Reduction Potential Question. Calculating delta G

JL asked:

What is the value of DGo in kJ at 25 oC for the reaction between the pair:

Cu(s) and Cr3 (aq) to give Cr(s) and Cu2 (aq)

Use the reduction potentials for Cr3 (aq) is -0.74 V and for Cu2 (aq) is 0.34 V.

I know that Cu is being reduced since its reduction potential is higher but then that makes Cr 0.74 for the oxidation potential. So E is positive and doesn’t that make G negative since G = -nFE?

I keep getting -6E2. How would it be positive?

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One Response to “Reduction Potential Question. Calculating delta G”

  • Chem Man says:

    Cu —–> Cu2+ + 2e- Eo = -0.34 V
    Cr3+ + 3e- —-> Cr Eo = -.073 V
    3Cu + 2Cr3+ —-> 3Cu2+ + 2Cr Eocell = -1.07 V
    This reaction will not occur spontaneously because the Eocell value is negative.

    dGo = -nFEocell = -6 mol x 96,500 C/mol x (-1.07 J/C) = 620,000 J = 620 kJ

    dGo is positive, meaning that no useful work can be obtained from this cell. The reason is that it is a nonspontaneous reaction.

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