What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Sn(s) and Pb2 (aq) to give Pb(s) and Sn2 (aq) Use

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What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Sn(s) and Pb2 (aq) to give Pb(s) and Sn2 (aq) Use the reduction potential values for Sn2 (aq) of -0.14 V and for Pb2 (aq) of -0.13 V Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly

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Thông Đạt 4 years 2021-08-31T13:43:06+00:00 1 Answers 8 views 0

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  1. ngocdiep
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    2021-08-31T13:44:07+00:00
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    Answer:

    The value of the equilibrium constant at 25°C is 2.177

    Explanation:

    Based on the half-reactions:

    Sn(s) → Sn²⁺ + 2e⁻  E° = -(Reduction potential) = 0.14V

    Pb²⁺ + 2e⁻ → Pb(s) E° = -0.13V

    Sn(s) + Pb²⁺ → Pb(s) + Sn²⁺ E° = 0.14V – 0.13V = 0.01V

    Using:

    Eºcell = 0.0592/n (log K) At 25°C

    Where E° cell = 0.01V

    n are moles of electrons = 2

    K is equilibrium constant

    Replaing:

    0.01V = 0.0592/2 * (logK)

    0.3378 = log K

    2.177 = K

    The value of the equilibrium constant at 25°C is 2.177

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )