Predict the possibility of occurrence of the following reactions :
`(i) Zn+2H^(+) to Zn^(2+) +H_(2), (ii) 2Ag +2H^(+) to 2Ag^(+)+H_(2)`
`{:("Electrode","Standard reduction Potential"(E^(Theta)"in volts")Zn),(Zn^(2+)//Zn," "-0.76),(Ag^(+)//Ag," "+0.80),(2H^(+)//H_(2)," "0.00):}`

Explanation

To predict the possibility of occurrence of the given reactions, we will calculate the standard cell potential (E°cell) for each reaction using the standard reduction potentials provided.

### Step-by-Step Solution:

**(i) Reaction: Zn + 2H⁺ → Zn²⁺ + H₂**

1. **Identify the oxidation and reduction half-reactions:**
- Zinc (Zn) is oxidized from 0 to +2 (Zn²⁺), so it is the anode.
- Hydrogen ions (H⁺) are reduced from +1 to 0 (H₂), so it is the cathode.

2. **Write the half-reactions:**
- Oxidation: Zn → Zn²⁺ + 2e⁻
- Reduction: 2H⁺ + 2e⁻ → H₂

3. **Standard reduction potentials (E°):**
- E°(Zn²⁺/Zn) = -0.76 V (anode)
- E°(H⁺/H₂) = 0.00 V (cathode)

4. **Calculate E°cell:**
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
\[ E°_{cell} = 0.00 V - (-0.76 V) = 0.76 V \]

5. **Determine feasibility:**
Since E°cell is positive (0.76 V), the reaction is feasible.

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**(ii) Reaction: 2Ag + 2H⁺ → 2Ag⁺ + H₂**

1. **Identify the oxidation and reduction half-reactions:**
- Silver (Ag) is oxidized from 0 to +1 (Ag⁺), so it is the anode.
- Hydrogen ions (H⁺) are reduced from +1 to 0 (H₂), so it is the cathode.

2. **Write the half-reactions:**
- Oxidation: 2Ag → 2Ag⁺ + 2e⁻
- Reduction: 2H⁺ + 2e⁻ → H₂

3. **Standard reduction potentials (E°):**
- E°(Ag⁺/Ag) = +0.80 V (anode)
- E°(H⁺/H₂) = 0.00 V (cathode)

4. **Calculate E°cell:**
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
\[ E°_{cell} = 0.00 V - (+0.80 V) = -0.80 V \]

5. **Determine feasibility:**
Since E°cell is negative (-0.80 V), the reaction is not feasible.

### Summary of Results:
- **Reaction (i)**: Zn + 2H⁺ → Zn²⁺ + H₂ is feasible (E°cell = +0.76 V).
- **Reaction (ii)**: 2Ag + 2H⁺ → 2Ag⁺ + H₂ is not feasible (E°cell = -0.80 V).

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