`Zn^(2+) to Zn_((s)),E^(@)=-0.76V`
`Cu^(2+)toCu_((s)),E^(@)=-0.34V`
which of the following is spontaneous

Options

A. `Zn^(2+)toCutoZn+Cu^(2+)`
B. `Cu^(2+)+ZntoCu+Zn^(2+)`
C. `Zn^(2+)+Cu^(2+)toZn+Cu`
D. none of these

Explanation

To determine which reaction is spontaneous, we need to analyze the standard reduction potentials (E°) provided for the half-reactions involving zinc (Zn) and copper (Cu).

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials:**
- The half-reaction for zinc is:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}_{(s)} \quad E^\circ = -0.76 \, V \]
- The half-reaction for copper is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}_{(s)} \quad E^\circ = -0.34 \, V \]

2. **Determine the Oxidation and Reduction:**
- Zinc (Zn) will undergo oxidation because it has a more negative E° value. Thus, it will act as the anode:
\[ \text{Zn}_{(s)} \rightarrow \text{Zn}^{2+} + 2e^- \quad \text{(oxidation)} \]
- Copper (Cu) will undergo reduction because it has a less negative E° value. Thus, it will act as the cathode:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}_{(s)} \quad \text{(reduction)} \]

3. **Calculate the Cell Potential (E°cell):**
- The formula for calculating the cell potential is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.34 \, V) - (-0.76 \, V) = -0.34 + 0.76 = +0.42 \, V \]

4. **Determine Spontaneity:**
- A positive E°cell indicates that the reaction is spontaneous. Therefore, the overall reaction:
\[ \text{Cu}^{2+} + \text{Zn}_{(s)} \rightarrow \text{Cu}_{(s)} + \text{Zn}^{2+} \]
- is spontaneous.

5. **Conclusion:**
- The spontaneous reaction is:
\[ \text{Cu}^{2+} + \text{Zn}_{(s)} \rightarrow \text{Cu}_{(s)} + \text{Zn}^{2+} \]