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An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode.

(a) What is the potential of this cell at 25°C if the aluminum electrode is placed in a solution in which [Al3+] = 7.3 10-3 M?
(b) When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25°C is 1.54 V. Calculate [Al3+] in the unknown solution. (Assume Al is oxidized.)

Question ID

April 1, 2011 4:15am UTC





2 answers

  1. a) Use the Nernst equation to calculate the Ni/Ni^+2 half cell potential. Do the same for the Al^+3/Al half cell potential.
    Add the two to obtain Ecell. I get -1.45v if I didn't make an error. You need to confirm that. Also you need to write the balanced equation for the cell AS WRITTEN, keeping in mind that it is negative which means it will not proceed spontaneously as written and as drawn.
    For part b, the directions tell you to make Al the oxidized electrode which means the cell is rearranged and will proceed spontaneously; you will need to adjust the signs of the numbers (as well as the equation) to fit.
    Ecell = 1.54v
    n = 6 electrons.
    You can fill in the values for components in Q.
    Ecell = Eocell -(0.0592/n)log Q

    Answer ID

    April 2, 2011 3:28am UTC



  2. (a) Well, if you have a nickel metal electrode and aluminum metal electrode, and they're separated by a porous disk, then I guess you could say they've got a "charged" relationship. But seriously, to calculate the potential of the cell, you'll need to use the Nernst equation:

    Ecell = Eºcell - (RT/nF) * ln(Q)

    Where Ecell is the potential of the cell, Eºcell is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C = 298 K), n is the number of electrons transferred in the cell reaction, F is Faraday's constant (96500 C/mol), and Q is the reaction quotient.

    In this case, the overall cell reaction is:

    Al(s) + Ni2+(aq) -> Al3+(aq) + Ni(s)

    To simplify things, let's assume the reaction is balanced and that two electrons are transferred.

    Now, you'll need to calculate the reaction quotient Q using the concentrations given. And then plug it all into the Nernst equation to find the potential of the cell.

    (b) Oh, you've got an unknown solution with an unknown [Al3+], huh? Well, you'll still need to use the Nernst equation to find the potential of the cell. This time, however, you know the measured cell potential and you need to calculate the unknown [Al3+].

    So, you'll set up the Nernst equation again, but this time, you'll rearrange it to solve for Q:

    Q = exp((Eºcell - Ecell) * (nF/RT))

    Once you have Q, you'll use it to calculate the unknown [Al3+] using the balanced cell reaction and the number of electrons transferred.

    Now, go forth and calculate those cell potentials and concentrations. Just remember, even if the math gets a bit shocking, don't get too amped up!

    Answer ID

    September 26, 2023 6:46am UTC



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