In a concentration cell the same reagents are present in both the anode and the cathode compartments , but at different concentrations. Calculate the emf of a cell of a cell containing 0.040 M. `Cr^(3+)` in one compartment and 1.0 M `Cr^(3+)` in the other if Cr electrodes are used in both.

Options

A. 0.028 V
B. 0.249 V
C. 0.083 V
D. 0.125 V

Answer

To calculate the emf of a concentration cell containing different concentrations of `Cr^(3+)`, we can follow these steps:

### Step 1: Write the cell representation
The cell consists of chromium electrodes in two compartments with different concentrations of `Cr^(3+)`. The cell representation is:
\[ \text{Cr (s)} | \text{Cr}^{3+} (0.04 \, M) || \text{Cr}^{3+} (1.0 \, M) | \text{Cr (s)} \]

### Step 2: Identify the anode and cathode
In a concentration cell:
- The anode is where oxidation occurs (lower concentration).
- The cathode is where reduction occurs (higher concentration).

Thus:
- Anode: `Cr (s)` → `Cr^(3+) (aq)` + 3e⁻ (0.04 M)
- Cathode: `Cr^(3+) (aq)` + 3e⁻ → `Cr (s)` (1.0 M)

### Step 3: Write the half-reactions
The half-reactions are:
1. Oxidation (at anode):
\[ \text{Cr (s)} \rightarrow \text{Cr}^{3+} (aq) + 3e^- \]

2. Reduction (at cathode):
\[ \text{Cr}^{3+} (aq) + 3e^- \rightarrow \text{Cr (s)} \]

### Step 4: Write the overall cell reaction
Combining the half-reactions, we have:
\[ \text{Cr (s)} + \text{Cr}^{3+} (1.0 \, M) \rightarrow \text{Cr}^{3+} (0.04 \, M) + \text{Cr (s)} \]

### Step 5: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q \]

Where:
- \( E^0_{cell} \) is the standard cell potential (0 for the same metal),
- \( n \) is the number of moles of electrons transferred (3 for chromium),
- \( Q \) is the reaction quotient.

### Step 6: Calculate the reaction quotient (Q)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Cr}^{3+}]_{products}}{[\text{Cr}^{3+}]_{reactants}} = \frac{0.04}{1.0} = 0.04 \]

### Step 7: Substitute values into the Nernst equation
Since \( E^0_{cell} = 0 \):
\[ E_{cell} = 0 - \frac{0.0591}{3} \log(0.04) \]

### Step 8: Calculate \( \log(0.04) \)
Using logarithmic properties:
\[ \log(0.04) = \log(4) - \log(100) = 0.602 - 2 = -1.398 \]

### Step 9: Substitute \( \log(0.04) \) back into the equation
\[ E_{cell} = -\frac{0.0591}{3} \times (-1.398) \]

### Step 10: Calculate the final value
\[ E_{cell} = \frac{0.0591 \times 1.398}{3} \]
\[ E_{cell} = \frac{0.0826}{3} \approx 0.0275 \, V \]

### Step 11: Round the answer
Rounding gives:
\[ E_{cell} \approx 0.028 \, V \]

Thus, the emf of the cell is approximately **0.028 volts**.

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