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The standard reduction potential for the half cell $$\ce{FeCO3_{(s)} + 2e- <=> Fe_{(s)} + CO3^2-}\quad\quad(E^\circ=-0.756\ \mathrm V)$$ and the standard reduction potential for the half cell $$\ce{Fe^2+ + 2e- <=> Fe_{(s)}}\quad\quad\quad\quad\quad\quad(E^\circ=-0.44\ \mathrm V)$$ are as given.

From these, I calculated the $E^\circ$ for the reaction $$\ce{FeCO3_{(s)} <=> Fe^2+ + CO3^2-}\quad\quad(E^\circ=-0.31\ \mathrm V)$$ How do I interpret this number because the dissolution is not a redox reaction? How do I calculate $\Delta G^\circ$ for the reaction since $n=0$ in the dissolution reaction ($\Delta G^\circ=-R\cdot T\cdot\ln K$ will not work)?

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  • $\begingroup$ What? Where is Fe (0)? $\endgroup$
    – Alchimista
    Jan 1, 2020 at 9:23
  • $\begingroup$ @Alchimista In the dissolution reaction, Fe stays in the +2 oxidation state on both sides of the equation so n = 0. $\endgroup$ Jan 1, 2020 at 13:04
  • $\begingroup$ I don't get it then... It is like wanting the circumference of a square or even worse. $\endgroup$
    – Alchimista
    Jan 2, 2020 at 8:27

1 Answer 1

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The standard reduction potential for the half cell $$\ce{FeCO3_{(s)} + 2e- <=> Fe_{(s)} + CO3^2-}\quad\quad(E^\circ=-0.756\ \mathrm V)$$ and the standard reduction potential for the half cell $$\ce{Fe^2+ + 2e- <=> Fe_{(s)}}\quad\quad\quad\quad\quad\quad(E^\circ=-0.44\ \mathrm V)$$ are as given.

If these are half cells, you have elemental iron forming at the cathode and being used up at the anode. To express this, you could write the redox reaction as:

$$\ce{FeCO3(s)_{cat} + Fe(s)_{an} <=> Fe(s)_{cat} + Fe^2+(aq)_{an} + CO3^2-(aq)_{cat}}\tag{1}$$

with subscripts "cat" and "an" designating cathode and anode species, respectively.

$n$ will be equal to 2, and now calculating the Gibbs energy will make perfect sense.

From these, I calculated the $E^\circ$ for the reaction $$\ce{FeCO3_{(s)} <=> Fe^2+ + CO3^2-}\quad\quad(E^\circ=-0.31\ \mathrm V)$$

That does not make any sense. If you have the iron carbonate just dissociating in a one-pot reaction, there is no electron transport (where would you attach your voltmeter?) and assigning a reduction potential is wrong.

However, the Gibbs energy of reaction of the one-pot reaction will be the same as that of the electrochemical cell described in reaction (1).

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    $\begingroup$ One has to take the last reaction as the net reaction of the cell, based on the half-cell reactions above, not as a single place reaction. $\endgroup$
    – Poutnik
    Aug 30, 2021 at 7:01

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