The oxidation ptentials of `Zn, Cu, Ag, H_2` and Ni are `0.76, - 34, - 0.80 . 0. 00, 0. 25` volt, respectively. Which of the following reactions will provide maximum voltage ?

Options

A. `Zn +Cu^(2+) rarr Cu +Zn^(2+)`
B. `Zn +2Ag^(+) rarr 2Ag +Zn^(2+)`
C. `H_(2) +Cu^(2+) rarr 2H^(+) +Cu`
D. `H_(2) +Ni^(2+) rarr 2H^(+) +Ni`

Explanation

To determine which reaction will provide the maximum voltage, we need to calculate the standard cell potential (E°cell) for each of the given reactions using the standard reduction potentials.

### Step-by-Step Solution:

1. **Convert Oxidation Potentials to Reduction Potentials:**
- The oxidation potentials given are:
- Zn: 0.76 V
- Cu: -0.34 V
- Ag: -0.80 V
- H₂: 0.00 V
- Ni: 0.25 V
- To convert these to standard reduction potentials, we reverse the sign:
- E°(Zn) = -0.76 V
- E°(Cu) = +0.34 V
- E°(Ag) = +0.80 V
- E°(H₂) = 0.00 V
- E°(Ni) = -0.25 V

2. **Identify the Reactions:**
- **Option A:** Zn + Cu²⁺ → Cu + Zn²⁺
- **Option B:** Zn + 2Ag⁺ → 2Ag + Zn²⁺
- **Option C:** H₂ + Cu²⁺ → 2H⁺ + Cu
- **Option D:** H₂ + Ni²⁺ → 2H⁺ + Ni

3. **Calculate E°cell for Each Reaction:**
- **For Option A:**
- E°cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn)
- E°cell = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V

- **For Option B:**
- E°cell = E°(Ag⁺/Ag) - E°(Zn²⁺/Zn)
- E°cell = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 V

- **For Option C:**
- E°cell = E°(Cu²⁺/Cu) - E°(H⁺/H₂)
- E°cell = 0.34 - 0.00 = 0.34 V

- **For Option D:**
- E°cell = E°(Ni²⁺/Ni) - E°(H⁺/H₂)
- E°cell = -0.25 - 0.00 = -0.25 V (not spontaneous)

4. **Determine the Maximum Voltage:**
- The calculated E°cell values are:
- Option A: 1.10 V
- Option B: 1.56 V
- Option C: 0.34 V
- Option D: -0.25 V (not possible)
- The maximum voltage is from Option B, which is 1.56 V.

### Final Answer:
The reaction that will provide the maximum voltage is **Option B: Zn + 2Ag⁺ → 2Ag + Zn²⁺**.