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 \( E_{M^{+}//M} \) of a 0.1 M solution of \( M^{+} \) ions, we will use the Nernst equation. Here’s the step-by-step solution:

### Step 1: Write down the Nernst Equation
The Nernst equation is given by:

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

Where:
- \( E \) is the cell potential.
- \( E^{\circ} \) is the standard electrode potential.
- \( n \) is the number of electrons transferred in the half-reaction.
- \([M]\) is the concentration of the solid metal, which is 1 (since it is a solid).
- \([M^{+}]\) is the concentration of the ion in solution.

### Step 2: Identify the values
From the problem:
- \( E^{\circ} = -2.36 \, \text{V} \)
- \( n = 1 \) (since one electron is involved in the half-reaction)
- \([M^{+}] = 0.1 \, \text{M}\)
- \([M] = 1\) (for solids, we consider the activity to be 1)

### Step 3: Substitute the values into the Nernst Equation
Substituting the known values into the Nernst equation:

\[ E = -2.36 - \frac{0.0591}{1} \log \left( \frac{1}{0.1} \right) \]

### Step 4: Calculate the logarithm
Calculate the logarithm:

\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

### Step 5: Substitute the logarithm value back into the equation
Now substitute this value back into the equation:

\[ E = -2.36 - 0.0591 \cdot 1 \]

### Step 6: Perform the calculation
Now, calculate \( E \):

\[ E = -2.36 - 0.0591 = -2.4191 \, \text{V} \]

### Final Answer
Thus, the single electrode potential \( E_{M^{+}//M} \) is approximately:

\[ E \approx -2.42 \, \text{V} \]

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