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17 July, 11:38

What is the cell potential for the reaction mg (s) + fe2 + (aq) →mg2 + (aq) + fe (s) at 71 ∘c when [fe2+] = 3.30 m and [mg2+] = 0.310 m?

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  1. 17 July, 14:19
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    The cell potential for this reaction is 1.955 V

    Explanation:

    Step 1: Data given

    Molarity of Fe^2+] = 3.30 M

    Molarity of [Mg^2+] = 0.310 M

    Temperature = 71°C

    E∘ standard potential = 1.92 V

    Step 2: The balanced equation

    Mg (s) + Fe2 + (aq) → Mg^2 + (aq) + Fe (s)

    Step 3: Nernst equation

    E = Eo - RT/nF ln Q

    ⇒with E° = the standard potential = 1.92 V

    ⇒with R = 8.314 J/K*mol

    ⇒with T = the temperature = 344 K

    ⇒with n = the number of electrons transfered = 2

    ⇒with F = Constant of Faraday F = 96500 C/mol

    ⇒ with Q = [Mg^2+]/[Fe^2+] = 0.310 / 3.30 = 0.094

    E = 1.92 - 8.314*344 / (2*96500) * ln (0.094)

    E = 1.92 - 8.314*344 / (2*96500) * (-2.365)

    E = 1.955 V

    The cell potential for this reaction is 1.955 V
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