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9 November, 16:39

Given: logK=nE∘0.0592 What is the value K for this redox reaction? Zn2 + (aq) + 2 Cl - (aq) → Zn (s) + Cl2 (

g. Ecell = - 2.12 V

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  1. 9 November, 16:50
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    K = 2.4 * 10^ (-72)

    Step 1. Determine the value of n

    Zn^ (2+) + 2e^ (-) → Zn

    2Cl^ (-) → Cl_2 + 2e^ (-)

    Zn^ (2+) + 2Cl^ (-) → Zn + Cl_2

    ∴ n = 2

    Step 2. Calculate K

    logK = nE°/0.0592 V = [2 * (-2.12 V) ]/0.0592 V = - 71.62

    K = 10^ (-71.62) = 2.4 * 10^ (-72)
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