`E_(1)`, `E_(2)` and `E_(3)` are the emfs of the following three galvanic cells respectively
I. `Zn((s))`|`Zn^(2+) (0.1M)`| |`CU^(2+) (1M)`| `Cu((s))`
II. `Zn((s)) ZN^(2+) (1M)` ||`Cu^(2+)(1M)`|`Cu(s)`
III. `Zn(s)` | `Zn^(2+) (1M)`||`CU^(2+)(0.1M)CU(s)`

Options

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

Explanation

To solve the problem, we need to calculate the electromotive force (emf) for each of the three galvanic cells using the Nernst equation. The Nernst equation is given by:

\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \left( \frac{[Ox]}{[Red]} \right) \]

Where:
- \( E^{\circ}_{cell} \) is the standard cell potential (which we will assume to be 0 for simplicity in this case).
- \( n \) is the number of moles of electrons transferred in the redox reaction.
- \([Ox]\) is the concentration of the oxidized species (anode).
- \([Red]\) is the concentration of the reduced species (cathode).

### Step 1: Identify the half-reactions and the concentrations for each cell.

1. **Cell I:**
- Anode: \( Zn(s) \rightarrow Zn^{2+}(aq) \) (0.1 M)
- Cathode: \( Cu^{2+}(aq) \rightarrow Cu(s) \) (1 M)
- Concentration of anode = 0.1 M, concentration of cathode = 1 M.

2. **Cell II:**
- Anode: \( Zn(s) \rightarrow Zn^{2+}(aq) \) (1 M)
- Cathode: \( Cu^{2+}(aq) \rightarrow Cu(s) \) (1 M)
- Concentration of anode = 1 M, concentration of cathode = 1 M.

3. **Cell III:**
- Anode: \( Zn(s) \rightarrow Zn^{2+}(aq) \) (1 M)
- Cathode: \( Cu^{2+}(aq) \rightarrow Cu(s) \) (0.1 M)
- Concentration of anode = 1 M, concentration of cathode = 0.1 M.

### Step 2: Calculate \( E_{cell} \) for each galvanic cell.

1. **For Cell I:**
- \( n = 2 \) (2 electrons transferred)
- \( E_{cell} = 0 - \frac{0.0591}{2} \log \left( \frac{0.1}{1} \right) \)
- \( E_{cell} = -0.02955 \log(0.1) \)
- \( \log(0.1) = -1 \)
- \( E_{cell} = -0.02955 \times (-1) = 0.02955 \, V \)

2. **For Cell II:**
- \( n = 2 \)
- \( E_{cell} = 0 - \frac{0.0591}{2} \log \left( \frac{1}{1} \right) \)
- \( E_{cell} = -0.02955 \log(1) \)
- \( \log(1) = 0 \)
- \( E_{cell} = 0 \)

3. **For Cell III:**
- \( n = 2 \)
- \( E_{cell} = 0 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \)
- \( E_{cell} = -0.02955 \log(10) \)
- \( \log(10) = 1 \)
- \( E_{cell} = -0.02955 \times 1 = -0.02955 \, V \)

### Step 3: Compare the values of \( E_{cell} \).

- \( E_1 = 0.02955 \, V \)
- \( E_2 = 0 \, V \)
- \( E_3 = -0.02955 \, V \)

### Step 4: Determine the order of the emfs.

From the calculated values, we can conclude:
- \( E_1 > E_2 > E_3 \)

Thus, the order of the emfs is \( E_1 > E_2 > E_3 \).