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QUESTION: Calculate the standard free-energy change (deltaG°) for the following reaction at 25 °C in kJ.

2Ag^3+(aq) + 3Zn(s)<==> 2Au(s) + 3Zn^2+(aq)

here's what i have so far:
first i wrote out the half reactions:
Au^3+(aq) + 3e^- --> Au(s) = +1.52*2
Zn(s) --> Zn^2+(aq) + 2e^- = +0.76*3

so E°cell= (1.52 * 2) + (0.76 * 3)=5.32V

then i plugged it into the equation..
deltaG°= -nFE°cell, where F=96,485 C/mol

am i on the right track? how do i find n so i can solve for deltaG°?

  • chemistry -

    There is one error, you have to have each equation have the same number of electrons. You did, however, get the E values right.

    The correct equations should be like this:

    2Au^3+(aq) + 6e^- --> 2Au(s) = +1.52*2
    3Zn(s) --> 3Zn^2+(aq) + 6e^- = +0.76*3

    "n" refers to the amount of electrons in the reaction. Since now they are equal in both reactions, n = 6.

  • chemistry -

    There is a second error. The Ecell is not 5.32 v.
    Ecell = 1.52 + 0.76 = ?
    You do NOT multiply by the coefficients 2 and 3. Yes, n will be 6.

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