Consider the reaction `M_((aq))^(n+)+n e^(-)toM_((s))`. The standard reduction potential values of the element `M_(1),M_(2) and M_(3)` are -0.34V,-3.05 and -1.66V respectively. The order of their reducing power will be:-

Options

A. `M_(1) gt M_(2) gt M_(3)`
B. `M_(3) gt M_(2) gt M_(1)`
C. `M_(1) gt M_(3) gt M_(2)`
D. `M_(2) gt M_(3) gt M_(1)`

Explanation

To determine the order of reducing power for the elements \( M_1, M_2, \) and \( M_3 \) based on their standard reduction potentials, we can follow these steps:

### Step 1: Write down the standard reduction potentials
The standard reduction potentials for the elements are given as follows:
- \( E^\circ(M_1) = -0.34 \, \text{V} \)
- \( E^\circ(M_2) = -3.05 \, \text{V} \)
- \( E^\circ(M_3) = -1.66 \, \text{V} \)

### Step 2: Understand the relationship between reduction potential and reducing power
The reducing power of a species is inversely related to its standard reduction potential. A more negative reduction potential indicates a stronger reducing agent. Thus, the order of reducing power can be determined by comparing the values of the standard reduction potentials.

### Step 3: Compare the standard reduction potentials
- The most negative potential is \( -3.05 \, \text{V} \) for \( M_2 \).
- The next most negative is \( -1.66 \, \text{V} \) for \( M_3 \).
- The least negative is \( -0.34 \, \text{V} \) for \( M_1 \).

### Step 4: Determine the order of reducing power
Based on the comparison:
- \( M_2 \) has the highest reducing power (most negative potential).
- \( M_3 \) has a moderate reducing power.
- \( M_1 \) has the lowest reducing power (least negative potential).

Thus, the order of reducing power is:
\[ M_2 > M_3 > M_1 \]

### Final Answer
The order of reducing power is \( M_2 > M_3 > M_1 \).

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