`Cu^(+)` ions reacts with `Fe^(+)` ion according to the following reaction `Cu^(+) + 2Fe^(2+)hArrCu+2Fe^(3+)` At equilibrium the concentration of `Cu^(2+)` ions is not changed by the addition of

Options

A. `Cu^(2+)`
B. `Fe^(2+)`
C. `Cu`
D. `Fe^(3+)`

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ \text{Cu}^+ + 2 \text{Fe}^{2+} \rightleftharpoons \text{Cu}^{2+} + 2 \text{Fe}^{3+} \]

### Step 1: Understand the Equilibrium Expression
The equilibrium constant \( K_c \) for the reaction can be expressed as:

\[ K_c = \frac{[\text{Cu}^{2+}][\text{Fe}^{3+}]^2}{[\text{Cu}^+][\text{Fe}^{2+}]^2} \]

In this expression, the concentrations of solids and pure liquids do not appear.

### Step 2: Identify the Effect of Adding Different Species
To determine what addition will not change the concentration of \(\text{Cu}^{2+}\) at equilibrium, we need to consider how the equilibrium shifts according to Le Chatelier's principle.

- If we add \(\text{Cu}^+\) (the reactant), it will shift the equilibrium to the right, increasing \(\text{Cu}^{2+}\).
- If we add \(\text{Fe}^{2+}\) (another reactant), it will also shift the equilibrium to the right, increasing \(\text{Cu}^{2+}\).
- If we add \(\text{Fe}^{3+}\) (the product), it will shift the equilibrium to the left, decreasing \(\text{Cu}^{2+}\).

### Step 3: Identify the Correct Addition
The only addition that will not affect the concentration of \(\text{Cu}^{2+}\) is the addition of solid copper (Cu). Since solids do not appear in the equilibrium expression, adding solid copper will not change the concentrations of the aqueous species at equilibrium.

### Conclusion
Thus, the concentration of \(\text{Cu}^{2+}\) ions will not change if we add solid copper.

### Final Answer
The concentration of \(\text{Cu}^{2+}\) ions is not changed by the addition of solid copper.

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