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. 26 V`
B. ` 0. 339 V`
C. ` -0 .399 V`
D. ` -0 . 26 V`

Explanation

To find the potential for the cell given the standard electrode potentials and concentrations, we can follow these steps:

### Step 1: Identify the half-reactions and their standard potentials
We are given:
- \( E^\circ_{Cr^{3+}/Cr} = -0.72 \, V \)
- \( E^\circ_{Fe^{2+}/Fe} = -0.42 \, V \)

### Step 2: Write the balanced cell reaction
The balanced cell reaction for the given cell can be written as:
\[ 2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe \]
In this reaction, chromium (Cr) is oxidized to chromium ions (\(Cr^{3+}\)), and iron ions (\(Fe^{2+}\)) are reduced to iron (Fe).

### Step 3: Determine the standard cell potential (\(E^\circ_{cell}\))
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Here, \(Fe^{2+}/Fe\) is the cathode (reduction) and \(Cr^{3+}/Cr\) is the anode (oxidation).

Substituting the values:
\[ E^\circ_{cell} = (-0.42) - (-0.72) \]
\[ E^\circ_{cell} = -0.42 + 0.72 = 0.30 \, V \]

### Step 4: Use the Nernst equation to find the cell potential (\(E_{cell}\))
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where:
- \(n\) is the number of moles of electrons transferred in the balanced equation.
- \(Q\) is the reaction quotient.

From the balanced reaction, we can see that 6 electrons are transferred (2 moles of Cr oxidized and 3 moles of \(Fe^{2+}\) reduced).

### Step 5: Calculate the reaction quotient \(Q\)
The reaction quotient \(Q\) is given by:
\[ Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \]
Substituting the concentrations:
- \([Cr^{3+}] = 0.1 \, M\)
- \([Fe^{2+}] = 0.01 \, M\)

Thus,
\[ Q = \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{0.000001} = 10000 \]

### Step 6: Substitute values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \log(10000) \]
Since \(\log(10000) = 4\):
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \times 4 \]
\[ E_{cell} = 0.30 - \frac{0.0591 \times 4}{6} \]
\[ E_{cell} = 0.30 - \frac{0.2364}{6} \]
\[ E_{cell} = 0.30 - 0.0394 \]
\[ E_{cell} = 0.2606 \, V \]

### Final Answer
The potential for the cell is approximately \(0.26 \, V\).

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