Question

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.)

2 K(s) + 12(s) → 2 K⁺(aq) + 2 I⁻ (aq)
K⁺(aq) + e⁻ → K(s) E°= -2.93 V
I2(s) + e⁻ → 2 I⁻ (aq) E° = +0.54 V

Answer 1

The standard cell potential for the reaction 2 K(s) + I2(s) → 2 K+(aq) + 2 I−(aq) is calculated using the provided half-cell potentials, and with the reaction being spontaneous under standard conditions, the calculated cell potential is 3.47 V.

Step-by-step explanation:

To calculate the standard cell potential for the reaction 2 K(s) + I2(s) → 2 K+(aq) + 2 I−(aq) using the given half-cell potentials, we must first identify the oxidation and reduction reactions.

The given half-cell reactions are:

• K+(aq) + e− → K(s) with an E° = -2.93 V (oxidation)
• I2(s) + 2e− → 2 I−(aq) with an E° = +0.54 V (reduction)

Since the potassium is going from the solid state to ionized in solution, it is being oxidized, and thus the sign for its E° value will be reversed when it functions as the anode. The iodine is being reduced from molecular iodine to iodide ions, and will act as the cathode.

Following the equation Ecell = Ecathode - Eanode, the cell potential would be:

Ecell = (+0.54 V) - (-2.93 V) = 3.47 V

The calculated standard cell potential is positive, indicating that the reaction is spontaneous under standard conditions.