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28 June, 18:43

Using standard electrode potentials calculate δg∘rxn and use its value to estimate the equilibrium constant for each of the reactions at 25 ∘c.

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  1. 28 June, 18:47
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    The value of K for the equation Pb2 + (aq) + Mg (s) - >Pb (s) + Mg2 + (aq) is K = 2.35 X10^75 The value of k for the equation Br2 (l) + 2Cl - (aq) - >2Br - (aq) + Cl2 (g) is

    K = 1.26 X10^-10

    The value of k for the equation MnO2 (s) + 4H + (aq) + Cu (s) - > Mn2 + (aq) + 2H2O (l) + Cu2 + (aq) Eo is K = 1.17 X10^30

    Explanation:

    We have the formula,

    By the Nernst equation:

    E = Eo - (0.0592/n) (log Q)

    Put E=0which at equilibrium becomes

    0 = Eo - (0.0592/n) (log K)

    which re-write the equation as

    log K = Eo / (0.0592/n)

    1. Pb2 + (aq) + Mg (s) - >Pb (s) + Mg2 + (aq)

    Finding the value of Eo we get,

    Pb2 + (aq) - -> Pb (s) Eo = - 0.125

    Mg (s) - > Mg2 + (aq) Eo = + 2.356

    Pb2 + (aq) + Mg (s) - >Pb (s) + Mg2 + (aq) Eo = 2.231

    Subsitute in this formula we get

    log K = Eo / (0.0592/n)

    log K = 2.231 / (0.0592/2)

    log K = 75.37

    K = 2.35 X10^75

    The value of K for the equation Pb2 + (aq) + Mg (s) - >Pb (s) + Mg2 + (aq) is K = 2.35 X10^75

    2. Br2 (l) + 2Cl - (aq) - >2Br - (aq) + Cl2 (g) Eo = - 0.293

    Similarly, for the other equation follow the same procedure

    The equation is written in the form of non-spontaneous

    2Cl - (aq) - > Cl2 Eo = - 1.358

    Br2 (l) - -> 2Br - (aq Eo = 1.065

    Br2 (l) + 2Cl - (aq) - >2Br - (aq) + Cl2 (g) Eo = - 0.293

    log K = - 0.293 / (0.0592/2)

    log K = - 9.90

    K = 1.26 X10^-10

    The value of k for the equation Br2 (l) + 2Cl - (aq) - >2Br - (aq) + Cl2 (g) is

    K = 1.26 X10^-10

    3. MnO2 (s) + 4H + (aq) + Cu (s) - > Mn2 + (aq) + 2H2O (l) + Cu2 + (aq)

    Similarly, for the other equation follow the same procedure

    The equation is written in the form of non-spontaneous

    MnO2 (s) - > Mn2 + Eo = 1.23

    Cu (s) - > Cu2 + Eo = - 0.340

    MnO2 (s) + 4H + (aq) + Cu (s) - > Mn2 + (aq) + 2H2O (l) + Cu2 + (aq) Eo = 0.89

    log K = 0.89 / (0.0592/2)

    log K = 30.067

    K = 1.17 X10^30

    The value of k for the equation MnO2 (s) + 4H + (aq) + Cu (s) - > Mn2 + (aq) + 2H2O (l) + Cu2 + (aq) Eo is K = 1.17 X10^30
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