The next cell contains the following reaction: (6 pts) 2 Ag(s) + Cu(s) → Cu2+ (ac) + 2 Ag + (ac) Determine if the reaction takes place spontaneously. Ag+ (ac) + e − → Ag (s) E∘= 0.80 V Cu2+ (ac) +2 e −→Cu (s) E∘= 0.34 V If it does not occur spontaneously, write the half oxidation reaction and the half reduction reaction with its new voltages. In addition to this, calculate the new voltage of the cell.

To determine if the reaction takes place spontaneously, we can use the standard cell potential formula:

E_{cell} = E_{cathode} - E_{anode}

Given the standard reduction potentials:
E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V}
E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \text{ V}

The standard cell potential for the given reaction can be calculated as follows:

E_{cell} = E_{\text{cathode}} - E_{\text{anode}}
E_{cell} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Ag}^+/\text{Ag}}
E_{cell} = 0.34 \text{ V} - 0.80 \text{ V}
E_{cell} = -0.46 \text{ V}

Since the calculated cell potential is negative (-0.46 V), the reaction does not occur spontaneously in the given direction.

To find the half-reactions for the new spontaneous reaction, we need to reverse the original half-reactions and the sign of their standard reduction potentials:

The half-oxidation reaction would be:
{\text{Ag} (s) \rightarrow Ag^+ (ac) + e^-}
E^\circ_{\text{new, Ag}^+/\text{Ag}} = -0.80 \text{ V}

The half-reduction reaction would be:
{\text{Cu}^{2+} (ac) + 2e^- \rightarrow Cu (s)}
E^\circ_{\text{new, Cu}^{2+}/\text{Cu}} = -0.34 \text{ V}

The new voltage of the cell for the spontaneous reaction can be calculated by subtracting the new anode half-reaction from the new cathode half-reaction:

E_{\text{cell, new}} = E_{\text{cathode}} - E_{\text{anode}}
E_{\text{cell, new}} = E^\circ_{\text{new, Cu}^{2+}/\text{Cu}} - E^\circ_{\text{new, Ag}^+/\text{Ag}}
E_{\text{cell, new}} = -0.34 \text{ V} - (-0.80 \text{ V})
E_{\text{cell, new}} = 0.46 \text{ V}

\boxed{E_{\text{cell, new}} = 0.46 \text{ V}}