An iron wire is immersed in a solution containing `ZnSO_(4) and NiSO_(4)`. When the concentration of each salt is 1 M, predict giving reactions which of the following reactions is likely to proceed : (i) Iron reduces `Zn^(+)` ions (ii) Iron reduces `Ni^(2+)` ions.
Given
`E^(Theta)= -0.76 V " for "Zn^(2+)//Zn`
`E^(Theta)= -0.44 V" for "Fe^(2+)//Fe`
`E^(Theta)= -0.25V " for "Ni^(2+)//Ni`

Explanation

To solve the problem of whether iron can reduce \( Zn^{2+} \) ions or \( Ni^{2+} \) ions when immersed in a solution of \( ZnSO_4 \) and \( NiSO_4 \), we will analyze the standard reduction potentials given for each half-reaction.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard reduction potentials**:
- For \( Zn^{2+} + 2e^- \rightarrow Zn \), \( E^\circ = -0.76 \, V \)
- For \( Ni^{2+} + 2e^- \rightarrow Ni \), \( E^\circ = -0.25 \, V \)
- For \( Fe^{2+} + 2e^- \rightarrow Fe \), \( E^\circ = -0.44 \, V \)

2. **Determine the oxidation and reduction reactions**:
- When iron (Fe) is immersed in the solutions, it can either oxidize (lose electrons) or reduce (gain electrons).
- The oxidation half-reaction for iron is:
\[ Fe \rightarrow Fe^{2+} + 2e^- \]
- The reduction half-reactions are:
- For zinc:
\[ Zn^{2+} + 2e^- \rightarrow Zn \]
- For nickel:
\[ Ni^{2+} + 2e^- \rightarrow Ni \]

3. **Calculate the cell potentials for both reactions**:
- **For the reaction with \( Zn^{2+} \)**:
- Anode (oxidation): \( Fe \rightarrow Fe^{2+} + 2e^- \) (oxidation potential = \( +0.44 \, V \))
- Cathode (reduction): \( Zn^{2+} + 2e^- \rightarrow Zn \) (reduction potential = \( -0.76 \, V \))
- Calculate \( E^\circ_{cell} \):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.76) - (-0.44) = -0.32 \, V \]
- Since \( E^\circ_{cell} < 0 \), this reaction is not feasible.

- **For the reaction with \( Ni^{2+} \)**:
- Anode (oxidation): \( Fe \rightarrow Fe^{2+} + 2e^- \) (oxidation potential = \( +0.44 \, V \))
- Cathode (reduction): \( Ni^{2+} + 2e^- \rightarrow Ni \) (reduction potential = \( -0.25 \, V \))
- Calculate \( E^\circ_{cell} \):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.25) - (-0.44) = 0.19 \, V \]
- Since \( E^\circ_{cell} > 0 \), this reaction is feasible.

4. **Conclusion**:
- Iron cannot reduce \( Zn^{2+} \) ions because the cell potential is negative.
- Iron can reduce \( Ni^{2+} \) ions because the cell potential is positive.

### Final Answer:
- (i) Iron does not reduce \( Zn^{2+} \) ions.
- (ii) Iron reduces \( Ni^{2+} \) ions.