The oxidation electrode potential E, of a `0.1` M solution of `M^(+)` ions `(E_(RP)^(o) = - 2.36 V)` is :

Options

A. `-2.41`V
B. `+0.241` V
C. `-4.82 `V
D. ` +2.41 `V

Answer

To find the oxidation electrode potential \( E \) of a \( 0.1 \, M \) solution of \( M^+ \) ions, we will use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The oxidation reaction can be represented as:
\[ M^+ + e^- \rightarrow M \]
This indicates that \( M^+ \) ions gain an electron to form \( M \).

### Step 2: Identify Standard Electrode Potential
The standard reduction potential \( E^\circ_{RP} \) given is \( -2.36 \, V \). To find the oxidation potential, we need to reverse the sign:
\[ E^\circ_{ox} = -E^\circ_{RP} = 2.36 \, V \]

### Step 3: Use the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
For our reaction, \( n = 1 \) (one electron is transferred).

### Step 4: Substitute Values into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E = 2.36 - \frac{0.0591}{1} \log \left( \frac{[M]}{[M^+]} \right) \]
Since we are considering the concentration of \( M^+ \) ions which is \( 0.1 \, M \), and the concentration of \( M \) is taken as \( 1 \) (as it is a solid and not included in the equilibrium expression):
\[ E = 2.36 - 0.0591 \log \left( \frac{1}{0.1} \right) \]

### Step 5: Calculate the Logarithm
Calculating the logarithm:
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

### Step 6: Substitute Back and Calculate
Now substituting back into the equation:
\[ E = 2.36 - 0.0591 \times 1 \]
\[ E = 2.36 - 0.0591 = 2.3009 \, V \]

### Step 7: Final Result
Thus, the oxidation electrode potential \( E \) of the \( 0.1 \, M \) solution of \( M^+ \) ions is approximately:
\[ E \approx 2.30 \, V \]