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26 February, 11:17

A voltaic cell employs the following redox reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate the cell potential at 25 ∘C under each of the following conditions. A voltaic cell employs the following redox reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate the cell potential at 25 ∘C under each of the following conditions.

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  1. 26 February, 12:19
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    1st condition: E = 2.25 V

    2nd condition = E = 2.41 V

    Explanation:

    A voltaic cell employs the following reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate cell potential at 25 ∘C under each condition?

    1st condition: [Fe3+] = 1.6 * 10^-3 M; [Mg2+] = 2.05 M

    2nd condition: [Fe3+] = 2.05 M; [Mg2+] = 1.6*10^-3 M

    Step 1: Data given

    Temperature = 25 °C

    The reduction potential for Fe+3 → Fe (s) = - 0.037

    The reduction potential for Mg+2 + 2e - → Mg (s) = + 2.356

    Step 2: The balanced equation

    2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq)

    Step 3:

    Fe+3 + 3e-→ Fe (s) - 0.037

    Mg (s) → Mg+2 + 2e - + 2.356

    We have to multiply the first equation by 2 and thesecond equation by 3

    2Fe+3 + 6e-→ 2Fe (s) (-0.037)

    3Mg (s) → 3Mg+2 + 6e - (+2.356)

    2Fe+3 + 3Mg → 2Fe (s) + 3Mg+2 E° = + 2.319

    Step 4: Calculate the cell potential for the first condition

    Nernst Equation: E = E° - 0.0592/n * log [Mg+2]³ / [Fe+3]²

    ⇒with n = the number of electrons = 6

    ⇒with [Mg+2] = 2.05 M

    ⇒with [Fe+3] = 1.6 * 10^-3 M

    E = + 2.319 - 0.0592/6 * log ((2.05) ³ / (1.6 x 10^-3) ²)

    E = + 2.319 - 0.0592/6 * 6.53

    E = 2.319 - 0.0099 * 6.53

    E = 2.25 V

    Step5 : Calculate the cell potential for the second condition

    Nernst Equation: E = E° - 0.0592/n * log [Mg+2]³ / [Fe+3]²

    ⇒with n = the number of electrons = 6

    ⇒with [Fe^3+] = 2.05 M

    ⇒with [[Mg+2] = 1.6 * 10^-3 M

    E = + 2.319 - 0.0592/6 * log ((1.6 x 10^-3) ³ / (2.05) ²)

    E = + 2.319 - 0.0592/6 * (-9.0)

    E = 2.41 V
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