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Complete the equation:

$$\ce{Cu + FeCl3 -> ? + ?}$$

I wrote $\ce{CuCl2}$ and $\ce{Fe}$, but it turned out to be $\ce{Cu}$ and $\ce{FeCl2}$. I really wonder why.

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closed as off-topic by Mithoron, Todd Minehardt, airhuff, Jon Custer, Jannis Andreska Dec 23 '17 at 15:32

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  • $\begingroup$ Reading this q it looks like a typo was made. Should the answer be $\ce{CuCl_2}$ and $\ce{FeCl_2}$? $\endgroup$ – Oscar Lanzi Dec 22 '17 at 13:36
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Take a look at standard reduction potentials at $\pu{25 ^\circ C}$:

$$ \begin{array}{clr} \hline \text{Reaction} & E^\circ~(\pu{V}) \\ \hline \begin{align} \ce{Fe^3+ + e- &-> Fe^2+} \\ \ce{Cu^2+ + 2e- &-> Cu} \\ \ce{Fe^3+ + 3e- &-> Fe} \\ \ce{Fe^2+ + 2e- &-> Fe} \end{align} & \begin{array}{r} +0.77 \\ +0.34 \\ -0.04 \\ -0.41 \end{array} \\ \hline \end{array} $$

A more positive potential means a greater tendency to be reduced, so $\ce{Cu(II)}$ is easier to reduce to copper metal than either $\ce{Fe(II)}$ or $\ce{Fe(III)}$ being reduced to iron metal. So, copper metal can't force the formation of iron metal.

But $\ce{Fe(III)}$ can be reduced just to $\ce{Fe(II)}$ at a relatively high potential, high enough to allow the copper to be in its oxidized form as $\ce{Cu(II)}$. So, we can have $\ce{Cu(II)}$ and $\ce{Fe(II)}$ at equilibrium and thus we can get the corresponding chlorides $\ce{CuCl2}$, $\ce{FeCl2}$.

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    $\begingroup$ Are you comfortable with the edit? $\endgroup$ – Nilay Ghosh Dec 22 '17 at 5:05
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    $\begingroup$ It should be $E^0$ in table. Otherwise very nice. $\endgroup$ – MaxW Dec 22 '17 at 5:06
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    $\begingroup$ Yup my mistake. But geeze, super picky with the markup. $\endgroup$ – MaxW Dec 22 '17 at 13:27
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    $\begingroup$ @andselisk - Nope. If you believe in entropy, then disorder rules. ;-) $\endgroup$ – MaxW Dec 22 '17 at 15:51
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    $\begingroup$ Iron (III) is reduced to iron(II) at a potential well below the 1.36 V listed for the chlorine/chloride equilibrium. The chlorine thus remains reduced. $\endgroup$ – Oscar Lanzi Dec 10 '18 at 19:32

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