Write the following redox reactions using half equations:
`2Fe^(3+)(aq)+2I^(ө)(aq)rarrI_(2)(aq)+2Fe^(2+)(aq)`

Explanation

To write the given redox reaction using half equations, we will break the overall reaction into two half-reactions: one for oxidation and one for reduction.

### Step 1: Identify the oxidation and reduction processes
In the reaction:
\[ 2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(aq) + 2Fe^{2+}(aq) \]

- **Oxidation**: Iodide ions (\(I^{-}\)) are oxidized to iodine (\(I_{2}\)).
- **Reduction**: Iron ions (\(Fe^{3+}\)) are reduced to iron ions (\(Fe^{2+}\)).

### Step 2: Write the half-equation for the oxidation process
The oxidation half-reaction can be written as:
\[ 2I^{-} \rightarrow I_{2} + 2e^{-} \]
This shows that two iodide ions lose two electrons to form one molecule of iodine.

### Step 3: Write the half-equation for the reduction process
The reduction half-reaction can be written as:
\[ 2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+} \]
This shows that two iron ions gain two electrons to form two iron ions in the +2 oxidation state.

### Step 4: Combine the half-equations
Now we can combine the two half-equations to represent the overall redox reaction:
1. Oxidation: \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)
2. Reduction: \( 2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+} \)

Combining these gives us:
\[ 2Fe^{3+} + 2I^{-} \rightarrow I_{2} + 2Fe^{2+} \]

### Final Answer
The half-equations for the given redox reaction are:
1. Oxidation half-equation: \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)
2. Reduction half-equation: \( 2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+} \)

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