The potential associated. with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gås appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction is carried out at 298 K, then the potential of each electrode is said to be the standard electrode potential. By convention, the standard electrode potential of hydrogen electrode is 0:0 volt. The electrode potential value for each electrode process is a measure, of relative tendency of the active species in the process to remain in the oxidized / reduced form. A negative `E^@` means that the redox couple is a stronger reducing agent than the `H^(+)//H_2` couple. A positive E mears that the redox couple is a weaker reducing agent than. the `H^(+)//H` couple. The metal with greater positive value of standard reduction potentlal forms the oxide of greater thermal stability:
Given the standard reduction potentials.
`E_(K^(+)//K)^(@)=-2.93V, E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Hg^(+)//Hg)^(@)=0.79V`
`E_(Mg^(+)//Mg)^(@)=-2.37V, E_(Cr^(3+)//Cr)^(@)=-0.74V`
The correct increasing order of reducing power is:

Options

A. `K gt Mg gt Cr gt Hg gt Ag`
B. `Ag gt Hg gt Cr gt Mg gt K`
C. `Mg gt K gt Cr gt Hg gt Ag`
D. `Cr gt Hg gt K gt Mg gt Ag`

Answer

To determine the increasing order of reducing power based on the given standard reduction potentials, we need to analyze the provided values and understand the relationship between electrode potential and reducing power.

### Step-by-Step Solution:

1. **Understand Reducing Power**:
- A substance with a more negative standard reduction potential (E°) is a stronger reducing agent. This is because it is more likely to lose electrons (be oxidized) compared to the hydrogen electrode, which has a standard potential of 0.0 V.

2. **List the Given Standard Reduction Potentials**:
- \( E_{K^+//K}^\circ = -2.93 \, V \)
- \( E_{Ag^+//Ag}^\circ = +0.80 \, V \)
- \( E_{Hg^+//Hg}^\circ = +0.79 \, V \)
- \( E_{Mg^{2+}//Mg}^\circ = -2.37 \, V \)
- \( E_{Cr^{3+}//Cr}^\circ = -0.74 \, V \)

3. **Convert Reduction Potentials to Oxidation Potentials**:
- The oxidation potential is the negative of the reduction potential:
- \( E_{K//K^+} = +2.93 \, V \)
- \( E_{Ag//Ag^+} = -0.80 \, V \)
- \( E_{Hg//Hg^+} = -0.79 \, V \)
- \( E_{Mg//Mg^{2+}} = +2.37 \, V \)
- \( E_{Cr//Cr^{3+}} = +0.74 \, V \)

4. **Rank the Oxidation Potentials**:
- The higher the oxidation potential, the stronger the reducing agent:
- \( K \) (2.93 V)
- \( Mg \) (2.37 V)
- \( Cr \) (0.74 V)
- \( Hg \) (0.79 V)
- \( Ag \) (0.80 V)

5. **Determine the Increasing Order of Reducing Power**:
- Based on the oxidation potentials, we can arrange the reducing power from weakest to strongest:
- \( Ag < Hg < Cr < Mg < K \)

### Final Answer:
The correct increasing order of reducing power is:
\[ \text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K} \]