`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 regarding the EMFs (E1, E2, E3) of the three galvanic cells, we will use the Nernst equation, which relates the cell potential to the concentrations of the reactants and products involved in the electrochemical reaction.

### Step-by-Step Solution:

1. **Identify the Reactions in Each Cell:**
- For all three cells, the anode reaction (oxidation) is:
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+} + 2e^- \]
- The cathode reaction (reduction) is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \]

2. **Write the Nernst Equation:**
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
where \( n \) is the number of electrons transferred (which is 2 in this case).

3. **Calculate E1:**
For cell I:
- \([\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- \([\text{Cu}^{2+}] = 1 \, \text{M}\)

Substituting into the Nernst equation:
\[ E_1 = E^\circ - \frac{0.0591}{2} \log \left( \frac{0.1}{1} \right) \]
\[ E_1 = E^\circ - \frac{0.0591}{2} \log(0.1) \]
\[ E_1 = E^\circ - \frac{0.0591}{2} (-1) = E^\circ + 0.02955 \]

4. **Calculate E2:**
For cell II:
- \([\text{Zn}^{2+}] = 1 \, \text{M}\)
- \([\text{Cu}^{2+}] = 1 \, \text{M}\)

Substituting into the Nernst equation:
\[ E_2 = E^\circ - \frac{0.0591}{2} \log \left( \frac{1}{1} \right) \]
\[ E_2 = E^\circ - \frac{0.0591}{2} (0) = E^\circ \]

5. **Calculate E3:**
For cell III:
- \([\text{Zn}^{2+}] = 1 \, \text{M}\)
- \([\text{Cu}^{2+}] = 0.1 \, \text{M}\)

Substituting into the Nernst equation:
\[ E_3 = E^\circ - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]
\[ E_3 = E^\circ - \frac{0.0591}{2} (1) = E^\circ - 0.02955 \]

6. **Compare the EMFs:**
- From the calculations:
- \( E_1 = E^\circ + 0.02955 \)
- \( E_2 = E^\circ \)
- \( E_3 = E^\circ - 0.02955 \)

Therefore, the order of EMFs is:
\[ E_1 > E_2 > E_3 \]

### Final Answer:
The order of EMFs is \( E_1 > E_2 > E_3 \).