When an aqueous solution of `CuSO_(4)` is stirred with a silver spoon then :

Options

A. (a) `Cu^(+)` will be formed
B. (b) `Ag^(+)` will be formed
C. (c) `Cu^(2+)` will be deposited
D. (d) None of these

Answer

To solve the question regarding the reaction between an aqueous solution of `CuSO4` and a silver spoon, we can follow these steps:

### Step 1: Identify the components in the reaction
The components involved are:
- Copper sulfate (`CuSO4`), which dissociates into `Cu^2+` and `SO4^2-` ions in solution.
- Silver (`Ag`), which is the material of the spoon.

### Step 2: Determine the possible reaction
Since silver is a more noble metal than copper, we can consider a displacement reaction where silver might displace copper from the copper sulfate solution:
\[ \text{Cu}^{2+} + \text{Ag} \rightarrow \text{Ag}^+ + \text{Cu} \]

### Step 3: Check the standard reduction potentials
We need to check the standard reduction potentials for the half-reactions:
- For copper: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\) with \(E^\circ = 0.34 \, \text{V}\)
- For silver: \(\text{Ag}^+ + e^- \rightarrow \text{Ag}\) with \(E^\circ = 0.80 \, \text{V}\)

### Step 4: Calculate the standard cell potential
The overall cell potential \(E^\circ_{\text{cell}}\) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In this case:
- The cathode reaction (reduction) is the silver reaction: \(E^\circ_{\text{cathode}} = 0.80 \, \text{V}\)
- The anode reaction (oxidation) is the copper reaction: \(E^\circ_{\text{anode}} = 0.34 \, \text{V}\)

Thus,
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]

### Step 5: Analyze the feasibility of the reaction
To determine if the reaction is feasible, we can use the Gibbs free energy change (\(\Delta G\)):
\[ \Delta G = -nFE^\circ_{\text{cell}} \]

Where:
- \(n\) = number of moles of electrons transferred (1 for this reaction)
- \(F\) = Faraday's constant (approximately \(96500 \, \text{C/mol}\))

Since \(E^\circ_{\text{cell}} = 0.46 \, \text{V}\) is positive, \(\Delta G\) will be negative, indicating that the reaction is feasible.

### Step 6: Conclusion
Since the reaction is feasible, we can conclude that:
- Silver will displace copper from the solution, resulting in the formation of silver ions and solid copper.

### Final Answer
The correct answer is that a reaction occurs, and the products formed are silver ions and solid copper.

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