Based on data given below for the electrode potential , reducing power of `Fe^(2+)` , Al and `Br^(-)` will increase in the order
`Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq),E^@=+1.08V`
`Al^(3+)(aq)+3e^(-)rarrAl(s),E^@=-1.66V`
`Fe^(3+)(aq)+e^(-2)rarrFe^(+2) (aq) , E^(@) =+0.77V`

Options

A. `Br^(-) lt Fe^(2+) lt Al`
B. `Al lt Br^(-) lt Fe^(2+)`
C. `Fe^(2+) lt Al lt Br^(-)`
D. `Al lt Fe ^(2+) lt Br ^(-)`

Answer

To determine the order of reducing power of \( Fe^{2+} \), \( Al \), and \( Br^{-} \) based on their electrode potentials, we will analyze the given reduction reactions and their respective standard electrode potentials.

### Step 1: Identify the reduction potentials
We have the following half-reactions and their standard electrode potentials (E°):

1. \( Br_2(aq) + 2e^- \rightarrow 2Br^-(aq), \quad E° = +1.08 \, V \)
2. \( Al^{3+}(aq) + 3e^- \rightarrow Al(s), \quad E° = -1.66 \, V \)
3. \( Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq), \quad E° = +0.77 \, V \)

### Step 2: Determine the reducing power
The reducing power of a species is inversely related to its reduction potential. A more negative reduction potential indicates a stronger reducing agent because it is more willing to lose electrons.

- **For \( Br^{-} \)**: The reduction potential is +1.08 V, which means it is the least willing to lose electrons and thus has the weakest reducing power.
- **For \( Fe^{2+} \)**: The reduction potential is +0.77 V, indicating it has moderate reducing power.
- **For \( Al \)**: The reduction potential is -1.66 V, which indicates it is the most willing to lose electrons and thus has the strongest reducing power.

### Step 3: Order the reducing power
Based on the analysis above, we can rank the reducing power from weakest to strongest:

1. \( Br^{-} \) (weakest reducing power)
2. \( Fe^{2+} \) (moderate reducing power)
3. \( Al \) (strongest reducing power)

### Final Order
The order of increasing reducing power is:
\[ Br^{-} < Fe^{2+} < Al \]