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 find the standard reduction potential (E°) for the half-reaction of the couple \( M^{2+}/M \) given the overall cell reaction and its potential.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell reaction given is:
\[ M(s) + 2H^+(1M) \rightarrow H_2(g)(1atm) + M^{2+}(0.1m) \]

2. **Determine the Cell Potential (E_cell)**:
The potential of the cell for the reaction is given as:
\[ E_{cell} = 1.500 \, V \]

3. **Use the Nernst Equation**:
The Nernst equation relates the cell potential to the standard reduction potential and the concentrations of the reactants and products:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q \]
where \( n \) is the number of electrons transferred and \( Q \) is the reaction quotient.

4. **Determine n (Number of Electrons)**:
In the reaction, \( M \) is oxidized to \( M^{2+} \) (losing 2 electrons), and \( 2H^+ \) is reduced to \( H_2 \) (gaining 2 electrons). Thus, \( n = 2 \).

5. **Calculate the Reaction Quotient (Q)**:
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[M^{2+}]}{[H^+]^2 \cdot P_{H_2}} \]
Substituting the values:
\[ Q = \frac{0.1}{(1)^2 \cdot 1} = 0.1 \]

6. **Substitute Values into the Nernst Equation**:
Rearranging the Nernst equation gives:
\[ E^{\circ}_{cell} = E_{cell} + \frac{0.059}{n} \log Q \]
Substituting the known values:
\[ E^{\circ}_{cell} = 1.500 + \frac{0.059}{2} \log(0.1) \]

7. **Calculate the Logarithm**:
The logarithm of 0.1 is:
\[ \log(0.1) = -1 \]
Therefore:
\[ E^{\circ}_{cell} = 1.500 + \frac{0.059}{2} \cdot (-1) \]

8. **Calculate E°_cell**:
\[ E^{\circ}_{cell} = 1.500 - 0.0295 = 1.4705 \, V \approx 1.47 \, V \]

9. **Determine Standard Reduction Potential (E°) for M²⁺/M**:
The standard reduction potential for the half-reaction \( M^{2+}/M \) can be calculated using:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Here, the cathode is \( H^+/H_2 \) with \( E^{\circ} = 0 \, V \) and the anode is \( M^{2+}/M \):
\[ 1.47 = 0 - E^{\circ}_{M^{2+}/M} \]
Thus:
\[ E^{\circ}_{M^{2+}/M} = -1.47 \, V \]

### Final Answer:
The standard reduction potential for the \( M^{2+}/M \) couple is:
\[ E^{\circ}_{M^{2+}/M} = -1.47 \, V \]