Question

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.

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Esmeralda

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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}}

Given the standard reduction potentials:

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

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:

The half-reduction reaction would be:

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:

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