}(aq)\]

    \[Fe_{2}S(s)+4H_{2}O(l)+10Ce^{4+}(aq)\rightarrow 2Fe^{2+}(aq)+SO_{4}^{2-}(aq)+8H^{+}(aq)+10Ce^{3+}(aq)\]

     

    Balancing Redox Reactions in Basic Solutions

    Exercise \(\PageIndex{2.6}\)

    Balance the reaction in a basic solution:

    \(Fe(s)+NO_{3}^{-}(aq)\rightarrow FeO_{2}^{2-}(aq)+NH_{3}(aq)\)

    Answer

    \[Fe(s)+NO_{3}^{-}(aq)\rightarrow FeO_{4}^{2-}(aq)+NH_{3}(aq)\]

     

    \[Fe(s)+2H_{2}O(l)\rightarrow FeO_{2}^{2-}(aq)+4H^{+}(aq)+2e^{-}\]

    \[NO_{3}^{-}(aq)+9H^{+}(aq)+8e^{-}\rightarrow NH_{3}(aq)+3H_{2}O(l)\]

     

    \[Fe(s)+2H_{2}O(l)+4OH^{-}(aq)\rightarrow FeO_{2}^{2-}(aq)+4H^{+}(aq)+2e^{-}\]

    \[NO_{3}^{-}(aq)+9H^{+}(aq)+8e^{-}\rightarrow NH_{3}(aq)+3H_{2}O(l)+9OH^{-}(aq)\]

     

    \[4Fe(s)+7OH^{-}(aq)+NO_{3}^{-}(aq)\rightarrow 4FeO_{2}^{2-}(aq)+NH_{3}(aq)+2H_{2}O(l)\]

    Exercise \(\PageIndex{2.7}\)

    Balance the reaction in a basic solution:

    \(Cr\left ( OH \right )_{3}(s)+BrO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+Br_{2}(l)\)

    Answer

    \[Cr\left ( OH \right )_{3}(s)+BrO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+Br_{2}(l)\]

    \[Cr\left ( OH \right )_{3}(s)+H_{2}O(l)\rightarrow CrO_{4}^{2-}(aq)+5H^{+}(aq)+3e^{-}\]

    \[2BrO^{-}(aq)+4H^{+}(aq)+2e^{-}\rightarrow Br_{2}(l)+2H_{2}O(l)\]

     

    \[Cr\left ( OH \right )_{3}(s)+5OH^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+4H_{2}O(l)+3e^{-}\]

    \[2BrO^{-}(aq)+2H_{2}O(l)+2e^{-}\rightarrow Br_{2}(l)+4OH^{-}(aq)\]

     

    \[2Cr\left ( OH \right )_{3}(s)+6BrO^{-}(aq)\rightarrow 2CrO_{4}^{2-}(aq)+2H_{2}O(l)+3Br_{2}(l)+2OH^{-}(aq)\]

    Exercise \(\PageIndex{2.8}\)

    Balance the reaction in a basic solution:

    \(Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)\rightarrow PbO_{2}(s)+Cl^{-}(aq)\)

    Answer

    \[Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)\rightarrow PbO_{2}(s)+Cl^{-}(aq)\]

     

    \[Pb\left ( OH \right )_{4}^{2-}(aq)\rightarrow PbO_{2}(s)+2H_{2}O(l)+2e^{-} ClO^{-}(aq)+2H^{+}(aq)+2e^{-}\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

     

    \[ClO^{-}(aq)+H_{2}O(l)(aq)+2e^{-}\rightarrow Cl^{-}(aq)+2OH^{-}(aq)\]

     

    \[ClO^{-}(aq)+Pb\left ( OH \right )_{4}^{2-}(aq)\rightarrow PbO_{2}(s)+H_{2}O(l)+Cl^{-}(aq)+2OH^{-}(aq)\]

    Exercise \(\PageIndex{2.9}\)

    Balance the reaction in a basic solution:

    \(Cl^{-}(aq)+MnO_{4}^{-}(aq)\rightarrow Cl_{2}(g)+MnO_{2}(s)\)

    Answer

    \[Cl^{-}(aq)+MnO_{4}^{-}(aq)\rightarrow Cl_{2}(g)+MnO_{2}(s)\]

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

    \[MnO_{4}^{-}(aq)+2H_{2}O(l)+3e^{-}\rightarrow MnO_{2}(s)+4OH^{-}(aq)\]

    \[6Cl^{-}(aq)+2MnO_{4}^{-}(aq)+4H_{2}O(l)\rightarrow 3Cl_{2}(g)+2MnO_{2}(s)+8OH^{-}(aq)\]

    Exercise \(\PageIndex{2.10}\)

    Balance the reaction in a basic solution:

    \(IO_{3}^{-}(aq)+NO_{2}^{-}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\)

    Answer

    \[I O_{3}^{-}(a q)+N O_{2}^{-}(a q) \rightarrow I^{-}(a q)+N O_{3}^{-}(a q)\]

    \[I O_{3}^{-}(a q)+3 H_{2} O(l)+6 e^{-} \rightarrow I^{-}(a q)+6 O H^{-}(a q)\]

    \[N O_{2}^{-}(a q)+2 O H^{-}(a q) \rightarrow N O_{3}^{-}(a q)+H_{2} O(l)+2 e^{-}\]

    \[I O_{3}^{-}(a q)+3 N O_{2}^{-}(a q) \rightarrow I^{-}(a q)+3 N O_{3}^{-}(a q)\]

     

    19.3: Electrochemical Cells

    Electrochemical Cell Notation

    Exercise \(\PageIndex{3.1}\)

    Write the following reactions in standard electrochemical cell notation at standard state conditions.

    1. \(C u(s)+C d^{2+}(a q) \rightarrow C u^{2+}(a q)+C d(s)\)
    2. \(2 N a(s)+F e^{2+}(a q) \rightarrow 2 N a^{+}(a q)+F e(s)\)
    3. \(3Z n(s)+2 F e^{3+}(a q) \rightarrow 3 Z n^{2+}(a q)+2 F e(s)\)
    4. \(2 A l(s)+3 M n^{2+}(a q) \rightarrow 2 A l^{3+}(a q)+3 M n(s)\)
    5. \(2 H^{+}(a q)+Z n(s) \rightarrow H_{2}(g)+Z n^{2+}(a q)\)
    6. \(H g(l)+S n^{4+}(a q) \rightarrow H g_{2}^{2+}+Sn(s)\)
    Answer a.

    \[Cu(s)|Cu^{2+}(1 M)||Cd^{2+}(1 M)|Cd(s)\]

    Answer b.

    \[Na(s)|Na^{2+}(1 M)||Fe^{2+}(1 M)|Fe(s)\]

    Answer c.

    \[Zn(s)|Zn^{2+}(1 M)||Fe^{3+}(1 M)|Fe(s)\]

    Answer d.

    \[Al(s)|A l^{3+}(1 M)||Mn^{2+}(1 M)|Mn(s)\]

    Answer e.

    \[Zn(s)|Zn^{2+}(1 M)||H^{+}(1 M)|H_{2}(1 atm)|Pt(s)\]

    Answer f.

    \[Hg(l)|Hg_{2}^{2+}(1 M)||Sn^{4+}(1 M)|Sn(s)\]

     

    Voltaic Cell

    Exercise \(\PageIndex{3.2}\)

    What is the half-reaction at the anode in the voltaic reaction?

    \(2Cu^{+}(aq)+I_{2}(l)\rightarrow 2Cu^{2+}(aq)+2I^{-}(aq)\)

    Answer

    \[Cu^{+}\rightarrow Cu^{2+}+e^{-}\]

    Exercise \(\PageIndex{3.3}\)

    What is the reaction occurring at the cathode from Question 19.3.2?

    Answer

    \[I_{2}+2e^{-}\rightarrow 2I^{-}\]

    Exercise \(\PageIndex{3.4}\)

    Which one can occur at the cathode of an electrochemical cell?

    1. \(NO\rightarrow NO_{2}^{-}\)
    2. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)
    3. \(I_{2}\rightarrow I^{-}\)
    4. none of the above
    Answer

    c. \(I_{2}\rightarrow I^{-}\)

    Exercise \(\PageIndex{3.5}\)

    Which one can occur at the anode of an electrochemical cell?

    1. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)
    2. \(Fe^{2+} \rightarrow Fe\)
    3. \(I_{2}\rightarrow I^{-}\)
    4. none of the above
    Answer

    a. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)

    Exercise \(\PageIndex{3.6}\)

    Using the information given, determine which reaction occurs at the anode.

    1. \(Al(s) \rightarrow Al^{3+}(aq)+3 ^{-}\)
    2. \(ClO_{3}^{-}(aq)+6H^{+}(aq)+6 e^{-} \rightarrow Cl^{-}(aq)+3H_{2} O(l)\)
     
    1. i.
    2. ii.
    3. neither
    4. both
    Answer

    a. i.

    Exercise \(\PageIndex{3.7}\)

    Using the same information in Question 19.3.6, which electrode is consumed?

    1. anode
    2. cathode
    3. neither
    4. both
    Answer

    a. anode

    Exercise \(\PageIndex{3.8}\)

    Using the same information in Question 19.3.6, which electrode is positive?

    1. anode
    2. cathode
    3. both
    4. neither
    Answer

    b. cathode

     

     

    19.4: Electrochemical Cell Fundamentals

    Half-reaction ϵred0
    \(I_{2}(s)+2e^{-}\rightarrow 2I^{-}(aq)\) 0.54 V
    \(Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)\) -0.44 V
    \(Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)\) 0.77 V
    \(Cr^{3+}(aq)+3e^{-}\rightarrow Cr(s)\) -0.74 V
     
    Figure \(\PageIndex{1}\): Use the following information to answer Questions 19.4.1-19.4.5

    Exercise \(\PageIndex{4.1}\)

    Determine the standard potential (V) for the cell reaction.

    \(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

    Answer

    \[2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\]

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

    Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-0.54=0.23 v\]

    Exercise \(\PageIndex{4.2}\)

    Determine the standard potential (V) for the cell reaction.

    \(Cr(s)+3Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\)

    Answer

    \[Cr(s)+3Fe^{3+}(aq) \rightarrow 3Fe^{2+}(a )+Cr^{3+}(aq)\]

    Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 e^{-}\]

    Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

    \[E^{0}=E_{\text {cathode}}^{o}-E_{\text {anode}}^{o}=0.77+0.74=1.51 v\]

    Exercise \(\PageIndex{4.3}\)

    Determine the standard potential (V) for the cell reaction.

    \(Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\)

    Answer

    \[Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\]

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

    Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

    \[E^{0}=E_{\text {cathode}}^{o}-E_{\text {anode}}=0.77-(-0.44)=1.21 v\]

    Exercise \(\PageIndex{4.4}\)

    Determine the standard potential (V) for the cell reaction.

    \(2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\)

    Answer

    \[2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\]

    Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 e^{-}\]

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

    \[E^{0}=E_{cathode}^{o}-E_{anode}=-0.44-(-0.74)=0.30 v\]

    Exercise \(\PageIndex{4.5}\)

    Determine the standard potential (V) for the cell reaction.

    \(3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6l^{-}(aq)\)

    Answer

    \[3l_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\]

    Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 ^{-}\]

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

    \[E^{0}=E_{\text {cathode} \omega \dot{e}}^{o}-E_{\text {anode}}^{o}=0.54-(-0.74)=1.28 \mathrm{v}\]

     

    19.5: Standard Electrochemical Potentials

    Exercise \(\PageIndex{5.1}\)

    What is the E°cell?

    \(Cu(s)+Cd^{2+}(aq) \rightarrow Cu^{2+}(aq)+Cd(s)\)

    Answer

    \[E^{0}_{cell}=E^{0}_{red}\left ( cathode \right )-E^{0}_{red}\left ( anode \right )\]

    \[Cu(s)|Cu^{2+}(1 M)|Cd^{2+}(1 M)|Cd(s)\]

    \[E^{0}_{cell}=-0.403V-\left ( 0.337V \right )\]

    \[E^{0}_{cell}=-0.740V\]

    Exercise \(\PageIndex{5.2}\)

    What is the expression for Q, in the equation:

    \(3Zn(s)+2Fe^{3+}(aq) \rightarrow 3Zn^{2+}(aq)+2Fe(s)\)

    Answer

    \[Q=\frac{\left [ Zn^{2+} \right ]^{3}}{\left [ Fe^{3+} \right ]^{2}}\]

    Exercise \(\PageIndex{5.3}\)

    What is the Ecell?

    \(3Zn(s)+2Au^{3+}(0.004 M) \rightarrow 3Zn^{2+}(0.0051 M)+2Au(s)\)

    Answer

    \[E_{cell}=E^{0}_{cell}-\frac{RT}{nF}lnQ\]

    \[E^{0}_{cell}=+1.50V-\left ( -0.763V \right )=+2.26V\]

    \[Q=\frac{\left [ Zn^{2+} \right ]^{3}}{\left [ Au^{3+} \right ]^{2}}=\frac{\left [ 0.0051 \right ]^{3}}{\left [ 0.004 \right ]^{2}}\]

    \[n=6\]

    \[E_{cell}=+2.26V-\frac{(8.314)298}{6(96,500)}ln\frac{\left [ 0.0051 \right ]^{3}}{\left [ 0.004 \right ]^{2}}=2.28V\]

    Exercise \(\PageIndex{5.4}\)

    cell is +2.28V, what is ∆G?

    \(3Zn(s)+2Au^{3+}(aq) \rightarrow 3Zn^{2+}(aq)+2Au(s)\)

    Answer

    \[\Delta G^{\circ}=-nFE^{\circ}\]

    \[\Delta G^{\circ}=-\left(6 {mol}{e}^{-}\right)\left(96,500{J}/{V} \bullet {mol}\,\,{e}^{-}\right)\right)(2.26 {V})\]

    \[\Delta{G}^{\circ}=-1.31 {MJ}\]

    Exercise \(\PageIndex{5.5}\)

    cell is -0.763, what is the ∆G?

    \(2H^{+}(aq)+Zn(s) \rightarrow H_{2}(g)+Zn^{2+}(aq)\)

    Answer

    \[\Delta G^{\circ}=-nFE^{\circ}\]

    \[\Delta G^{\circ}=-\left(2 {mol}{e}^{-}\right)\left(96,500{J}/{V} \bullet {mol}\,\,{e}^{-}\right)\right)(-0.763 {V})\]

    \[\Delta{G}^{\circ}=147 {kJ}\]

     

    19.6: Electrochemistry and Thermodynamics

    Half-reaction εred°
    \(I_{2}(s)+2e^{-} \rightarrow 2I^{-}(aq)\) 0.54 V
    \(Fe^{2+}(aq)+2e^{-} \rightarrow Fe(s)\) -0.44 V
    \(Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\) 0.77 V
    \(Cr^{3+}(aq)+3e^{-} \rightarrow Cr(s)\) -0.74 V
     
    Figure \(\PageIndex{2}\): Use the following table to answer questions 19.6.1-19.6.5

    Exercise \(\PageIndex{6.1}\)

    Using the given information, determine the ΔG0 in J for the following cell reaction. F=96500J/V mol

    \(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

    Answer

    \[2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-0.54=0.23 v\]

    \[\Delta G^{0}=-nFE^{0}=-96500 \times 2 \times 0.23=-44390{J}\]

    Exercise \(\PageIndex{6.2}\)

    Using the given information, determine the ΔG0 in J for the following cell reaction. F=96500J/V mol

    \(Cr(s)+3 Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\)

    Answer

    \[Cr(s)+3 Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77+0.74=1.51 v\]

    \[\Delta G^{0}=-nFE^{0}=-96500 \times 3 \times 1.51=-437145{J}\]

    Exercise \(\PageIndex{6.3}\)

    Using the given information, determine the ΔG0 in J for the following cell reaction. F=96500J/V mol

    \(Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\)

    Answer

    \[Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-(-0.44)=1.21 v\]

    \[\Delta G^{0}=-nFE^{0}=-96500 \times 2 \times 1.21=-233530{J}\]

    Exercise \(\PageIndex{6.4}\)

    Using the given information, determine the ΔG0 in J for the following cell reaction. F=96500J/V mol

    \(2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\)

    Answer

    \[2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=-0.44-(-0.74)=0.30 v\]

    \[\Delta G^{0}=-nFE^{0}=-96500 \times 6 \times 0.30=-173700{J}\]

    Exercise \(\PageIndex{6.5}\)

    Using the given information, determine the ΔG0 in J for the following cell reaction. F=96500J/V mol

    \(3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\)

    Answer

    \[3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\]

    \[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.54-(-0.74)=1.28 v\]

    \[\Delta G^{0}=-nFE^{0}=-96500 \times 6 \times 1.28=-741120{J}\]

     

    19.7: Electrochemical Cells under Nonstandard Conditions

    The Nernst Equation

    Exercise \(\PageIndex{7.1}\)

    Determine the cell voltage E at 25°C, when [Fe2+]=[I-]=0.01M, [Fe3+]=0.02M.

    \(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s), \quad E^{\circ}=0.23 v\)

    Answer

    \[E=E^{0}-\frac{0.0592}{n} \log Q=0.23-\frac{0.0592}{2} \log \frac{0.01^{2}}{0.01^{2} \times 0.02^{2}}=0.13 v\]

    Exercise \(\PageIndex{7.2}\)

    Determine the cell voltage E at 25°C, when [Fe2+]=0.01M, [Fe3+]=0.02M

    \(Fe(s)+2F e^{3+}(aq) \rightarrow 3 Fe^{2+}(aq), \quad E^{\circ}=1.21 v\)

    Answer

    \[E=E^{0}-\frac{0.0592}{n} \log Q=1.21-\frac{0.0592}{2} \log \frac{0.01^{3}}{0.02^{2}}=1.29v\]

    Exercise \(\PageIndex{7.3}\)

    Determine the cell voltage E at 25°C, when [Fe2+]=0.01M, [Cr3+]=0.005M

    \(2{Cr}(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq), \quad E^{\circ}=0.30 v\)

    Answer

    \[E=E^{0}-\frac{0.0592}{n} \log Q=0.30-\frac{0.0592}{6} \log \frac{0.005^{2}}{0.01^{3}}=0.31v\]

    Exercise \(\PageIndex{7.4}\)

    Calculate the equilibrium constant for the reaction at 25°C. R=8.314J/Kmol, F=96500 J/Vmol

    \(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

    Answer

    \[K_{eq}=e^{-\frac{\Delta G^{\circ}}{RT}}=e^{\frac{nFE^{0}}{RT}}=e^{\frac{2(9650)(0.23)}{8.314(288.15)}}=5.9\times 10^{7}\]

    or at 250C,

    \[K_{eq}=10^{\frac{nE^{0}}{0.0592}}=10^{\frac{2 \times 0.23}{0.0592}}=5.9 \times 10^{7}\]

    Exercise \(\PageIndex{7.5}\)

    For the reaction, if the Ecell is 1.50v and [Fe3+]=0.02M, determine the concentration of Fe2+.

    \(Fe(s)+2F e^{3+}(aq) \rightarrow 3 Fe^{2+}(aq)\)

    Answer

    \[E=E^{0}-\frac{0.0592}{n} \log Q\]

    \[1.50=1.21-\frac{0.0592}{2} \log \frac{x^{3}}{0.02^{2}}\]

    \[\log \frac{x^{3}}{0.02^{2}}=-9.797\]

    \[3 \log x-2 \log 0.02=-9.797\]

    \[x=4.0 \times 10^{-5} {M}\]

     

    Galvanic Cell Potentials

    Cell 1:

    \(Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\)
    \(Mg^{2+}(aq)+2e^{-}\rightarrow MG(s) \quad -2.37\,volts\)

    Cell 2:

    \(Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\)
    \(Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\)

    Cell 3:

    \(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
    \(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

    Figure \(\PageIndex{1}\): Use the following reactions to answer questions 19.7.6-19.7.11.

    Exercise \(\PageIndex{7.6}\)

    Find the following values for cell 1.

    1. What is the E°cell for cell 1?
    2. What is the ∆G°rxn for cell 1?
    Answer a.

    \[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

    \[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

    Answer b.

    \[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

    \[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(2\,mol\,e^{-})(96500C/mol\,e^{-})(1.93V)=-372490CV\]

    Exercise \(\PageIndex{7.7}\)

    What is the Ecell for cell 1(0.4M Mg2+ and 1.4M Fe2+)?

    Answer

    \[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

    \[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(2\,mol\,e^{-})(96500C/mol\,e^{-})(1.93V)=-372490CV\]

     

    \[E_{cell}={E}_

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    ^{\circ}-\left(\frac
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    \right) \ln {Q}\]

    \[E_{cell}=1.93 {V}-\left(\frac{8.314 {J} / {K} \cdot {m} {ol}(298.15 {K})}{2 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{0.4 {M} {Mg}^{2+}}{1.4 {M} {Fe}^{2+}}\right)\]

    \[{E}_

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    =1.95 {V}\]
     

    Exercise \(\PageIndex{7.8}\)

    Find the following values for cell 2.

    1. What is the E°cell for cell 2?
    2. What is the ∆G°rxn for cell 2?
    Answer a.

    \[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
    \[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

    Answer b.

    \[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
    \[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(0.65V)=-376350CV\]

    Exercise \(\PageIndex{7.9}\)

    What is the Ecell for cell 2(0.5M Mn2+ and 1.5M Ga3+)?

    Answer

    \[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
    \[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(0.65V)=-376350CV\]

     

    \[E_{cell}={E}_

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    \right) \ln {Q}\]

    \[E_{cell}=0.65 {V}-\left(\frac{8.314 {J} / {K} \cdot {mol}(298.15 {K})}{6 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{0.5 {M} {Mn}^{2+}}{1.5 {M} Gae}{3+}}\right)\]

    \[{E}_

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    =0.65 {V}\

    Exercise \(\PageIndex{7.10}\)

    Find the following values for cell 3.

    1. What is the E°cell for cell 3?
    2. What is the ∆G°rxn for cell 3?
    Answer a.

    \(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
    \(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=1.50V+2.89V=4.39V\]

    Answer b.

    \(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
    \(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=1.50V+2.89V=4.39V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(4.39V)=-2541810CV\

    Exercise \(\PageIndex{7.11}\)

    What is the Ecell for cell 3(1.2M Sr2+ and 0.6M Au3+)?

    Answer

    \(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
    \(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

     

    \[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

    \[E^{0}_{cell}=1.50V+2.89V=4.39V\]

     

    \[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

    \[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(4.39V)=-2541810CV\

     

    \[E_{cell}={E}_

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    \right) \ln {Q}\]

    \[E_{cell}=4.39 {V}-\left(\frac{8.314 {J} / {K} \cdot {m} {ol}(298.15 {K})}{6 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{1.2 {M} {Sr}^{2+}}{0.6{M} {Au}^{3+}}\right)\]

    \[{E}_

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    =4.39 {V}\]

     

    19.9: Electrolysis

    Exercise \(\PageIndex{9.1}\)

     How many minutes will it take to plate out 2.50g of chromium metal from Cr3+ solution using a current of 32.0 amps?

    Answer

    \[Cr^{3+}(aq)+3 e^{-} \rightarrow Cr(s)\]

    \[\frac{2.50 g}{52 g/ mol}=0.048 mol\]

    \[e^{-}needed: 0.048 \times 3=0.144 mol\]

    \[0.144 mol \times 96500C / mo l=1.39 \times 10^{4}C\]

    \[\frac{1.39 \times 10^{4}C}{32.0}=435 \sec =7.25 {min}\]

    Exercise \(\PageIndex{9.2}\)

    What current (in A) is needed to plated out 1.50g of chromium metal from Cr3+ solution in 30sec?

    Answer

    \[\frac{1.50 g}{52 g / m o l}=2.88 \times 10^{-2} {mol}\]

    \[e^{-}\, needed : 2.88 \times 10^{-2} {mol} \times 3=8.65 \times 10^{-2} {mol}\]

    \[8.65 \times 10^{-2} {mol} \times 96500 {C} / {mol}=8.35 \times 10^{3} {C}\]

    \[\frac{8.35 \times 10^{3} {C}}{30 {sec}}=2.78 \times 10^{2} {A}\]

    Exercise \(\PageIndex{9.3}\)

    How many grams of copper can be obtained if a current of 10amps passes through a solution of copper (II) sulfate for 20min?

    Answer

    \[10 A \times 20 \times 60 \sec =1.2 \times 10^{4} C \]

    \[\frac{1.2 \times 10^{4} C}{96500 C / m o l}=0.124 m o l \]

    \[\frac{0.124 m o l}{2} \times 63.55 g / m o l=3.948\]

    Exercise \(\PageIndex{9.4}\)

     In an electrolytic cell, 0.05g of copper has produced from the copper (II) sulfate solution. How many grams of potassium would be plated out if the same current was applied through molten KCl?

    Answer

    \[\frac{0.05 g}{63.55 g / m o l} \times 2 \times 39 g / m o l=0.06 g\]

    Exercise \(\PageIndex{9.5}\)

    Calculate the number of kilowatt-hrs of electricity required to produce 1.00kg of Al by electrolysis of Al3+ if 5.00V of emf is applied.

    Answer

    \[\frac{1000 g}{27.0 g / m o l} \times 3 \times 96500 C / m o l=1.072 \times 10^{7} C\]

    \[\frac{1.072 \times 10^{7} C \times 5.00 v}{3.6 \times 10^{6}}=14.89 k W h\]

     

    General Questions

    Exercise \(\PageIndex{1}\)

    In the voltaic cell that is represented as

    Cd | Cd2+ || (Pt) Fe3+ | Fe2+

    the electron flow will be from

    1. Pt to Cd2+
    2. Pt to Cd
    3. Fe2+ to Cd2+
    4. Cd2+ to Fe2+
    5. Cd to Fe3+
    Answer

    e. Cd to Fe3+

    Exercise \(\PageIndex{2}\)

    In the voltaic cell that is represented as

    Zn(s) | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu(s)

    Which of the following statements is false?

    1. The mass of the zinc electrode decreases during discharge.
    2. The copper electrode is the anode.
    3. Electrons flow through the external circuit from the zinc electrode to the copper electrode.
    4. Reduction occurs at the copper electrode during discharge.
    5. The concentration of Cu2+ decreases during discharge.
    Answer

    b. The copper electrode is the anode.

    Exercise \(\PageIndex{3}\)

    For a galvanic cell using Fe|Fe2+(1.0 M) and Pb|Pb2+(1.0 M) half-cells, which of the following statements is correct?

    \(Fe^{2+}+2e^{-}\rightarrow Fe \quad -0.44V\)
    \(Pb^{2+}+2e^{-}\rightarrow Pb \quad -0.13V\)

    1. The mass of the iron electrode decreases during discharge.
    2. Electrons leave the lead electrode to pass through the external circuit during discharge.
    3. The concentration of Pb2+ increases during discharge.
    4. The iron electrode is the cathode.
    5. When the cell has completely discharged (to zero voltage), the concentration of Pb2+ is zero.
    Answer

    a. The mass of the iron electrode decreases during discharge.

    Exercise \(\PageIndex{4}\)

    What is the cell reaction of the voltaic cell Cr(s)|Cr3+(aq)||Cl-(aq)|Cl2(g)|Pt?

    Answer

    \[2Cr(s)+3Cl_{2}(g)\rightleftharpoons 2Cr^{3+}(aq)+6Cl^{-}(aq)\]

    Exercise \(\PageIndex{5}\)

    When Au is obtained by electrolysis from NaAuCl4, what is the minimum number of coulombs required to produce 1.00 mol of gold?

    Answer

    \[Na(+1) \quad Au(+3) \quad 4CN(-1) \]

    \[Au^{3+}+3e^{-}\rightarrow Au(s) \]

    \[3*96500=2.90*10^{5}\,coulombs\]

    Exercise \(\PageIndex{6}\)

    How many faradays are required to convert a mole of Cr2O72- to Cr3+?

    Answer

    \[Cr_{2}O_{7}^{2}+14H^{+}+6e^{-} \rightarrow 2Cr^{3+}+7H_{2}O\]

    \[1e^{-}=1F \therefore 6e^{-}=6F\]

    Exercise \(\PageIndex{7}\)

    Which of the following cell reactions would require the use of an inert electrode?

    1. \(Zn(s)+2MnO_{2}(s)+2NH_{4}^{+}(aq) \rightarrow Zn^{2+}(aq)+Mn_{2}O_{3}(s)+2NH_{3}(aq)+H_{2}O(l)\)
    2. \(Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)\)
    3. \(3Cu(s)+2Au^{3+}(aq) \rightarrow 3Cu^{2+}(aq)+2Au(s)\)
    4. \(Cl_{2}(g)+2I^{-}(aq) \rightarrow 2Cl^{-}(aq)+I_{2}(s)\)
     
    1. 1 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1 and 4 only
    5. 3 and 4 only
    Answer

    d. 1 and 4 only

    Exercise \(\PageIndex{8}\)

    How many moles of electrons are produced form a current of 15.0 A in 1.00 hr?

    Answer

    \[x*(6.023*10^{23})*(1.6*10^{-19})=15.0*1.00*60*60 \]

    \[x=0.560\,moles\,e^{-}\]

    Exercise \(\PageIndex{9}\)

    The following has a potential of 0.92 V:

    \(2Hg^{2+}(aq)+H_{2}(g)\rightarrow 2H^{+}(aq)+Hg_{2}^{2+}(aq)\)

    If the concentration of the ions were 1.0 molar and the pressure of H2 were 1.0 atmosphere, then what would be the E° for the half-reaction?

    \(2Hg^{2+}(aq)+2e^{-}\rightarrow Hg_{2}^{2+}\)

    Answer

    \[2Hg^{2+}(aq)+H_{2}(g)\rightarrow 2H^{+}(aq)+Hg_{2}^{2+}(aq)\]

    \[E_{cell}=E^{0}_{cell}-\frac{0.0591}{n}log\frac{\left [ Hg_{2}^{2+} \right ]\left [ H^{+} \right ]^{2}}{\left ( P_{H_{2}} \right )\left [ Hg^{2+} \right ]} \]

    \[0.92=E^{0}_{cell}-\frac{0.0591}{2}log\frac{\left [ 1.0 \right ]\left [ 1 \right ]^{2}}{\left ( 1 \right )\left [ 1.0 \right ]} \]

    \[E^{0}_{cell}=0.92V\]

    Exercise \(\PageIndex{10}\)

    A cell with the potential of 0.74 V has the cell reaction

    \(2Cr+6H^{+}\rightarrow 2Cr^{3+}+3H_{2}\)

    If the concentrations of the ions were 1.0 molar and the pressure of H2 were 1.0 atmosphere, then what would be the E° of the half-reaction?

    \(Cr^{3+}+3e^{-}\rightarrow Cr\)

    Answer

    Cell reaction: \(2Cr+6H^{+} \longrightarrow 2Cr^{3+}+3H_{2} \)

    At anode: \(2Cr \longrightarrow 2Cr^{3+}+6e^{-} \)

    At cathode: \(6H^{+}+6e^{-} \longrightarrow 3H_{2}\)

    \[E_{cell}=E_{cell}^{0}-\frac{0.0591}{n}logQ \]

    \[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log\frac{\left [ Cr^{3+} \right ]^{2}\left ( P_{H_{2}} \right )^{3}}{\left [ H^{+} \right ]^{6}}\]

    \[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log\frac{\left [ 1.0 \right ]^{2}\left ( 1.0 \right )^{3}}{\left [ 1.0 \right ]^{6}} \]

    \[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log(1) \]

    \[0.74V=0.00V-E_{Cr^{3+}/Cr}^{0} \]

    \[E_{Cr^{3+}/Cr}^{0}=-0.74V \]

    Exercise \(\PageIndex{11}\)

    A standard cell that consisted of a strip of zinc dipped into 1.0 M zinc ion and a strip of copper dipped into 1.0 M copper ion and which was connected by a salt bridge had a potential of 1.10 V

    \(Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu\)

    If a potential of 0.60 V were assigned to the half-reaction

    \(Cu^{2+}+2e^{-}\rightarrow Cu\)

    instead of 0.34 V, which is the potential given in a standard reduction potential table, the potential for the reaction would be

    \(Zn^{2+}+2e^{-}\rightarrow Zn\)

    Answer

    \[E_{cell}=E_{cathode}^{0}-E_{anode}^{0} \]

    \[E_{cell}=0.60V-1.10 \]

    \[E_{cell}=-0.50V\]

    Exercise \(\PageIndex{12}\)

    Consider the following electrode potentials:

    \(Mg^{2+}+ 2e^{-} \rightarrow Mg \quad E^{0}=-2.37V\)

    \(V^{2+}+2e^{-} \rightarrow V \quad E^{0}=-1.18V\)

    \(Cu^{2+}+e^{-} \rightarrow Cu^{+} \quad E^{0}=0.15V\)
     

    Which one of the reactions below will proceed spontaneously from left to right?

    1. \(Mg^{2+}+V \rightarrow V^{2+}+Mg\)
    2. \(Mg^{2+}+2Cu^{+} \rightarrow 2Cu^{2+}+Mg\)
    3. \(V^{2+}+2Cu^{+} \rightarrow V+Cu^{2+}\)
    4. \(V+2Cu^{2+} \rightarrow V^{2+}+2Cu^{+}\)
    5. none of these
    Answer

    d. \(V+2Cu^{2+} \rightarrow V^{2+}+2Cu^{+}\)

    Exercise \(\PageIndex{13}\)

    Consider the following standard reduction potentials:

    \(2H^{+}(aq)+2e^{-} \rightarrow H_{2}(g) \quad E^{0}=0.00V\)

    \(Sn^{2+}(aq)+2e^{-} \rightarrow Sn(s) \quad E^{0}=-0.15V\)

    \(Cd^{2+}(aq)+2e^{-} \rightarrow Cd(s) \quad E^{0}=-0.40V\)

    which pair of substances react spontaneously

    1. Sn2+ with Cd2+
    2. Sn with Cd2+
    3. Sn2+ with H+
    4. Cd with Sn
    5. Cd with Sn2+
    Answer

    e. Cd with Sn2+

    Exercise \(\PageIndex{14}\)

    Calculate the maximum electrical work obtained when 7.10 grams of Cl2 gas are consumed in the reaction Cd(s) + Cl2(g) ⇌ Cd2+(aq) +2 Cl-(aq). (Ecell=1.76V)

    Answer

    \[W_{\max }=-n \times E_{cell} \times F\]

    where F=96,500kJ/V mole of electron

    Ecell=1.76V(given)

    gram of Cl2=7.10g molar mass of Cl2=70.906 g/mole

    than mole of Cl2=7.10/70.906=0.10013 mole

    Half reaction of cell as

    \[Cl_{2}+2e^{-}\rightarrow 2Cl^{-}\]

    1 mole Cl2 required 2 mole electrons

    than 0.10013 mole Cl2 required, mole of electron (n)=0.10013 * 2=0.20026 mole e-

    Now put all value

    \[W_{max }=-\left(0.20026 \,mole \,e^{-}\right) \times(1.76V) \times\left(96.500 kJ / \mole \,e^{-} \,V\right)=34.0121584\,kJ\]

    Exercise \(\PageIndex{15}\)

    Calculate E° for the cell reaction 2Cr + 3Sn4+ → 3Sn2+ + 2Cr3+.

    \(Cr^{3+}+3e^{-} \rightarrow Cr \quad E^{0}=-0.74V\)

    \(Sn^{4+}+2e^{-} \rightarrow Sn^{2+} \quad E^{0}=0.15V\)

    Answer

    \[E^{0}_{cell}=E_{cathode}^{0}-E_{anode}^{0}\]

    \[E^{0}_{cell}=0.15-(-0.74)\]

    \[E^{0}_{cell}=0.89V\]

    Exercise \(\PageIndex{16}\)

    At 25°C, calculate the voltage of the cell

    \(2Ag^{+}(aq)(0.16 M)+Cu(s)\rightarrow Cu^{2+}(aq)(0.015 M)+2Ag(s)\)

    if E°cell = 0.460 V.

    Answer

    \[E_{cell}=E^{0}-(0.596/n)log(A_{red}/A_{oxd})\]

    [Ared=conc. of reductant, Aoxd=conc. of oxidant]

    \[E_{cell}=0.460-(0.0596 / 2) \log (0.015 / 0.16)\]

    [n=no. of electrons that occur in reaction, since copper is reduced to 2, change in oxidation state =2]

    \[E_{cell}=0.46-(0.0298)*(-1.028028724) \]

    \[E_{cell}=0.46+0.030635255 \]

    \[E_{cell}=0.491V\]

    \(Cr^{3+}+3e^{-}\rightarrow Cr \quad -0.74V\)
    \(Pb^{2+}+2e^{-}\rightarrow Pb \quad -0.13V\)
    Figure \(\PageIndex{4}\): Use the following information for questions 19.17-19.18

    Exercise \(\PageIndex{17}\)

    What is the standard cell potential for the reaction?

    \(2Cr+3Pb^{2+}\rightarrow 3Pb+2Cr^{3+}\)

    Answer

    \[E_{cell}=E_{cathode}^{0}-E_{anode}^{0} \]

    \[E_{cell}=-0.13V-(-0.74V) \]

    \[E_{cell}=0.61V\]

    Exercise \(\PageIndex{18}\)

    Calculate the Gibbs free energy change for the reaction above the initial concentration of Cr3+ and Pb2+ are 1.00M.

    Answer

    \[\Delta G=-nFE^{0} \]

    \[\Delta G=-(6) \times (96,500) \times (0.61) \]

    \[\Delta G=-353190\,J \]

    \[\Delta G=-353\,kJ\]

    Exercise \(\PageIndex{19}\)

    What is the value of the reaction quotient, Q, for the cell that is constructed from the two half-reactions?

    \(Zn^{2+}+2e^{-}\rightarrow Zn \quad -0.76V\)

    \(Ag^{+}+e^{-}\rightarrow Ag \quad 0.80V\)
     

    when the Zn2+ concentration is 0.0100 M and the Ag+ concentration is 1.25M?

    Answer

    \[Q= \frac{\left [ products \right ]}{\left [ reactants \right ]} \]

    \[Q=\frac{\left[Zn^{2+}\right]}{\left[Ag^{+}\right]^{2}} \]

    \[Q=\frac{[0.0100]}{[1.25]^{2}} \]

    \[Q=6.40 \times 10^{-3}\]

    Exercise \(\PageIndex{20}\)

    Calculate ∆G° (in joules) for the reaction 2AlI3(aq) ⇌ 2Al(s) + 3I2(s).

    \(Al^{3+}+3e^{-} \rightleftharpoons Al \quad E^{0}=-1.66V \)

    \(I_{2}(s)+2e^{-} \rightleftharpoons 2I^{1} \quad E^{0}=0.54V\)
     

    Answer

    \[2Al^{3+}+6e^{-}\rightleftharpoons 2Al(s) \quad E^{0}=-1.66V \]

    \[6I^{-}\rightleftharpoons 3I_{2}(s)+6e^{-} \quad E^{0}=-0.54V \]

    \[E^{0}=E_{cat}+E_{ano}=-1.66+(-0.54)=-2.20V \]

    \[\Delta G=-nFE^{0}=-(6) \times (96,500) \times (-2.20)=1.3*10^{6}\]

    Exercise \(\PageIndex{21}\)

    What is the reduction potential for the half-reaction Al3+(aq) + 3e- → Al(s) + at 25°C is [Al+] = 0.10 M and E° = -1.66 V?

    Answer

    \[E_{Al^{3+}/Al}=E^{0}_{Al^{3+}/Al}-\frac{RT}{nF}ln\frac{1}{\left [ Al^{3+} \right ]} \]

    \[E_{Al^{3+}/Al}=E^{0}_{Al^{3+}/Al}-\frac{0.0591}{n}log\frac{1}{\left [ Al^{3+} \right ]} \]

    \[E_{Al^{3+}/Al}=-1.66-\frac{0.0591}{3}log\frac{1}{0.10} \]

    \[E_{Al^{3+}/Al}=-1.66-0.0197 \]

    \[E_{Al^{3+}/Al}=-1.68\]

    Exercise \(\PageIndex{22}\)

    What is the emf at 25°C for the following cell?

    \(Cr\left|Cr^{3+}(0.010 M) \| Ag^{+}(0.00010 M)\right| Ag\)

    \(Cr^{3+}+3e^{-} \rightleftharpoons Cr \quad E^{0}=-0.74 V\)

    \(Ag^{+}+e^{-} \rightleftharpoons Ag\quad E^{0}=0.80V\)
     

    Answer

    \[Cr^{3+}(aq)+3e^{-} \rightarrow Cr(s) \quad E^{0}=-0.74V \]

    \[Ag^{+}(a q)+e^{-} \rightarrow Ag(s)  \quad E^{0}=+0.80V \]

    \[3Ag^{+}(aq)+Cr(s) \rightarrow Ag(s)+Cr^{3+}(aq) \]

    \[E_{0}=E_{cathode}+E_{anode}=+0.80V+0.74V=1.54V \]

    \[E_{cell}=E_{0}-\frac{0.0591}{n} \log \frac{[\text { products }]}{\left[\text { reactants }\right]} \]

    \[E_{cell}=1.54-\frac{0.0591}{3} \log \frac{\left[Cr^{3+}\right]}{\left[Ag^{+}\right]^{3}} \]

    \[E_{cell}=1.54-\frac{0.0591}{3} \log \frac{0.01}{(0.0001)^{3}} \]

    \[E_{cell}=1.54-\frac{0.0591}{3} \times 10 \]

    \[E_{cell}=1.54-0.197\]

    \[E_{cell}=1.34V\]

    Exercise \(\PageIndex{23}\)

    What is the Cu2+ concentration at 25°C in the cell Zn(s) | Zn2+ (1.0 M) || Cu2+(aq) | Cu(s)? The cell emf is 1.03 V. The standard cell emf is 1.10 V.

    Answer

    \[E_{cell}=E_{0}-\frac{0.0591}{n} \log \frac{[Zn^{2+}]}{\left[Cu^{2+}\right]} \]

    \[1.03=1.10-\frac{0.0591}{2} \log \frac{1.0}{\left[Cu^{2+}\right]} \]

    \[-0.07=-\frac{0.0591}{2} \log \frac{1.0}{\left[Cu^{2+}\right]} \]

    \[2.369=\log \frac{1.0}{\left[Cu^{2+}\right]} \]

    \[233.81= \frac{1.0}{\left[Cu^{2+}\right]} \]

    \[\left[Cu^{2+}\right]}=0.004M\]

    Exercise \(\PageIndex{24}\)

    Molten magnesium chloride is electrolyzed using insert electrodes and reactions represented by the following equations

    \(2Cl^{-} \rightarrow Cl_{2}+2e^{-}\)

    \(Mg^{2+}+2e^{-} \rightarrow Mg\)

    concerning this electrolysis which of the following statements is true

    1. oxidation occurs at the cathode
    2. Mg2+ ions are reduced at the anode
    3. electrons pass through the metallic part of the circuit from Mg2+ ions to the Cl- ion
    4. Cl- ions are oxidizing agents
    5. the cations and the electrolyte undergo reduction
    Answer

    e. the cations and the electrolyte undergo reduction

    Exercise \(\PageIndex{25}\)

    A piece of iron half immerse in a sodium chloride solution will corrode more rapidly than a piece of iron half immersed in pure water because

    1. the sodium iron oxidized the iron atoms
    2. the chloride ions oxidized the iron atoms
    3. the chloride ions form a precipitate with iron
    4. the chloride ions increase the pH of the solution
    5. the sodium ions and chloride ions carry a current through the solution
    Answer

    e. the sodium ions and chloride ions carry a current through the solution

    Exercise \(\PageIndex{26}\)

    A reaction is spontaneous win

    1. ∆G° is negative or E° is positive
    2. ∆G° is negative or E° is negative
    3. ∆G° is positive or E° is negative
    4. ∆G° is positive or E° is positive
    5. ∆G° is negative, ∆H° is negative, and E° is negative
    Answer

    a. ∆G° is negative or E° is positive

    Exercise \(\PageIndex{27}\)

    What reaction occurs at the cathode during electrolysis of aqueous CuSO4?

    1. \(2H_{2}O+2e^{-} \rightarrow H_{2}+2OH^{-}\)
    2. \(Cu \rightarrow Cu^{2+}+2e^{- \)
    3. \(2H_{2}O \rightarrow O_{2}+4H^{+}+4e^{-}\)
    4. \(2H^{+}+2e^{-} \rightarrow H_{2}\)
    5. \(Cu^{2+}+2e^{-} \rightarrow Cu\)
    Answer

    e. \(Cu^{2+}+2e^{-} \rightarrow Cu\)

    Exercise \(\PageIndex{28}\)

    For a certain oxidation-reduction reaction, he is positive. this means that

    1. ∆G° is negative and K is less than 1
    2. ∆G° is negative and K is greater than 1
    3. ∆G° is zero and K is greater than 1
    4. ∆G° is positive and K is greater than 1
    5. ∆G° is positive and K is less than 1
    Answer

    b. ∆G° is negative and K is greater than 1

    Exercise \(\PageIndex{29}\)

    Cathodic protection results when

    1. iron is attached to a more active metal
    2. iron is amalgamated with Mercury
    3. iron is tin plated for use as a tin can
    4. iron is painted to protect it from corrosion
    5. iron is made amphoteric
    Answer

    a. iron is attached to a more active metal

    Exercise \(\PageIndex{30}\)

    What mass of chromium could be deposited by electrolysis of an aqueous solution fo Cr2(SO4)2 for 60.0 minutes using constant current of 10.0 amperes? (One faraday = 96,500 coulombs.)

    Answer

    \[Cr^{3+}+3e^{-}=Cr \]

    \[Charge = current + time=10A *\left(60^{+} 60\right)s=36000C \]

    No of moles of electron flowed

    \[=36000 / 96500 =0.373 \,moles \]

    As 1 mole of Cr3+ is reduced by 3 moles of electrons, so 0.124 moles of Cr3+ is reduced by 0.373 moles of electrons. 

    Mass of Cr deposited 

    \[=0.124 * molar mass of \,Cr \]

    \[=0.124 * 52.01 \,g \\

    \[=6.47 \,g\]

    Exercise \(\PageIndex{31}\)

    In a galvanic cell porn electrolysis cell, the cathode is always

    1. the positive electrode
    2. the negative electrode
    3. the positive electrode and the negative electrode, respectively
    4. the electrode at which some species gain electrons
    5. the electrode at which some species lose electrons
    Answer

    d. the electrode at which some species gain electrons

    Exercise \(\PageIndex{32}\)

    The voltage of the cell at 25°C would be

    \(Zn^{2+}+2e^{-} \rightarrow Zn \quad E^{0}=-0.7V \)

    \(Cu^{2+}+2e^{-} \rightarrow Cu \quad E^{0}=0.34V\)
     

    1. between 0.76 and 1.10V
    2. between 0.34 and 0.76 V
    3. between 0.00 and 0.76 speak
    4. less than 0.42 V
    5. greater than 1.10 V
    Answer

    e. greater than 1.10 V

    Exercise \(\PageIndex{33}\)

    On the basis afford going standard electrode potentials, determine which of the following is the strongest oxidizing agent.

    \(Zn^{2+}+2e^{-} \rightarrow Zn \quad E^{0}=-0.76V \)

    \(Cu^{2+}+2e^{-} \rightarrow Cu \quad E^{0}=+0.34V \)

    \(Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-} \rightarrow 2Cr^{3+}+7H_{2}O \quad E^{0}=+1.33V\)
     

    1. Zn2+
    2. Zn
    3. Cu
    4. Cr2O72-
    5. Cr3+
    Answer

    d. Cr2O72-

    Exercise \(\PageIndex{34}\)

    Which of the following species is the best reducing agent?

    1. Cd
    2. Mg2+
    3. I-
    4. Sn
    5. F-
    Answer

    a. Cd

     

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