For the above cell

Options

A. `E_(cell)lt0,DeltaGgt0`
B. `E_(cell)lt0, DeltaG lt0`
C. `E_(cell) lt 0, DeltaG =^(@) gt 0`
D. `E_(cell)gt 0,DeltaG^(@) lt 0`

Explanation

To solve the problem regarding the electrochemical cell, we need to calculate the cell potential (E_cell) and the Gibbs free energy change (ΔG) based on the given concentrations of the ions involved. Let's break it down step by step.

### Step 1: Identify the Cell Reactions
In an electrochemical cell, we have two half-reactions: one at the anode and one at the cathode. The anode is where oxidation occurs, and the cathode is where reduction occurs.

Assuming the cell is represented as:
\[ \text{M}^{+} (0.05 \, \text{M}) | \text{M}^{+} (1 \, \text{M}) \]

The half-reactions can be represented as:
- Anode (oxidation): \( \text{M} \rightarrow \text{M}^{+} + e^{-} \)
- Cathode (reduction): \( \text{M}^{+} + e^{-} \rightarrow \text{M} \)

### Step 2: Calculate the Standard Cell Potential (E°)
Given that the standard reduction potential (E°) for the half-reaction is 0 V (as stated in the video), we can use the Nernst equation to find the cell potential (E_cell).

### Step 3: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]

Where:
- \( E \) = cell potential
- \( E^{\circ} \) = standard cell potential (0 V in this case)
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin (assume 298 K if not provided)
- \( n \) = number of moles of electrons transferred (1 for this reaction)
- \( F \) = Faraday's constant (96485 C/mol)
- \( Q \) = reaction quotient

### Step 4: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{M}^{+}]_{\text{anode}}}{[\text{M}^{+}]_{\text{cathode}}} = \frac{0.05}{1} = 0.05 \]

### Step 5: Substitute Values into the Nernst Equation
Now substituting the values into the Nernst equation:
\[ E = 0 - \frac{(8.314)(298)}{(1)(96485)} \ln(0.05) \]

Calculating the logarithm:
\[ \ln(0.05) \approx -2.9957 \]

Now substituting this back into the equation:
\[ E = - \frac{(8.314)(298)}{96485} \times (-2.9957) \]

Calculating the value:
\[ E \approx \frac{(8.314)(298)(2.9957)}{96485} \]

### Step 6: Calculate ΔG
The relationship between ΔG and E_cell is given by:
\[ \Delta G = -nFE \]

Substituting the values:
\[ \Delta G = -1 \times 96485 \times E \]

### Step 7: Determine the Signs of E_cell and ΔG
From the calculations:
- If \( E \) is positive, then \( E_{cell} > 0 \)
- If \( \Delta G \) is negative, then \( \Delta G < 0 \)

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

Thus, the correct option is D.