If a small Cu rod is placed in an aqueous solution of ferrous salt, then which of the following will be observed? (`E_((Cu^(2+)) /(Cu))^0 = -0.34 V`,`E_((Fe^(2+)) /(Fe))^0 = -0.44 V`)

Options

A. Copper will be oxidised
B. `Fe^(2+)` will be reduced
C. No reaction will take place
D. `Fe^(2+)`will be oxidised

Answer

To solve the problem, we need to analyze the electrochemical behavior of copper (Cu) and iron (Fe) in the given scenario. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We are given the standard reduction potentials for the half-reactions:
- \( E^{\circ}_{Cu^{2+}/Cu} = -0.34 \, V \)
- \( E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V \)

### Step 2: Identify the Half-Reactions
The half-reactions for the reduction of copper and iron can be written as:
1. \( Cu^{2+} + 2e^- \rightarrow Cu \) (reduction of copper)
2. \( Fe^{2+} + 2e^- \rightarrow Fe \) (reduction of iron)

### Step 3: Compare the Reduction Potentials
The reduction potential indicates the tendency of a species to gain electrons (be reduced). A more positive (or less negative) reduction potential means a greater tendency to be reduced. Here:
- Copper has a higher reduction potential (-0.34 V) compared to iron (-0.44 V).

### Step 4: Determine the Reactivity
Since copper has a higher (less negative) reduction potential, it is less reactive than iron. This means that copper cannot reduce iron ions (\( Fe^{2+} \)) to iron (\( Fe \)).

### Step 5: Conclusion on the Reaction
Since copper cannot reduce \( Fe^{2+} \) ions, no reaction will take place when a copper rod is placed in an aqueous solution of ferrous salt.

### Final Answer
The correct observation is that **no reaction will take place**.