The potential of the cell for the reaction, `M(s)+2H^(+) (1M)rightarrow H_(2) (g) (1atm)+M^(2+) (0.1m)`' is 1.500 V. The standard reduction potential for `M^(2+)` / M(s) couple is :

Options

A. `(0.170V)`
B. (-1.470V)
C. `(14.70)`
D. None of these

Explanation

To solve the problem, we need to determine the standard reduction potential (E°) for the M²⁺/M(s) couple using the given cell potential and the Nernst equation.

### Step-by-step Solution:

1. **Identify the Reaction**:
The cell reaction is:
\[ M(s) + 2H^+(1M) \rightarrow H_2(g)(1atm) + M^{2+}(0.1M) \]
Here, M(s) is oxidized to M²⁺, and H⁺ is reduced to H₂.

2. **Determine Anode and Cathode**:
- **Anode**: The oxidation occurs at the anode, where M(s) loses electrons to form M²⁺.
- **Cathode**: The reduction occurs at the cathode, where H⁺ gains electrons to form H₂.

3. **Write the Nernst Equation**:
The Nernst equation for the cell can be expressed as:
\[ E_{cell} = E°_{cathode} - E°_{anode} \]
Here, we know that:
- \(E_{cell} = 1.500 \, V\)
- \(E°_{cathode} = 0 \, V\) (for the H⁺/H₂ couple)
- Let \(E°_{anode} = E°\) (for the M²⁺/M couple)

Thus, we can write:
\[ 1.500 = 0 - E° \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ E° = -1.500 \, V \]

5. **Using the Nernst Equation**:
The Nernst equation can also be expressed in terms of concentrations:
\[ E_{cell} = E° - \frac{0.0591}{n} \log \frac{[M^{2+}]}{[H^+]^2} \]
Where:
- \(n = 2\) (number of electrons transferred)
- \([M^{2+}] = 0.1 \, M\)
- \([H^+] = 1 \, M\)

Substituting the values:
\[ 1.500 = E° - \frac{0.0591}{2} \log \frac{0.1}{1^2} \]

6. **Calculating the Logarithm**:
\[ \log(0.1) = -1 \]
Thus:
\[ 1.500 = E° - \frac{0.0591}{2} \times (-1) \]
\[ 1.500 = E° + 0.02955 \]

7. **Solving for E°**:
\[ E° = 1.500 - 0.02955 \]
\[ E° = 1.47045 \, V \]

### Final Answer:
The standard reduction potential for the M²⁺/M(s) couple is approximately:
\[ E° \approx -1.47 \, V \]