Let's break down how to determine the overall voltage for the redox reaction involving the half-reactions Ag⁺ + e^- → Ag and Cu²⁺ + 2e^- → Cu.

1. Identify the Half-Reactions

* Reduction: Ag⁺ + e^- → Ag (This half-reaction gains electrons)

* Oxidation: Cu → Cu²⁺ + 2e^- (This half-reaction loses electrons)

2. Look Up Standard Reduction Potentials

You'll need a table of standard reduction potentials (E°) to find the values for each half-reaction. Here's a typical example:

* Ag⁺ + e^- → Ag E° = +0.80 V

* Cu²⁺ + 2e^- → Cu E° = +0.34 V

3. Determine the Cell Potential

* The reaction with the higher E° value is the reduction. In this case, Ag⁺ + e^- → Ag is the reduction.

* The reaction with the lower E° value is the oxidation. In this case, Cu → Cu²⁺ + 2e^- is the oxidation.

To calculate the overall cell potential (E°_cell), subtract the standard reduction potential of the oxidation half-reaction from the standard reduction potential of the reduction half-reaction.

E°_cell = E°_reduction - E°_oxidation

E°_cell = +0.80 V - (+0.34 V)

E°_cell = +0.46 V

Therefore, the overall voltage for this redox reaction is +0.46 V.

Important Notes:

* Spontaneity: A positive cell potential (like we calculated) indicates that the reaction is spontaneous under standard conditions.

* Standard Conditions: This calculation assumes standard conditions of 25°C and 1 atm pressure.

* Balancing: Make sure the number of electrons gained in the reduction half-reaction equals the number of electrons lost in the oxidation half-reaction. You may need to multiply one or both half-reactions by a factor to achieve this.