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19 June, 02:24

Calculate the e. m. f of a cell if 0.01M solution of each of Cd2 + and Ag + are used?

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  1. 19 June, 02:28
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    1.20 V

    Explanation:

    The concentration of both solutions as stated in the question is 0.01M

    However, based of reduction potential; Cd^2 + = - 0.40V and Ag^ + = 0.80V. Silver will be the cathode while cadmium will be the anode.

    Hence

    E°cell = E°cathode - E°anode

    E°cell = 0.80 - (-0.40) = 1.20 V

    But n=2 moles of electrons

    From Nernst Equation:

    E = E°cell - 0.0592/n log [Red]/[Ox]

    E = 1.20 - 0.0592/2 log (0.01/0.01)

    E = 1.20 V
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