Calculate `E_(cell)` for `Cr|Cr^(3+)(0.04M)||Cr^(3+)(1M)|Cr`:

Options

A. `(0.028V)`
B. `(0.083V)`
C. `(0V)`
D. `(0.125V)`

Explanation

To calculate the cell potential \( E_{\text{cell}} \) for the given electrochemical cell \( \text{Cr} | \text{Cr}^{3+} (0.04 \, \text{M}) || \text{Cr}^{3+} (1 \, \text{M}) | \text{Cr} \), we will follow these steps:

### Step 1: Identify the Anode and Cathode
In the cell notation, the left side represents the anode and the right side represents the cathode. Since both electrodes are chromium, we can identify the reactions occurring at each electrode.

**Anode Reaction**:
At the anode, chromium is oxidized:
\[ \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \]
The concentration of \(\text{Cr}^{3+}\) at the anode is \(0.04 \, \text{M}\).

**Cathode Reaction**:
At the cathode, \(\text{Cr}^{3+}\) is reduced:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \]
The concentration of \(\text{Cr}^{3+}\) at the cathode is \(1 \, \text{M}\).

### Step 2: Write the Overall Reaction
The overall cell reaction can be written as:
\[ \text{Cr}^{3+} (1 \, \text{M}) + \text{Cr} (s) \rightarrow \text{Cr}^{3+} (0.04 \, \text{M}) + \text{Cr} (s) \]

### Step 3: Use the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
For this concentration cell, the standard cell potential \( E^{\circ}_{\text{cell}} \) is \(0\) because both electrodes are the same.

Here, \( n = 3 \) (the number of electrons transferred).

### Step 4: Substitute Values into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{3} \log \frac{[0.04]}{[1]} \]
This simplifies to:
\[ E_{\text{cell}} = - \frac{0.0591}{3} \log (0.04) \]

### Step 5: Calculate the Logarithm
Calculating the logarithm:
\[ \log (0.04) = \log \left( \frac{4}{100} \right) = \log (4) - \log (100) = 0.602 - 2 = -1.398 \]

### Step 6: Substitute the Logarithm Value
Now substituting this value back into the equation:
\[ E_{\text{cell}} = - \frac{0.0591}{3} \times (-1.398) \]
Calculating this gives:
\[ E_{\text{cell}} = \frac{0.0591 \times 1.398}{3} \approx \frac{0.0826}{3} \approx 0.0275 \, \text{V} \]

### Final Answer
Thus, the cell potential \( E_{\text{cell}} \) is approximately:
\[ E_{\text{cell}} \approx 0.028 \, \text{V} \]