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Since Gibbs potential is a state function, it should be additive, regardless of the path chosen to go from a state ($\ce{Cu^{2+}}$) to another ($\ce{Cu}$).

While providing a solution to another question, I noticed that the Gibbs energies are not adding up. Consider the following:

$$ \begin{align} \ce{Cu^{2+} + 2e- &-> Cu} \hspace{20pt} &\Delta G\\ \ce{Cu^{2+} + 2e- &-> Cu+} \hspace{20pt} &\Delta G_1\\ \ce{Cu^{2+} + 2e- &-> Cu} \hspace{20pt} &\Delta G_2 \end{align} $$

I calculated the Gibbs potential changes:

On-step and serial reduction of cupric ion

I verified the reduction potentials from ChemLibreTexts, which seem close to the accurate values. What is causing this discrepancy?

ananta
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1 Answers1

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I had actually misplaced a decimal point and, upon correcting for the accurate values, the discrepancy was resolved.

one-step and serial reduction of cupric ion

ananta
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    You may also want to consider significant figures. Since the potentials with copper(II) are given to the nearest 10 mV, the free energies are required to agree only to that level of accuracy; the remaining difference between 680 and 677 mV is not significant. – Oscar Lanzi May 07 '23 at 09:14
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    +3.4 V should catch your eyes, such strong oxidants are not even in fluorine chemistry. :-) – Poutnik May 07 '23 at 09:20