To determine which reaction is spontaneous based on the given electrode potential data, we will analyze the reactions and their standard electrode potentials (E°).
### Given Reactions and Electrode Potentials:
1. **Reaction 1**:
\[
Cl_2 + 2H_2O \rightarrow 2ClO^{-} + 4H^{+} + 2e^{-}, \quad E^{\circ} = -1.61 \, \text{V}
\]
2. **Reaction 2**:
\[
ClO^{-} + 2H_2O \rightarrow ClO_3^{-} + 4H^{+} + 4e^{-}, \quad E^{\circ} = -0.50 \, \text{V}
\]
### Step 1: Understand Spontaneity
A reaction is spontaneous if the change in Gibbs free energy (ΔG) is negative. The relationship between ΔG and the standard electrode potential (E°) is given by:
\[
\Delta G = -nFE
\]
Where:
- \( n \) = number of moles of electrons transferred
- \( F \) = Faraday's constant (approximately 96485 C/mol)
- \( E \) = cell potential
For ΔG to be negative, \( E \) must be positive.
### Step 2: Analyze the Reactions
Both reactions have negative E° values, indicating that they are non-spontaneous in their given directions. To find a spontaneous reaction, we can reverse one of the reactions and combine them.
### Step 3: Consider the Options
1. **Option 1**: \( Cl_2 + ClO^{-} \rightarrow ClO_3^{-} \)
- This combines the two reactions as they are, which will keep the overall E° negative. Thus, this reaction is non-spontaneous.
2. **Option 2**: \( ClO^{-} \rightarrow Cl_2 + ClO_3^{-} \)
- Here, we reverse the first reaction:
\[
2ClO^{-} + 4H^{+} + 2e^{-} \rightarrow Cl_2 + 2H_2O, \quad E^{\circ} = +1.61 \, \text{V}
\]
- The second reaction remains the same:
\[
ClO^{-} + 2H_2O \rightarrow ClO_3^{-} + 4H^{+} + 4e^{-}, \quad E^{\circ} = -0.50 \, \text{V}
\]
- If we add these reactions, we can calculate the overall E°:
\[
E_{\text{cell}} = E^{\circ}_{\text{reversed}} + E^{\circ}_{\text{forward}} = 1.61 - 0.50 = 1.11 \, \text{V}
\]
- Since \( E_{\text{cell}} \) is positive, this reaction is spontaneous.
3. **Option 3**: \( ClO_3^{-} \rightarrow Cl_2 + ClO^{-} \)
- This is the reverse of the second reaction, which would yield a negative E° and thus be non-spontaneous.
4. **Option 4**: \( ClO^{-} + Cl_2 \rightarrow ClO_3^{-} \)
- This is similar to Option 1 and would also yield a negative E°.
### Conclusion
The only spontaneous reaction based on the given electrode potentials is **Option 2**:
\[
ClO^{-} \rightarrow Cl_2 + ClO_3^{-}
\]
