Given that the standard elctrode potentials `(E^(Theta))` of metals are:
`K^(+)//K= -2.93V, Ag^(+)//g=0.80V`,
`Cu^(2+)//Cu=0.34V, Mg^(2+)//Mg=-2.37V`,
`Cr^(3+)//Cr= -0.74V, Fe^(2+)//Fe= -0.44V`.
Arrange these metals in an increasing order of their reducing power.

Explanation

To arrange the metals in increasing order of their reducing power based on their standard electrode potentials, we can follow these steps:

### Step 1: Understand Reducing Power
The reducing power of a metal is inversely related to its standard electrode potential (E°). A more negative E° indicates a stronger tendency to lose electrons (be oxidized), thus making it a better reducing agent.

### Step 2: List the Given Standard Electrode Potentials
We have the following standard electrode potentials:
- \( K^+ // K = -2.93 \, V \)
- \( Ag^+ // Ag = 0.80 \, V \)
- \( Cu^{2+} // Cu = 0.34 \, V \)
- \( Mg^{2+} // Mg = -2.37 \, V \)
- \( Cr^{3+} // Cr = -0.74 \, V \)
- \( Fe^{2+} // Fe = -0.44 \, V \)

### Step 3: Arrange the Metals by Their E° Values
To find the increasing order of reducing power, we will arrange the metals from the most positive E° to the most negative E°:
1. \( Ag^+ // Ag = 0.80 \, V \) (most positive)
2. \( Cu^{2+} // Cu = 0.34 \, V \)
3. \( Fe^{2+} // Fe = -0.44 \, V \)
4. \( Cr^{3+} // Cr = -0.74 \, V \)
5. \( Mg^{2+} // Mg = -2.37 \, V \)
6. \( K^+ // K = -2.93 \, V \) (most negative)

### Step 4: Determine the Order of Reducing Power
Now, we can determine the order of reducing power by reversing the order of E° values:
- Strongest reducing agent (most negative E°): \( K \)
- Next: \( Mg \)
- Next: \( Cr \)
- Next: \( Fe \)
- Next: \( Cu \)
- Weakest reducing agent (most positive E°): \( Ag \)

### Final Order of Reducing Power
Thus, the metals arranged in increasing order of their reducing power are:
1. \( Ag \)
2. \( Cu \)
3. \( Fe \)
4. \( Cr \)
5. \( Mg \)
6. \( K \)