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29 April, 04:15

An electrochemical cell is constructed such that on one side a pure lead electrode is in contact with a solution containing Pb2 + ions at a concentration of 0.002 M. The other cell half consists of a pure nickel electrode that is immersed in a solution of Ni2 + ions having a concentration of 0.4 M. Given that the standard electrode potentials for lead and nickel are - 0.126 and - 0.250 V, respectively, at what temperature will the potential between the two electrodes be - 0.012 V?

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  1. 29 April, 06:58
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    -490.7 K

    Explanation:

    Given:

    [Ni^2+] = 0.4 M

    [Pb^2+]=0.002 M

    ∆V = - 0.012 V

    VNi = - 0.250V

    VPb = - 0.126V

    F = 96500 C

    R = 8.314 JK-1 mol-1

    n = 2

    From

    T = - nF/R [∆V - (VNi-VPb) / ln [Pb2+]/[Ni2+]]

    T = 2 (96500) / 8.314[ (-0.012) - (-0.250) - (-0.126)) / ln[0.002]/[0.4]

    T = 23213.856 (0.112 / (-5.298))

    T = - 490.7 K
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