Calculate the standard cell potential (E^ext{cell}) at 25^ext{o}C for the following reaction: Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s). Given: E^ext{o}_{ext{Zn}^{2+}/ext{Zn}} = -0.76 V, E^ext{o}_{ext{Cu}^{2+}/ext{Cu}} = +0.34 V.

• A. 1.10 V
• B. 0.42 V
• C. 0.76 V
• D. -1.10 V

Answer: (A)

Verified step by step guidance

1. Identify the half-reactions involved in the overall redox reaction. The oxidation half-reaction is Zn(s) → Zn^{2+}(aq) + 2e^{-}, and the reduction half-reaction is Cu^{2+}(aq) + 2e^{-} → Cu(s).
2. Write down the standard reduction potentials given: E^{\(\circ\)}_{\(\text{Zn}\)^{2+}/\(\text{Zn}\)} = -0.76\,\(\text{V}\) and E^{\(\circ\)}_{\(\text{Cu}\)^{2+}/\(\text{Cu}\)} = +0.34\,\(\text{V}\).
3. Determine which species is oxidized and which is reduced. Zinc is oxidized (loses electrons), so use the oxidation potential for Zn, which is the negative of its reduction potential: E^{\(\circ\)}_{\(\text{oxidation}\)} = +0.76\,\(\text{V}\). Copper ion is reduced, so use its reduction potential as given: +0.34\,\(\text{V}\).
4. Calculate the standard cell potential using the formula: \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\), where the cathode is the reduction half-reaction and the anode is the oxidation half-reaction. Alternatively, since you have the oxidation potential for Zn, you can add the oxidation and reduction potentials: \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{oxidation}} + E^{\circ}_{\text{reduction}}\).
5. Substitute the values into the equation and sum them to find the standard cell potential. This value will indicate the voltage generated by the cell under standard conditions.