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$\pu{Cu^{2+}}$ ions react with $\pu{Fe^{2+}}$ ions according to the following reaction: $$\ce{Cu^+2 + 2Fe^2+ <=> Cu + 2Fe^3+}$$ At equilibrium, the concentration of $\pu{Cu^{2+}}$ ions is not changed by the addition of which ion $\pu{(Cu/Cu^{2+}/Fe^{2+}/Fe^{3+})}$.
Explain the result.

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  • $\begingroup$ This appears to be a homework question, please share your thoughts and attempts towards the solution. If you receive useful answers, consider accepting one. The homework tag is deprecated, please don't use it. $\endgroup$ – Martin - マーチン Nov 27 '18 at 21:23
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Obviously $\ce{Cu}$, as it is a solid and has no effect on Equilibrium constant and thus no effect on altering equilibrium concentration of $\ce{Cu^2+}$.

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    $\begingroup$ While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy. $\endgroup$ – Martin - マーチン Nov 27 '18 at 16:41
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Cu is the answer. As the concentration of any solid component is fixed and does not effect the Kc value at all.

let concentration = $C$$ , mass = $$m$$ , molar mass = $$M$$ , volume= $$V$$ , density=$$d$$ and no. of moles = $$n$;

then$$C=\frac{n}{V}$$ and $$n=\frac{m}{M}$$ and by the 2 equations we get

$$C=\frac{m}{MV}$$ which equals $$C=\frac{d}{M}$$ and as $$d,V$$ are constant for a solid as Cu, adding more substance would not change its concentration.

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    $\begingroup$ While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy. $\endgroup$ – Martin - マーチン Nov 27 '18 at 16:41

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