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6 July, 18:53

The voltage generated by the zinc concentration cell described by, zn (s) |zn2 (aq, 0.100 m) ||zn2 (aq,? m) |zn (s), is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

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  1. 6 July, 19:07
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    The concentration cell is:

    Zn (s) / Zn²⁺ (aq, 0.100 M) / / Zn²⁺ (aq, x M) / Zn (s)

    voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V

    - In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.

    - Oxidation takes place at the anode and reduction takes place at the cathode.

    so [Zn²⁺]oxidation = 0.100 M

    [Zn²⁺] reduction = x M

    From Nernst equation:

    Ecell = - 0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction

    Number of electrons, n = 2. Substitute and solve for x:

    16 x 10⁻³ V = - 0.0592 / 2 log (0.100 / x)

    log 0.100 / x = - 0.54

    0.100 / x = 0.288

    x = 0.347

    So the concentration of Zn²⁺ at the cathode = 0.406
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