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19 February, 12:16

What is the value (in V) of Eºcell for the following reaction? Zn2 + (aq) + Pb (s) → Zn (s) + Pb2 + (aq)

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  1. 19 February, 15:58
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    - 0.63 V.

    Explanation:

    From the question, the reaction is as follows -

    Zn²⁺ (aq) + Pb (s) → Zn (s) + Pb²⁺ (aq)

    From the above reaction,

    Zn²⁺ is reduced to Zn, as the oxidation state changes from + 2 to 0

    and,

    Pb is oxidized to Pb²⁺, as the oxidation state changes from 0 to + 2.

    In a cell, the process of oxidation takes place in the Anode, and reduction takes place in Cathode.

    Hence, the half cell reaction taking place at anode and cathode is as follows -

    Cathode reaction: Zn²⁺ (aq) + 2e⁻ → Zn (s); E⁰ Zn²⁺ / Zn = - 0.76 V.

    Anode reaction: Pb (s) → Pb²⁺ (aq) + 2e⁻; E⁰ Pb²⁺ / Pb = - 0.13 V

    The E⁰cell is calculated as the difference in the E⁰cathode and E⁰anode.

    E⁰cell = E⁰cathode - E⁰anode = - 0.76 - (-0.13) = - 0.63 V.
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