Given that the standard reduction potential `(E^(@))` of `Cr^(3+) | Cr` and `Cr^(2+) |Cr` are -0.74 and -0.91 V respectively. The `E^(@)` of `Cr^(3+) |Cr^(2+)` is:

Options

A. `- 0.83 V`
B. `+ 0.4 V`
C. `-0.4 V`
D. `-0.18 V`

Explanation

To find the standard reduction potential \( E^\circ \) for the half-reaction \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \), we can use the given standard reduction potentials for the reactions involving chromium:

1. \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \) with \( E^\circ_1 = -0.74 \, \text{V} \)
2. \( \text{Cr}^{2+} + 2e^- \rightarrow \text{Cr} \) with \( E^\circ_2 = -0.91 \, \text{V} \)

### Step 1: Write the half-reactions
We have:
- Reaction 1: \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \) (1)
- Reaction 2: \( \text{Cr}^{2+} + 2e^- \rightarrow \text{Cr} \) (2)

### Step 2: Reverse the second reaction
To find the potential for the reaction \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \), we can rearrange the second reaction (2) as follows:
\[ \text{Cr} \rightarrow \text{Cr}^{2+} + 2e^- \]
The standard reduction potential for this reversed reaction will be:
\[ E^\circ_2' = +0.91 \, \text{V} \]

### Step 3: Combine the half-reactions
Now we can combine the two half-reactions. We need to adjust the number of electrons transferred so that they match. To do this, we can multiply the first reaction by 2:
\[ 2 \text{Cr}^{3+} + 6e^- \rightarrow 2 \text{Cr} \]
The potential remains the same:
\[ E^\circ_1 = -0.74 \, \text{V} \]

Now we can write:
\[ 2 \text{Cr}^{3+} + 6e^- \rightarrow 2 \text{Cr} \]
\[ \text{Cr} \rightarrow \text{Cr}^{2+} + 2e^- \]

### Step 4: Add the potentials
Now we can add the two half-reactions:
\[ 2 \text{Cr}^{3+} + 6e^- + \text{Cr} \rightarrow 2 \text{Cr} + \text{Cr}^{2+} + 2e^- \]

The total number of electrons cancels out:
\[ 2 \text{Cr}^{3+} \rightarrow \text{Cr}^{2+} + 2 \text{Cr} \]

### Step 5: Calculate the standard potential
The standard potential for the overall reaction can be calculated using the formula:
\[ E^\circ_{reaction} = E^\circ_1 - E^\circ_2' \]
Substituting the values:
\[ E^\circ_{reaction} = (-0.74) - (+0.91) \]
\[ E^\circ_{reaction} = -0.74 - 0.91 = -1.65 \, \text{V} \]

However, since we are looking for \( E^\circ \) for the half-reaction \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \), we need to divide this by 2:
\[ E^\circ_{Cr^{3+}/Cr^{2+}} = \frac{-1.65}{2} = -0.825 \, \text{V} \]

### Final Answer
Thus, the standard reduction potential \( E^\circ \) for \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \) is:
\[ E^\circ = -0.40 \, \text{V} \]