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Anonymous asked in Science & MathematicsChemistry · 1 month ago

Cadmium metal questions help?

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  • 1 month ago
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    a. ∆G° = - nF E°

    ∆G° = -2 (96485 J/V mol) (-0.40 V) = +77 kJ/mol

    b. i. Cd(s) + Pb2+(aq) --> Cd2+(aq) + Pb(s)

    ii. E° = 0.40 + (-0.13) = 0.27 V

    iii. E = E° - RT/nF ln [Pb2+]/[Cd2+]

    iv. E = 0.27 V - (8.314 J/molK)(298.15 K) / 2mol0 (96485 J/V mol) ln (0.5 / 0.01)

    E = 0.22 V

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