The single electrode potential `E_(M^(+)//M)` of `0.1 `M solution of `M^(+)` ions for the half cell `M^(+)+e^(-)rarr M(s)` with `E^(o)=-2.36V`is :

Options

A. `+ 2.41 V`
B. `-2.41 V`
C. `-4.82V`
D. `+4.82 V`

Answer

To find the single electrode potential \( E_{M^{+}//M} \) for a 0.1 M solution of \( M^{+} \) ions, we will use the Nernst equation:

\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{1}{[M^{+}]} \right) \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Standard electrode potential \( E^{\circ} = -2.36 \, V \)
- Concentration of \( M^{+} \) ions = 0.1 M
- Number of electrons transferred \( n = 1 \)

2. **Write the Nernst equation:**
\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{1}{[M^{+}]} \right) \]

3. **Substitute the known values into the Nernst equation:**
\[ E = -2.36 - \frac{0.0591}{1} \log \left( \frac{1}{0.1} \right) \]

4. **Calculate the logarithm:**
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

5. **Substitute the logarithm value back into the equation:**
\[ E = -2.36 - 0.0591 \times 1 \]

6. **Perform the multiplication:**
\[ 0.0591 \times 1 = 0.0591 \]

7. **Final calculation:**
\[ E = -2.36 - 0.0591 = -2.4191 \, V \]

8. **Round the answer to two decimal places:**
\[ E \approx -2.42 \, V \]

### Final Answer:
The single electrode potential \( E_{M^{+}//M} \) for the 0.1 M solution of \( M^{+} \) ions is approximately \( -2.42 \, V \).