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20 February, 10:55

If the concentration of sn2 + in the cathode compartment is 1.50 m and the cell generates an emf of 0.24 v, what is the concentration of pb2 + in the anode compartment?

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  1. 20 February, 14:43
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    E° cell = E° cathode - E° anode

    =-0.14 + 0.13

    = - 0.01V

    Q = [pb⁺²]/[Sn⁺²] = Q = [pb⁺²]/1.50]

    E = E° = - 0.05916/n log Q

    0.24 = - 0.01 = 0.05916/2 log Q

    Log Q = - 8.452

    Q = + 3.53 * 10⁻⁹

    (pb + 2) = 3.53 * 10⁻⁹ * 1.5

    =5.297 * 10⁻⁹M

    [pb⁺²] = 5.3 * 10⁻⁹M
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