The \(E^0 _{Zn^{2+} /Zn} + E^0_{Cu^{2+} /Cu}\) cell gives a positive value of 1.10 V; therefore the process

\(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu\) is spontaneous

The correct answer is option 4. \(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu\) is spontaneous.

To determine the spontaneity of the reaction in the voltaic cell, let's examine the standard reduction potentials and the cell potential given.

Standard Reduction Potentials

For zinc: \( \text{Zn}^{2+} + 2e^{-} \rightarrow \text{Zn} \) (standard reduction potential \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \) V)

For copper: \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \) (standard reduction potential \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \) V)

Cell Potential Calculation

The cell potential \( E^\circ_{\text{cell}} \) is given by the difference between the reduction potentials of the cathode and the anode:

\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)

In this case:

Copper (Cu) is the cathode: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \) V

Zinc (Zn) is the anode: \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \) V

\( E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \)

Since the cell potential \( E^\circ_{\text{cell}} = 1.10 \) V is positive, the reaction is spontaneous.

Spontaneous Reaction

The spontaneous reaction in this voltaic cell can be written as:

\(\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)

This indicates that zinc is oxidized (loses electrons) and copper is reduced (gains electrons).

Therefore, the correct option is: \(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu\) is spontaneous.