What is the potential for the cell
`Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
`E^(@)Cr^(3+)// Cr=-0.74V`,
`E^(@)Fe^(2+)//Fe=-0.44V`

Options

A. `+0.2606V`
B. `+0.5212V`
C. `+0.1303V`
D. `-0.2606V`.

Explanation

To find the potential for the cell given by the notation `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard potentials
The cell notation indicates that chromium is being oxidized and iron is being reduced. The standard reduction potentials are given as:
- For Chromium: \( E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \)
- For Iron: \( E^\circ_{Fe^{2+}/Fe} = -0.44 \, V \)

### Step 2: Determine the oxidation potential for chromium
Since the reduction potential for chromium is given, we need to convert it to oxidation potential:
- Oxidation potential for \( Cr \):
\[ E^\circ_{oxidation} = -E^\circ_{reduction} = +0.74 \, V \]

### Step 3: Write the half-reactions
The half-reactions can be written as follows:
- Anode (oxidation):
\[ Cr \rightarrow Cr^{3+} + 3e^- \]
- Cathode (reduction):
\[ Fe^{2+} + 2e^- \rightarrow Fe \]

### Step 4: Balance the electrons
To balance the number of electrons transferred, we need to multiply the half-reactions:
- Multiply the chromium half-reaction by 2:
\[ 2Cr \rightarrow 2Cr^{3+} + 6e^- \]
- Multiply the iron half-reaction by 3:
\[ 3Fe^{2+} + 6e^- \rightarrow 3Fe \]

### Step 5: Write the overall cell reaction
Combining the balanced half-reactions gives us the overall cell reaction:
\[ 2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe \]

### Step 6: Calculate the standard EMF of the cell
The standard EMF of the cell can be calculated using:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = (-0.44 \, V) - (-0.74 \, V) = 0.30 \, V \]

### Step 7: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
Where:
- \( n = 6 \) (total electrons transferred)
- \( Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{0.000001} = 10000 \)

### Step 8: Calculate the logarithm and substitute into the Nernst equation
Calculating \( \log Q \):
\[ \log(10000) = 4 \]
Now substituting into the Nernst equation:
\[ E = 0.30 \, V - \frac{0.0591}{6} \cdot 4 \]
Calculating:
\[ E = 0.30 \, V - \frac{0.0591 \cdot 4}{6} = 0.30 \, V - 0.0394 \, V = 0.2606 \, V \]

### Final Answer
The potential for the cell is:
\[ E \approx 0.2606 \, V \]