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26 September, 19:32

Calculate the equilibrium constant at 25 ∘C for the reaction Fe (s) + 2Ag + (aq) →Fe2 + (aq) + 2Ag (s)

Standard Reduction Potentials at 25 ∘C Fe2 + (aq) + 2e-→Fe (s) E∘ = - 0.45 V Ag + (aq) + e-→Ag (s) E∘ = 0.80 V

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  1. 26 September, 19:43
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    1.7 * 10 ^42

    Explanation:

    Using Nernst equation

    E°cell = RT/nF Inq

    at equilibrium

    Q=K

    E°cell = 0.0257 / n Ink = 0.0592/n log K

    Fe2 + (aq) + 2e-→Fe (s) E∘ = - 0.45 V

    Ag+aq) + e-→Ag (s) E∘ = 0.80 V

    Fe (s) + 2Ag + (aq) →Fe2 + (aq) + 2Ag (s)

    balance the reaction

    Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v

    2 Ag⁺ + 2e⁻ → 2Ag

    n = 2 moles and K = equilibrium constant

    E° cell = 0.80 + 0.45 = 1.25 V

    E° cell = (0.0592 / n) log K

    substitute the value into the equations and solve for K

    (1.25 * 2) / 0.0592 = log K

    42.23 = log K

    k = 10^ 42.23

    K = 1.7 * 10 ^42
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