The standard reduction potential of normal calomel electrode and reduction potential of saturated calomel electrode are 0.27 V and 0.33 V, respectively. What is the concentration of Cl⁻ in a saturated solution of KCl?

To solve the problem, we need to find the concentration of \(\text{Cl}^-\) in the saturated solution of \(\text{KCl}\).

### Given:
- Standard reduction potential of the normal calomel electrode (\(E_{\text{NCE}}\)): 0.27 V
- Reduction potential of the saturated calomel electrode (\(E_{\text{SCE}}\)): 0.33 V
- The Nernst equation will be used to relate the electrode potential to the concentration of \(\text{Cl}^-\).

### Step 1: Write the Nernst equation

For the reduction half-cell reaction of the calomel electrode:
\[
\text{Hg}_2^{2+} + 2e^- \rightarrow 2\text{Hg} + 2\text{Cl}^-
\]

The Nernst equation for the calomel electrode is:
\[
E = E^\circ - \frac{0.0591}{2} \log \left[\frac{1}{[\text{Cl}^-]^2}\right]
\]
Where:
- \(E\) is the reduction potential (0.33 V for the saturated calomel electrode),
- \(E^\circ\) is the standard reduction potential of the normal calomel electrode (0.27 V),
- \([\text{Cl}^-]\) is the concentration of chloride ions.

### Step 2: Rearrange the Nernst equation

Rearranging the Nernst equation to solve for \([\text{Cl}^-]\):
\[
E - E^\circ = -\frac{0.0591}{2} \log \left[\frac{1}{[\text{Cl}^-]^2}\right]
\]
\[
E - E^\circ = \frac{0.0591}{2} \log \left([\text{Cl}^-]^2\right)
\]
Substitute \(E = 0.33\) V and \(E^\circ = 0.27\) V:
\[
0.33 - 0.27 = \frac{0.0591}{2} \log \left([\text{Cl}^-]^2\right)
\]
\[
0.06 = 0.02955 \log \left([\text{Cl}^-]^2\right)
\]
\[
\log \left([\text{Cl}^-]^2\right) = \frac{0.06}{0.02955}
\]
\[
\log \left([\text{Cl}^-]^2\right) \approx 2.03
\]
Now take the antilog on both sides:
\[
[\text{Cl}^-]^2 = 10^{2.03}
\]
\[
[\text{Cl}^-]^2 \approx 107.15
\]
\[
[\text{Cl}^-] = \sqrt{107.15}
\]
\[
[\text{Cl}^-] \approx 10.35 \, \text{mol/L}
\]

### Final Answer:
The concentration of \(\text{Cl}^-\) in the saturated solution of \(\text{KCl}\) is approximately 10.35 mol/L.