Given `E_(Cr^(3+)//cr)^@ =- 0.72 V, E_(Fe^(2+)//Fe)^@ =- 0.42 V`. The potential for the cell
`Cr | Cr^(3+) (0.1 M) || FE^(2+) (0.01 M) |` Fe is .

Options

A. `0.26V`
B. `0.339V`
C. `-0.339V`
D. `-0.26V`

Explanation

To solve the problem, we need to calculate the cell potential for the electrochemical cell given the standard reduction potentials for chromium and iron.

### Step 1: Identify the half-reactions and their standard reduction potentials
We have the following standard reduction potentials:
- For chromium:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \quad E^\circ = -0.72 \, \text{V} \]
- For iron:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad E^\circ = -0.42 \, \text{V} \]

### Step 2: Determine the anode and cathode
In an electrochemical cell, the half-reaction with the higher (less negative) reduction potential will occur at the cathode, while the one with the lower (more negative) reduction potential will occur at the anode.

- The reduction potential for iron (-0.42 V) is higher than that for chromium (-0.72 V), so:
- **Anode (oxidation)**: Cr → Cr³⁺ + 3e⁻
- **Cathode (reduction)**: Fe²⁺ + 2e⁻ → Fe

### Step 3: Balance the half-reactions
To combine the half-reactions, we need to balance the number of electrons transferred. The least common multiple of 2 and 3 is 6. Therefore, we multiply the half-reactions by appropriate coefficients:

- For chromium:
\[ 2 \text{Cr} \rightarrow 2 \text{Cr}^{3+} + 6e^- \]
- For iron:
\[ 3 \text{Fe}^{2+} + 6e^- \rightarrow 3 \text{Fe} \]

### Step 4: Write the overall cell reaction
Combining the balanced half-reactions gives us:
\[ 2 \text{Cr} + 3 \text{Fe}^{2+} \rightarrow 2 \text{Cr}^{3+} + 3 \text{Fe} \]

### Step 5: Calculate the standard cell potential
The standard cell potential \(E^\circ_{\text{cell}}\) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.42 \, \text{V}) - (-0.72 \, \text{V}) = -0.42 + 0.72 = 0.30 \, \text{V} \]

### Step 6: Adjust for non-standard conditions using the Nernst equation
Since the concentrations of the ions are not standard (0.1 M for Cr³⁺ and 0.01 M for Fe²⁺), we need to use the Nernst equation:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \(R\) = 8.314 J/(mol·K)
- \(T\) = 298 K (assuming standard temperature)
- \(n\) = total number of moles of electrons transferred (6 in this case)
- \(F\) = 96485 C/mol
- \(Q\) = reaction quotient = \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{0.000001} = 10000

Substituting these values into the Nernst equation:
\[ E = 0.30 - \frac{(8.314)(298)}{(6)(96485)} \ln(10000) \]
Calculating the second term:
\[ \frac{(8.314)(298)}{(6)(96485)} \approx 0.0005 \]
Now, calculate \(\ln(10000) \approx 9.21\):
\[ E = 0.30 - 0.0005 \times 9.21 \approx 0.30 - 0.0046 \approx 0.2954 \, \text{V} \]

### Final Answer
The potential for the cell \(Cr | Cr^{3+} (0.1 M) || Fe^{2+} (0.01 M) | Fe\) is approximately:
\[ E \approx 0.295 \, \text{V} \]