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 question, we need to analyze the electrochemical behavior of copper (Cu) when placed in an aqueous solution of ferrous salt (Fe²⁺).

### Step-by-Step Solution:

1. **Understanding the Electrochemical Series**:
- The electrochemical series ranks metals based on their standard reduction potentials. The higher the reduction potential, the more likely the species will be reduced (gain electrons).
- Given values:
- \( E^\circ (Cu^{2+}/Cu) = -0.34 \, V \)
- \( E^\circ (Fe^{2+}/Fe) = -0.44 \, V \)

2. **Comparing the Reduction Potentials**:
- Since copper has a higher reduction potential (-0.34 V) compared to iron (-0.44 V), copper is more likely to remain in its elemental form rather than oxidizing to Cu²⁺.
- Conversely, iron is more likely to be reduced from Fe²⁺ to Fe.

3. **Determining the Reaction**:
- When a copper rod is placed in a ferrous sulfate solution, we need to check if any reaction will occur based on the potentials.
- For a reaction to occur, the metal with the lower reduction potential (in this case, iron) must be able to oxidize (lose electrons) while the metal with the higher reduction potential (copper) remains unchanged.

4. **Conclusion**:
- Since copper cannot displace iron from its salt solution (as it is lower in the electrochemical series), no reaction will occur. Therefore, the copper rod will remain unchanged in the ferrous sulfate solution.

### Final Answer:
- **No reaction will occur** when a small copper rod is placed in an aqueous solution of ferrous salt.