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8 April, 08:32

A cell consists of a gold wire and a saturated calomel electrode (S. C. E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire is connected to the positive end of a potentiometer, and the S. C. E. is connected to the negative end of the potentiometer. What is the half‑reaction that occurs at the Au electrode? Include physical states.

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  1. 8 April, 11:45
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    The half reaction that occurs at the Au electrode is 1.64

    Explanation:

    Half cell reaction at 'Au' electrode

    We have the equation,

    Au + (aq) + e - - --> Au (s)

    Given the concenration of AuNO3=0.150 M

    [Au + ] = 0.150 m

    From the equation,

    Au + (aq) + e - - --> Au (s)

    Standard electric potential = Eo = 1.69 volt

    Solving the problem using the Nerst equation

    E cell = E0 cell - 2.303 RT / nF log Qc

    Where,

    T = 298 K

    n = no of electron lost or gained

    F = faraday's constant = 965000/mole

    R = universal constant = 8.314 J / K / Mole

    Substitue the values we get

    E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

    E cell = 1.69 volt - 0.05 g/1 0.824

    E cell = 1.64

    The half reaction that occurs at the Au electrode is 1.64
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