Standard electrode potential data is given below :
`Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V `
`Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V `
`Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V`
Based on the data given above, reducing power of `Fe^(2+)` Al and `Br^(-)` will increase in the order :

Options

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

Answer

To determine the increasing order of reducing power for the ions \( Fe^{2+} \), \( Al \), and \( Br^- \), we can analyze the standard electrode potentials provided. Here's a step-by-step solution:

### Step 1: Understand the Concept of Reducing Power
Reducing power refers to the ability of a species to donate electrons and reduce another species. A species with a lower (more negative) standard electrode potential is a stronger reducing agent because it is more likely to lose electrons.

### Step 2: List the Given Standard Electrode Potentials
The standard electrode potentials provided are:
- \( E^{\circ} \) for \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) is \( +0.77 \, V \)
- \( E^{\circ} \) for \( Al^{3+} + 3e^- \rightarrow Al \) is \( -1.66 \, V \)
- \( E^{\circ} \) for \( Br_2 + 2e^- \rightarrow 2Br^- \) is \( +1.08 \, V \)

### Step 3: Identify the Reducing Agents
To find the reducing power, we need to consider the reverse reactions (oxidation reactions) since reducing agents are oxidized:
- For \( Fe^{2+} \): \( Fe^{2+} \rightarrow Fe^{3+} + e^- \) (reverse of the given reaction)
- For \( Al \): \( Al \rightarrow Al^{3+} + 3e^- \) (reverse of the given reaction)
- For \( Br^- \): \( Br^- \rightarrow Br_2 + 2e^- \) (reverse of the given reaction)

### Step 4: Determine the Reducing Power Based on Electrode Potentials
- The reducing power is inversely related to the standard electrode potential. The more negative the potential, the stronger the reducing agent.
- The potentials for the oxidation reactions are:
- \( Fe^{2+} \) has a potential of \( -0.77 \, V \)
- \( Al \) has a potential of \( +1.66 \, V \) (more negative)
- \( Br^- \) has a potential of \( -1.08 \, V \)

### Step 5: Compare the Reducing Powers
Now we can compare the potentials:
- \( Al \) (most negative potential, strongest reducing agent)
- \( Fe^{2+} \) (less negative than \( Al \))
- \( Br^- \) (least negative potential, weakest reducing agent)

### Step 6: Write the Increasing Order of Reducing Power
Based on the analysis:
- The increasing order of reducing power is:
\[ Br^- < Fe^{2+} < Al \]

### Final Answer
Thus, the reducing power of \( Fe^{2+} \), \( Al \), and \( Br^- \) will increase in the order:
\[ Br^- < Fe^{2+} < Al \]