The equilibrium, $2C{{u}^{I}}\rightleftharpoons C{{u}^{0}}+C{{u}^{II}}$, in aqueous medium at ${{25}^{\circ }}C$ shifts towards the left in the presence of:
(A) $NO_{3}^{-}$
(B) $C{{l}^{-}}$
(C) $SC{{N}^{-}}$
(D) $C{{N}^{-}}$
Answer
606.6k+ views
Hint: For the reaction to occur in the left or backward direction the concentration of the product must be increased or the concentration of reactant must be decreased, so the ions must be used that can form a precipitate with copper ion having +1 oxidation state.
Complete step by step solution:
The reaction given in the question is:
$2C{{u}^{I}}\rightleftharpoons C{{u}^{0}}+C{{u}^{II}}$
The copper in oxidation state $+1$ is the reactant and the copper is oxidation state $0$ and $+2$ are the products.
The reaction will be forward when the product formed is more than the reactant. And the reaction will be back when the reactant is more than the product.
According to Le Chatlier’s principle, some factors decide the direction of the reaction. For the reaction to occur in a forward direction, either the concentration of the reactant must be increased or the concentration of the product must be decreased. And for reaction to occur in a backward direction, either the concentration of the reactant must be decreased or the concentration of the product must be increased.
So, from the option given above $C{{N}^{-}},\text{ }SC{{N}^{-}},\text{ }C{{l}^{-}}$ forms precipitate with $C{{u}^{+}}$, thus decreases the concentration of the reactant. So, as the precipitate forms, more and more $C{{u}^{+}}$ will be separated and the concentration decreases. The reaction will occur in a backward direction.
Therefore, the correct answers are an option (B), (C), and (D).
Note: Other factors that can be used to decide the direction of the reaction are, low pressure will cause the reaction to occur in a backward reaction which means that high pressure will shift the reaction towards forward or right and high temperature cause the reaction to occur in a backward reaction which means that low temperature will shift the reaction towards forward or right.
Complete step by step solution:
The reaction given in the question is:
$2C{{u}^{I}}\rightleftharpoons C{{u}^{0}}+C{{u}^{II}}$
The copper in oxidation state $+1$ is the reactant and the copper is oxidation state $0$ and $+2$ are the products.
The reaction will be forward when the product formed is more than the reactant. And the reaction will be back when the reactant is more than the product.
According to Le Chatlier’s principle, some factors decide the direction of the reaction. For the reaction to occur in a forward direction, either the concentration of the reactant must be increased or the concentration of the product must be decreased. And for reaction to occur in a backward direction, either the concentration of the reactant must be decreased or the concentration of the product must be increased.
So, from the option given above $C{{N}^{-}},\text{ }SC{{N}^{-}},\text{ }C{{l}^{-}}$ forms precipitate with $C{{u}^{+}}$, thus decreases the concentration of the reactant. So, as the precipitate forms, more and more $C{{u}^{+}}$ will be separated and the concentration decreases. The reaction will occur in a backward direction.
Therefore, the correct answers are an option (B), (C), and (D).
Note: Other factors that can be used to decide the direction of the reaction are, low pressure will cause the reaction to occur in a backward reaction which means that high pressure will shift the reaction towards forward or right and high temperature cause the reaction to occur in a backward reaction which means that low temperature will shift the reaction towards forward or right.
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