A simple model for a concentration cell involving a metal `M` is
`M(s)|M^(o+)(aq,0.05 ` molar`)||M^(o+)(aq,1` molar`)|M(s)`
For the abov electrolytic cell, the magnitude of the cell potential is `|E_(cell)|=70mV.`
If the `0.05` molar solution of `M^(o+)` is replaced by a `0.0025` molar `M^(o+)` solution, then the magnitude of the cell potential would be
(a)`35mV`
(b)`70mV`
(c)`140mV`
(d)`700mV`

Options

A. `35mV`
B. `70mV`
C. `140mV`
D. `700mV`

Explanation

To solve the problem of determining the new cell potential when the concentration of \( M^{o+} \) is changed from 0.05 M to 0.0025 M, we can follow these steps:

### Step 1: Understand the Cell Reaction
In a concentration cell, the reaction involves the oxidation of metal \( M \) at the anode and the reduction of \( M^{o+} \) ions at the cathode. The half-reactions can be written as:
- **Anode (Oxidation)**: \( M(s) \rightarrow M^{o+}(aq) + e^- \)
- **Cathode (Reduction)**: \( M^{o+}(aq) + e^- \rightarrow M(s) \)

### Step 2: Write the Nernst Equation
The Nernst equation for a concentration cell is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln \left( \frac{[M^{o+}]_{anode}}{[M^{o+}]_{cathode}} \right) \]
Where:
- \( E^\circ_{cell} = 0 \) for concentration cells.
- \( R \) is the universal gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin (assumed to be 298 K for standard conditions).
- \( n \) is the number of electrons transferred (1 for this reaction).
- \( F \) is Faraday's constant (96485 C/mol).

### Step 3: Substitute Known Values
Given:
- Initial concentration at anode: \( [M^{o+}]_{anode} = 0.05 \, \text{M} \)
- New concentration at anode: \( [M^{o+}]_{anode} = 0.0025 \, \text{M} \)
- Concentration at cathode: \( [M^{o+}]_{cathode} = 1 \, \text{M} \)
- The initial cell potential \( |E_{cell}| = 70 \, \text{mV} = 0.070 \, \text{V} \)

### Step 4: Calculate the New Cell Potential
Using the Nernst equation, we can calculate the new cell potential:
\[ E_{cell} = 0 - \frac{0.0592}{1} \log \left( \frac{0.0025}{1} \right) \]
Calculating the logarithm:
\[ \log(0.0025) = -2.602 \]
Now substituting this value into the equation:
\[ E_{cell} = -0.0592 \times (-2.602) = 0.154 \, \text{V} = 154 \, \text{mV} \]

### Step 5: Compare with Initial Cell Potential
The initial cell potential was 70 mV. When the concentration of \( M^{o+} \) is decreased from 0.05 M to 0.0025 M, the cell potential increases due to the logarithmic relationship in the Nernst equation.

### Final Step: Determine the Magnitude of the New Cell Potential
Since we calculated the new cell potential to be approximately 154 mV, we can round it to the nearest option provided:
- The closest option is \( 140 \, \text{mV} \).

Thus, the answer is:
**(c) 140 mV**