For `Zn^(2+) //Zn, E^(@) =- 0.76`, for `Ag^(+)//Ag, E^(@) = -0.799V`. The correct statement is

Options

A. the reaction `Zn` getting reduced `Ag` getting oxidized is spontaneous
B. `Zn` undergoes reduction and `Ag` is oxidized
C. `Zn` undergoes oxidation and `Ag^(+)` gets reduced
D. No suitable answer

Answer

To solve the problem, we need to analyze the given reduction potentials for zinc and silver and determine the correct statement based on their tendencies to undergo reduction or oxidation.

### Step-by-Step Solution:

1. **Identify the Given Reduction Potentials**:
- For Zinc (Zn²⁺/Zn), the standard reduction potential (E°) is -0.76 V.
- For Silver (Ag⁺/Ag), the standard reduction potential (E°) is -0.799 V.

2. **Compare the Reduction Potentials**:
- The reduction potential for Zn is -0.76 V, which is higher than that for Ag at -0.799 V.
- A higher reduction potential indicates a greater tendency for that species to be reduced.

3. **Determine Which Species Gets Reduced and Oxidized**:
- Since Zn has a higher reduction potential than Ag, Zn will be reduced (gaining electrons).
- Consequently, Ag, with the lower reduction potential, will be oxidized (losing electrons).

4. **Write the Half-Reactions**:
- Reduction half-reaction for Zinc:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]
- Oxidation half-reaction for Silver:
\[ \text{Ag} \rightarrow \text{Ag}^+ + e^- \]

5. **Determine the Cell Reaction**:
- The overall cell reaction can be represented as:
\[ \text{Zn}^{2+} + 2\text{Ag} \rightarrow \text{Zn} + 2\text{Ag}^+ \]

6. **Conclusion**:
- Based on the analysis, the correct statement is:
"Zn undergoes reduction while Ag is getting oxidized."