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N= 6 and E° = -1.428v for a redox reaction with E = 0.188v. Calculate pQ. Assume T = 298.15K.? HELP ME!!!?

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n = 6 and E° = -1.428v for a redox reaction with E = 0.188v. Calculate pQ. Assume T = 298.15K.

CAN ANYONE HELP ME PLZ!!!

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  1. Calculate it yourself.

    ΔE = ΔEo - (RT/nF)lnQ, where

    T= temperature = 298315

    E = 0.188

    Eo = -1.428

    R = 8.314JK-1mol-1 <-- constant

    n = 6

    F = 96500 Jmol-1 <-- constant

    Plug and chug.... you should get 163.8

    [Answer: see above]


  2. nernst:

    E= Eo - (0.0592 / n) (log Q)

    0.188 = - 1.428 - (0.0592/6) (logQ)

    1.428 + 0.188 = - (0.009867)(logQ)

    log Q = - 1.616 / 0.009867

    logQ = - 163.8

    pQ = - log Q = + 163.8

    your answer = pQ = + 163.8
You're reading: N= 6 and E° = -1.428v for a redox reaction with E = 0.188v. Calculate pQ. Assume T = 298.15K.? HELP ME!!!?

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