`E^(@)` for `Cl_2`(g) `+2e^(-)to2Cl^-`(aq) is 1.36 V, `E^(@)` for `Cl^-` (aq) `to1//2cl_2`(g)`+e^-` is:

Options

A. 1.36V
B. -1.36V
C. -0.68
D. 0.68V

Answer

To find the standard electrode potential \( E^\circ \) for the reaction:

\[ \text{Cl}^- (aq) \rightarrow \frac{1}{2} \text{Cl}_2 (g) + e^- \]

given that \( E^\circ \) for the reaction

\[ \text{Cl}_2 (g) + 2e^- \rightarrow 2\text{Cl}^- (aq) \]

is 1.36 V, we can follow these steps:

### Step 1: Identify the given reaction and its standard potential
The given reaction is:

\[ \text{Cl}_2 (g) + 2e^- \rightarrow 2\text{Cl}^- (aq) \]

with a standard potential \( E^\circ_1 = 1.36 \, \text{V} \).

### Step 2: Write the reverse reaction
To find the standard potential for the reaction involving \( \text{Cl}^- \), we need to consider the reverse of the given reaction:

\[ 2\text{Cl}^- (aq) \rightarrow \text{Cl}_2 (g) + 2e^- \]

### Step 3: Determine the standard potential for the reverse reaction
The standard potential for the reverse reaction is the negative of the standard potential of the forward reaction:

\[ E^\circ_2 = -E^\circ_1 = -1.36 \, \text{V} \]

### Step 4: Adjust for stoichiometry
Since the reaction we want is:

\[ \text{Cl}^- (aq) \rightarrow \frac{1}{2} \text{Cl}_2 (g) + e^- \]

we need to divide the entire reaction by 2. When we divide the reaction by 2, we also divide the standard potential by 2:

\[ E^\circ_2 = \frac{-1.36 \, \text{V}}{2} = -0.68 \, \text{V} \]

### Final Answer
Thus, the standard electrode potential \( E^\circ \) for the reaction

\[ \text{Cl}^- (aq) \rightarrow \frac{1}{2} \text{Cl}_2 (g) + e^- \]

is:

\[ E^\circ = -0.68 \, \text{V} \]

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