A cell is to be constructed to show a redox change `:`
`Cr+2Cr^(3+)hArr3Cr^(2+)`. The number of cells with different `E^(c-)` an `n` but same value of `DeltaG^(c-)` can be made `(` Given `E^(c-)._(Cr^(3+)|Cr^(2+))=-0.40V,E^(c-)._(Cr^(3+)|Cr)=-0.74V,` adn `E^(c-)._(Cr^(2+)|Cr)=-0.91V)`

Explanation

To solve the problem, we need to analyze the given redox reaction and calculate the number of different electrochemical cells that can be constructed with the same Gibbs free energy change (ΔG) but different standard electrode potentials (E°) and number of electrons transferred (n).

### Step 1: Identify the half-reactions
The given redox change is:
\[ \text{Cr} + 2\text{Cr}^{3+} \rightleftharpoons 3\text{Cr}^{2+} \]

We can break this down into two half-reactions:
1. Oxidation half-reaction:
\[ \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \]
2. Reduction half-reaction:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}^{2+} \]

### Step 2: Write the standard electrode potentials
We are given the following standard electrode potentials:
- \( E^\circ_{\text{Cr}^{3+}/\text{Cr}^{2+}} = -0.40 \, V \)
- \( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, V \)
- \( E^\circ_{\text{Cr}^{2+}/\text{Cr}} = -0.91 \, V \)

### Step 3: Calculate ΔG for each half-reaction
The relationship between Gibbs free energy change (ΔG) and standard electrode potential (E°) is given by the equation:
\[ \Delta G = -nFE^\circ \]
where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant (approximately 96500 C/mol), and \( E^\circ \) is the standard electrode potential.

#### For \( \text{Cr}^{3+}/\text{Cr}^{2+} \):
- \( n = 3 \)
- \( E^\circ = -0.40 \, V \)
\[ \Delta G = -3 \times 96500 \times (-0.40) = 115800 \, J/mol \]

#### For \( \text{Cr}^{3+}/\text{Cr} \):
- \( n = 3 \)
- \( E^\circ = -0.74 \, V \)
\[ \Delta G = -3 \times 96500 \times (-0.74) = 213300 \, J/mol \]

#### For \( \text{Cr}^{2+}/\text{Cr} \):
- \( n = 2 \)
- \( E^\circ = -0.91 \, V \)
\[ \Delta G = -2 \times 96500 \times (-0.91) = 175130 \, J/mol \]

### Step 4: Determine the number of cells with the same ΔG
We need to find the combinations of half-reactions that yield the same ΔG.

From the calculations:
- \( \Delta G \) for \( \text{Cr}^{3+}/\text{Cr}^{2+} \) is 115800 J/mol.
- \( \Delta G \) for \( \text{Cr}^{3+}/\text{Cr} \) is 213300 J/mol.
- \( \Delta G \) for \( \text{Cr}^{2+}/\text{Cr} \) is 175130 J/mol.

Since all three half-reactions yield different ΔG values, we cannot construct multiple cells with the same ΔG from these half-reactions.

### Conclusion
The number of cells with different \( E^\circ \) and \( n \) but the same value of \( \Delta G \) that can be made is **0**.