# Chemical Equilibrium in case of copper(II) ions [closed]

$$\pu{Cu^{2+}}$$ 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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Obviously $$\ce{Cu}$$, as it is a solid and has no effect on Equilibrium constant and thus no effect on altering equilibrium concentration of $$\ce{Cu^2+}$$.

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

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.

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