(i) `Cu+ 2HCl_(2) +H_(2)(g)` `[E_(Cu_(2+)//Cu^(@) = ).34V]`
(ii)`Zn + 2HClrightarrow ZNCl+ (2)+ H_(2) ((g))` `[E(ZN^(2+)//ZN^(@) = -0.76V]`
(iii) `Ag+ 2HCl rightarrow AgCl+(1)//(2) H_(2) (g)` `[E(Ag^(+)//Ag)^(@) =+0.80V]`
Which of the following reaction is feasible ?

Options

A. (ii)
B. (i)
C. (iii)
D. All of these

Explanation

To determine which of the given reactions is feasible, we need to analyze the standard electrode potentials (E°) for each half-reaction involved in the reactions. A reaction is considered feasible (spontaneous) if the overall cell potential (E°cell) is positive.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard electrode potentials (E°):**
- For the first reaction:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E° = +0.34 \, \text{V}) \]
- For the second reaction:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad (E° = -0.76 \, \text{V}) \]
- For the third reaction:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E° = +0.80 \, \text{V}) \]

2. **Write the overall reactions:**
- (i) \( \text{Cu} + 2\text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2 \)
- (ii) \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
- (iii) \( \text{Ag} + 2\text{HCl} \rightarrow \text{AgCl} + \text{H}_2 \)

3. **Determine the oxidation and reduction processes:**
- For (i): Cu is oxidized (0 to +2), and H⁺ is reduced (+1 to 0).
- For (ii): Zn is oxidized (0 to +2), and H⁺ is reduced (+1 to 0).
- For (iii): Ag is oxidized (0 to +1), and H⁺ is reduced (+1 to 0).

4. **Calculate E°cell for each reaction:**
- For (i):
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0 - 0.34 = -0.34 \, \text{V} \quad (\text{not feasible}) \]
- For (ii):
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0 - (-0.76) = +0.76 \, \text{V} \quad (\text{feasible}) \]
- For (iii):
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0 - 0.80 = -0.80 \, \text{V} \quad (\text{not feasible}) \]

5. **Conclusion:**
- The only feasible reaction is the second one:
\[ \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]

### Final Answer:
The feasible reaction is (ii) \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \).