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.`
For the above cell
(a)`E_(cell)lt0,DeltaGgt0`
(b)`E_(cell)gt0,DeltaGlt0`
(c)`E_(cell)lt0,DeltaG^(c-)gt0`
(d)`E_(cell)gt0,DeltaG^(c-)lt0`

Options

A. `E_(cell)lt0,DeltaGgt0`
B. `E_(cell)gt0,DeltaGlt0`
C. `E_(cell)lt0,DeltaG^(c-)gt0`
D. `E_(cell)gt0,DeltaG^(c-)lt0`

Explanation

To solve the problem regarding the concentration cell involving a metal \( M \), we will analyze the given information step by step.

### Step 1: Understanding the Cell Reaction
The concentration cell is represented as:
\[ M(s) | M^{o+}(aq, 0.05 \text{ molar}) || M^{o+}(aq, 1 \text{ molar}) | M(s) \]

In this cell:
- At the anode, the reaction can be written as:
\[ M(s) \rightarrow M^{o+}(aq) + e^- \]
- At the cathode, the reaction can be written as:
\[ M^{o+}(aq) + e^- \rightarrow M(s) \]

### Step 2: Writing the Overall Cell Reaction
Combining the two half-reactions, we have:
\[ M(s) + M^{o+}(aq, 1 \text{ molar}) \rightarrow M^{o+}(aq, 0.05 \text{ molar}) + M(s) \]

### Step 3: Applying the Nernst Equation
For a concentration cell, the standard cell potential \( E^\circ \) is zero:
\[ E_{cell} = E^\circ - \frac{0.0591}{n} \log \left( \frac{[M^{o+}]_{anode}}{[M^{o+}]_{cathode}} \right) \]
Here, \( n = 1 \) (since one electron is involved).

Substituting the values:
\[ E_{cell} = 0 - \frac{0.0591}{1} \log \left( \frac{0.05}{1} \right) \]

Calculating the logarithm:
\[ E_{cell} = -0.0591 \cdot \log(0.05) \]
Using the fact that \( \log(0.05) \approx -1.301 \):
\[ E_{cell} = -0.0591 \cdot (-1.301) \]
\[ E_{cell} \approx 0.0768 \text{ V} = 76.8 \text{ mV} \]

### Step 4: Analyzing the Sign of \( E_{cell} \) and \( \Delta G \)
Since \( E_{cell} \) is positive (approximately 76.8 mV), we can conclude:
- \( E_{cell} > 0 \)

Using the relationship between Gibbs free energy and cell potential:
\[ \Delta G = -nFE_{cell} \]
Since \( E_{cell} > 0 \), it follows that:
- \( \Delta G < 0 \)

### Conclusion
Based on the analysis:
- \( E_{cell} > 0 \)
- \( \Delta G < 0 \)

Thus, the correct option is:
**(b) \( E_{cell} > 0, \Delta G < 0 \)**