Skip to main content
Chemistry LibreTexts

Solutions

  • Page ID
    11051
  • 1. Write the cell diagram for the Cu+2/Cu and Al+3/Al half-cells and calculate the Eocell for the Cu+2/Cu and Al+3/Al half-cells.

    Write half-reactions with more negative on top:

    Al+3 (aq) + 3 e- -------> Al (s)

    Eo= -1.66 V

    Cu+2 (aq) + 2 e- -------> Cu (s)

    Eo= +0.34 V

    Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of Eo

    Al (s) -------> Al+3 (aq) + 3 e-

    Eo= +1.66 V (anode)

    Cu+2 (aq) + 2 e- -------> Cu (s)

    Eo= +0.34 V (cathode)

    Add cell potentials of half-reactions to determine Eocell

    Eocell= +2.00 V

    cell diagram (anode first):

    Al (s) | Al+3 (aq, 1M) || Cu+2 (aq, 1 M) | Cu (s)

    2. Write the cell diagram and calculate the Eo for the Zn/ Zn+2 and Ca/Ca+2 half-cells.

    Eo= 2.11 V
    cell diagram:
    Ca (s) | Ca+2 (aq, 1M) || Zn+2 (aq, 1 M) | Zn (s)

    3. a) Will I- react with Br2? Yes
    b) Will Ni+2 react with Br2? No
    c) Will Ni react with Br2? Yes
    Will Co+2 react with Br2 to give Co+3? No, reacts with Co+2 to give Co (s)

    4. Rank the following ions from strongest to weakest oxidizing agents:

    Eo , V
    Ag+
    strongest
    + 0.80
    Pb+2
    - 0.13
    Al+3
    - 1.66
    Sr+2
    - 2.89
    Li+
    weakest
    - 3.05

    5. Determine the E for the Ag+/Ag and Sn+2/Sn half-cells if the [Ag+] = 1.0 M and the [Sn+2] = 0.25 M.

    Write half-reactions with more negative on top:
    Sn+2 (aq) + 2 e- -------> Sn (s)
    Eo= -0.14 V
    Ag+ (aq) + 1 e- -------> Ag (s)
    Eo= +0.80 V
    Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of Eo
    Sn (s) -------> Sn+2 (aq) + 2 e-
    Eo= +0.14 V (anode)
    Ag+ (aq) + 1 e- -------> Ag (s)
    Eo= +0.80 V (cathode)
    Add cell potentials of half-reactions to determine Eocell Eocell= +0.94 V
    To determine E, use Nernst equation: E = Eo-(0.0591/n)log Q
    E = Eo-(0.0591/n)log Q; n = 2 (make e- lost = e- gained in the half-reactions);
    Q = [Sn+2]/ [Ag+]2([Sn+2] is on the product side, [Ag+] is on the reactant side after switching half-reactions)
    E = 0.94V -(0.0591 V/2)log {0.25M Sn+2/ (1.00 Ag+)2} E = 0.958 V

    6. Is the reaction below spontaneous? no Calculate the Eo. Eo = - 0.47 V

    Pb+2 + Cu ------> Pb + Cu+2

    7. Determine the \(\Delta{G}^o\) for the Ag+/Ag and Sn+2/Sn half-cells.

    missing...anyone want to do this?

    8. Calculate \(\Delta{G}^o\) for Cu+2/Cu and Mn+2/Mn half-cells. Is this reaction spontaneous?

    Eo = 1.52 V;

    \(\Delta{G}^o = -293,360 J\)

    or

    \[- 293.4 \, kJ\]

    Yes, this reaction is spontaneous as written.


    5. Calculate \(\Delta{G}\) for Ni+2/Ni and Fe+2/Fe half-cells if the [Ni+2] = 0.25 M and the [Fe+2] = 0.50 M.

    Eo = 0.19 V; E = 0.181 V; DG = - 34956 J or - 35.0 kJ

    6. Determine the \(\Delta{G}\) for the Al+3/Al and Ni+2/Ni half-cells.if the [Al+3] = 0.50 M and the [Ni+2] = 0.25 M.

    Eo = 1.41 V; E = 1.398 V; \(\Delta{G}\) = - 809442 J or - 809.4 kJ