The standard reduction potential of normal calomel electrode and reduction potential of saturated calomel electrodes are 0.27 and 0.33 volt respectively. What is the concentration of `Cl^-` in saturated solution of KCl?
(a)0.1M
(b)0.01M
(c)0.001M
(d)None

Options

A. 0.1M
B. 0.01M
C. 0.001M
D. None

Answer

To find the concentration of \( \text{Cl}^- \) in a saturated solution of KCl using the given standard reduction potentials, we can follow these steps:

### Step 1: Identify the Given Values
- Standard reduction potential of normal calomel electrode, \( E^\circ_{\text{NCE}} = 0.27 \, \text{V} \)
- Reduction potential of saturated calomel electrode, \( E^\circ_{\text{SCE}} = 0.33 \, \text{V} \)

### Step 2: Write the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
For the reduction of chlorine:
\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \]
Here, \( n = 2 \) (since 2 electrons are involved).

### Step 3: Set Up the Equation
Since we are comparing the normal calomel electrode and the saturated calomel electrode, we can equate their potentials:
\[ E^\circ_{\text{NCE}} = E^\circ_{\text{SCE}} - \frac{0.059}{n} \log \left( [\text{Cl}^-]^2 \right) \]
Substituting the known values:
\[ 0.27 = 0.33 - \frac{0.059}{2} \log \left( [\text{Cl}^-]^2 \right) \]

### Step 4: Rearrange the Equation
Rearranging gives:
\[ 0.27 - 0.33 = -\frac{0.059}{2} \log \left( [\text{Cl}^-]^2 \right) \]
\[ -0.06 = -\frac{0.059}{2} \log \left( [\text{Cl}^-]^2 \right) \]

### Step 5: Solve for the Logarithm
Multiply both sides by -1:
\[ 0.06 = \frac{0.059}{2} \log \left( [\text{Cl}^-]^2 \right) \]
Now, multiply both sides by \( \frac{2}{0.059} \):
\[ \frac{0.06 \times 2}{0.059} = \log \left( [\text{Cl}^-]^2 \right) \]
Calculating the left side:
\[ \frac{0.12}{0.059} \approx 2.0339 \]
Thus,
\[ \log \left( [\text{Cl}^-]^2 \right) \approx 2.0339 \]

### Step 6: Solve for \( [\text{Cl}^-] \)
Since \( \log \left( [\text{Cl}^-]^2 \right) = 2 \log \left( [\text{Cl}^-] \right) \):
\[ 2 \log \left( [\text{Cl}^-] \right) \approx 2.0339 \]
Dividing by 2:
\[ \log \left( [\text{Cl}^-] \right) \approx 1.01695 \]
Now, converting from logarithm to concentration:
\[ [\text{Cl}^-] \approx 10^{1.01695} \approx 10^{1} \approx 0.1 \, \text{M} \]

### Conclusion
The concentration of \( \text{Cl}^- \) in the saturated solution of KCl is approximately \( 0.1 \, \text{M} \).

### Answer
(a) 0.1M