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Solutions 1: Electrochemistry

  • Page ID
    10537
  • These are solutions to select problems from homework set #1.

    Q1.1

    Determine the oxidation states of the underlined elements. Hint: Read Module on Oxidation States

    1. Using rule #1, Cl isn't combine with any other element other than itself, therefore the oxidation state for Cl is 0.
    2. Using rule #2, we know that the sum of the oxidation states for Na and H is 0 because NaH is a neutral compound. Now, we might be tempted to say H has the oxidation state of +1 because of rule #5. However, recall that one must use the rules in order. Because Na is in Group 1A, according to rule #3, the oxidation state of Na is +1. Thus, H must have the oxidation state of -1 so that the sum of the oxidation state for Na and H is 0.
    3. Using rule #2, the sum of the oxidation state is 0. Using rule #5, H has the oxidation state of +1, and using rule #6, O has the oxidation state of -2. The equation for H2CO: 2(+1) + C + (-2) = 0. -----> 2 + C - 2 = 0 --------> C = 0
    4. Using rule #2, the sum of the oxidation state is -2. Using rule #6, O has the oxidation state of -2. We want to know the oxidation of S, so the equation forS2O3 2-: 2S + 3(-2) = -2 ------> 2S -6 = -2 ------> 2S = +4 ------> S = 4/2 -----> S = 2
    5. Using rule #2, the sum of the oxidation state is 0, and using rule #3, the oxidation state for K is +1. Using rule #6, O has the oxidation state of -2. The equation for KMnO4: 1 + Mn + 4(-2) = 0 -----> Mn = -1 + 8 -----> Mn = 7
    6. Using rule #2, the sume of oxidation state is 0. Using rule #7, Cl has a -1 charge. The equation for FeCl3 Fe + 3(-1) = 0 -----> Fe -3 = 0 -----> Fe = 3
    7. Using rule #1, N isn't combine with any other element other than itself, therefore the oxidation state for N is 0.
    8. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1. Using rule #6, O has the oxidation state of -2. The equation for H2SO4: 2(1) + S + 4(-2) = 0 -----> S = -2 + 8 -----> S= 6
    9. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1. Using rule #6, O has the oxidation state of -2. The equation for HClO2: 1 + Cl + 2(-2) = 0 -----> Cl = -1 + 4 -----> Cl = 3
    10. Using rule #2, the sum of the oxidation state is 0. Using rule #6, O has the oxidation state of -2. Cu only have one charge which is +2. The equation for CuSO4: 2 + S + 4(-2) = 0 -----> S = -2 + 8 -----> S = 6

    Q1.2

    Balance the following equations in both acidic and basic environments. Hint: Read Module on balancing REDOX reactions.

    1) \(H_2(g) + O_2(g) \rightarrow H_2O(l)\)

    • Acidic Answer: \(2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\)
    • Basic Answer: \(2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\)

    2) \(Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)\)

    • Acidic Answer: \(2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 16H^+(aq) \rightarrow 4Cr^{3+}(aq) + 2CO_2(g) + 11H_2O(l)\)
    • Basic Answer: \(2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 5 H_2O(l) \rightarrow 4Cr^{3+}(aq) + 2CO_2(g) + 16OH^-(aq)\)

    3) \(Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)\)

    • Acidic Answer: \(MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\)
    • Basic Answer: \(MnO_4^-(aq) + 5Fe^{2+}(aq) + 4H_2O(l) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 8OH^-(aq)\)

    4) \(Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)\)

    • Acidic Answer: \(3Zn(s) + 2NO_3^-(aq) + 8H^+(aq) \rightarrow 3Zn^{2+}(aq) + 2NO(g) + 4H_2O(l)\)
    • Basic Answer: \(3Zn(s) + 2NO_3^-(aq) + 4H_2O(l) \rightarrow 3Zn^{2+}(aq) + 2NO(g) + 8OH^-(aq)\)

    5) \(Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)\)

    • Acidic Answer: \(4Al(s) + 3O_2(g) + 10 H_2O(l) \rightarrow 4[Al(OH)_4]^-(aq) + 4H^+(aq)\)
    • Basic Answer: \(4Al(s) + 3O_2(g) + 6H_2O(l) + 4OH^-(aq) \rightarrow 4[Al(OH)4]^-(aq)\)

    Q1.3

    not given. But read Using Redox Potentials to Predict the Feasibility of Reactions if you are confused.

    Q1.4

    1. \(\mathrm{Al(s)| Al^{3+}|| Zn^{2+}(aq)| Zn(s)}\)

    Oxidation (anode): \(\mathrm{Al(s) \rightarrow Al^{3+} + 3e^- \quad E=-1.676\,V}\)

    Reduction (cathode): \(\mathrm{Zn^{2+} + 2e^- \rightarrow Zn(s) \quad E=-0.7618\,V}\)

    \(\begin{align}
    \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
    & = -0.7618 - (-1.676) \,\mathrm V \\
    & = +0.9142\,\mathrm{V}
    \end{align}\)

    1. \(\mathrm{Pt (s)| Fe^{2+}(aq), Fe^{3+}(aq)|| Cu^{2+}(aq)| Cu(s)}\)

    Oxidaiton (anode): \(\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+} \quad E=0.771\,V}\)

    Reduction (cathode): \(\mathrm{Cu(s)\rightarrow Cu^{2+} + 2e^- \quad E= 0.3419\,V}\)

    \(\begin{align}
    \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
    & = 0.3419 - 0.771\,\mathrm V \\
    & = 0.4291\,\mathrm{V}
    \end{align}\)

    1. \(\mathrm{Pt (s)| Cr^{3+}(aq), Cr_2O_7^{2-}(aq)|| Ag^+(aq)| Ag(s)}\)

    Oxidation (cathode): \(\mathrm{ 2 Cr^{3+} + 7H_2O(l) \rightarrow Cr_2O_7^{2−} + 14H^+ + 6e^- \quad E=1.36\,V}\)

    Reduction (anode): \(\mathrm{Ag(s) \rightarrow Ag^+ + e^- \quad E=0.7996\,V}\)

    \(\begin{align}
    \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
    & = \mathrm{0.7996\,V-1.36\,V} \\
    & = -0.56\,\mathrm{V}
    \end{align}\)

    1. \(\mathrm{O_2^-(aq)| O_2(g)|| H^+(aq)|H_2(g)| C(s)}\)

    Oxidation (anode): \(\mathrm{O_2^-+(aq) \rightarrow O_2(g) + e^- \quad E = -0.33\,V}\)

    Reduction (cathode): \(\mathrm{ 2H^+ + e^- \rightarrow H_2 \quad E=0\,V}\)

    \(\begin{align}
    \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\
    & = 0 - (-0.33)\,\mathrm V \\
    & = +0.33 \,\mathrm{V}
    \end{align}\)

    Q1.5

    1. The standard reduction potential of the reduction half reaction \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\) is 0.7996 V
      The standard reduction potential of the oxidation half reaction \(Cu^{2+} + 2e^− \rightleftharpoons Cu(s)\) is +0.3419 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = 0.7996 - 0.3419 = 0.46\] Since \(\mathrm{E^\circ_{cell}}\) is positive, the reaction will occur spontaneously as drawn.
    2. The standard reduction potential of the reduction half reaction \(\ce{Fe^3+(aq)}\) is 0.77 V
      The standard reduction potential of the oxidation half reaction \(\ce{Na(s)}\) is 2.71 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = 0.77 + 2.71 = 3.48\] \(\mathrm{E^\circ_{cell}}\) is positive, Yes it will happen spontaneously as drawn.
    3. The standard reduction potential of the reduction half reaction \(Zn^{2+}(aq) + 2e^− \rightleftharpoons Zn(s)\) is –0.7618 V
      The standard reduction potential of the oxidation half reaction \(I_2(aq) + 2e^− \rightleftharpoons 2I^-(s)\) is 0.5355 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} =-0.7618 - 0.5355 = -1.2973\] Since \(\mathrm{E^\circ_{cell}}\) is negative, the reaction will not happen spontaneously as drawn.

    See Using Redox Potentials to Predict the Feasibility of Reactions for help.

    Q1.6

    1. \(\ce{2Ag+(aq) + Cu(s) \rightarrow 2Ag(s) + Cu^2+(aq)}\)
      The standard reduction potential of the reduction half reaction \(\ce{Ag+(aq)}\) is +0.8 V
      The standard reduction potential of the oxidation half reaction \(\ce{Cu(s)}\) is +0.34 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = \mathrm{0.8 -0.34 = 0.46 V}\]
    2. \(\ce{2Fe^3+(aq) + Cu(s) \rightarrow 2Fe^2+(aq) + Cu^2+(aq)}\)
      The standard reduction potential of the reduction half reaction \(\ce{2Fe^3+(aq)}\) is 0.77 V
      The standard reduction potential of the oxidation half reaction \(\ce{Cu(s)}\) is +0.34 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = \mathrm{0.77 -0.34 = 0.43 V}\]
    3. \(\ce{PbO2(s) + Mn^2+(aq) \rightarrow Pb^2+(aq) + MnO2(s)}\).
      The standard reduction potential of the reduction half reaction \(\ce{PbO2(s)}\) is 1.46 V
      The standard reduction potential of the oxidation half reaction \(\ce{Mn^2+(aq)}\) is +1.229 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = \mathrm{1.229 - 1.46 = -0.231 V}\] As drawn, this cell is non-spontaneous. The cell diagram should be flipped to represent a galvanic cell.
    4. \(\ce{Cu(s) + O2(g) + 2H3O^+(aq) \rightarrow H2O2(aq) + Cu^2+(aq)}\)
      The standard reduction potential of the reduction half reaction \(\ce{O2(g)}\) is 0.695 V
      The standard reductio npotential of the oxidation half reaction \(\ce{Cu(s)}\) is +0.34 V
      \[\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\] \[\mathrm{E^\circ_{cell}} = \mathrm{0.695 -0.34 = 0.355 V}\]

    Q1.7

    not given

    Q1.8

    a) -3; [x + 3(1) = 0] b) 0 c) +4; [x +2(-2) = 0] d) +5; [x + 3(-2) = -1]

    Q1.9

    a) Br2 b) I c) Br d) Br2 e) I-

    Q1.10

    First determine whether substances may be oxidized or reduced, then look up Eoox for reducing agents and Eored for oxidizing agents and list in order.

    • Oxidizing agents (are reduced): H+ (Eored = 0 V).
    • Reducing agents (are oxidized): Ni (Eoox = +0.23 V), Au (Eoox = -1.50 V), Mg (Eoox = +2.37 V).

    Both: Sn2+ (Eored = -0.14 V; Eoox = -0.15 V), Fe2+ (Eored = -0.44 V, Eoox = -0.77 V), Cl2 (Eored = 1.36 V, Eoox = -1.47 V,).

    • Oxidizing agents (list in order of decreasing Eored): Cl2 > H+ > Sn2+ > Fe2+
    • Reducing agents (list in order of decreasing Eoox): Mg > Ni > Sn2+ > Fe2+ > Cl2 > Au