`E^(@)` for `CL_(2)//Cl^(-)=+1.36,I_(2)//I^(-)=+0.53,Ag^(+)//Ag=+0.79,Na^(+)//Na=-2.71 and Li^(+)//Li=-3.04V`. What is the order of decreasing reducing sterngth for these species?

Options

A. `Li gt Cl^(-) gt Ag gt I^(-) gt Na`
B. `Li gt Na gt I^(-) gt Ag gt Cl^-`
C. `Cl^(-) gt Ag gt I^(-) gt Na gt Li`
D. `Na gt Li gt Ag gt Cl^(-) gt I^(-)`

Explanation

To determine the order of decreasing reducing strength for the given species based on their standard reduction potentials (E°), we need to follow these steps:

### Step 1: List the Standard Reduction Potentials
We have the following standard reduction potentials:
- \( E^\circ \) for \( Cl_2/Cl^- = +1.36 \, V \)
- \( E^\circ \) for \( I_2/I^- = +0.53 \, V \)
- \( E^\circ \) for \( Ag^+/Ag = +0.79 \, V \)
- \( E^\circ \) for \( Na^+/Na = -2.71 \, V \)
- \( E^\circ \) for \( Li^+/Li = -3.04 \, V \)

### Step 2: Understand Reducing Strength
Reducing strength refers to the ability of a species to get oxidized (lose electrons) and, in turn, reduce other species. The species with the lowest standard reduction potential will have the highest tendency to get oxidized.

### Step 3: Arrange the Potentials in Increasing Order
To find the order of decreasing reducing strength, we need to arrange the standard reduction potentials from the lowest to the highest:
1. \( Li^+/Li = -3.04 \, V \) (lowest)
2. \( Na^+/Na = -2.71 \, V \)
3. \( I_2/I^- = +0.53 \, V \)
4. \( Ag^+/Ag = +0.79 \, V \)
5. \( Cl_2/Cl^- = +1.36 \, V \) (highest)

### Step 4: Determine the Order of Reducing Strength
Based on the arrangement, the order of decreasing reducing strength (from highest to lowest) is:
1. \( Li \) (highest reducing strength)
2. \( Na \)
3. \( I^- \)
4. \( Ag \)
5. \( Cl^- \) (lowest reducing strength)

### Final Answer
The order of decreasing reducing strength for the species is:
**Li > Na > I^- > Ag > Cl^-**