Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 2 5°C. (The equation is balanced.) Pb (s) + Br2 (l) → Pb2 + (aq) + 2 Br - (aq) Pb2 + (aq) + 2 e - → Pb (s) E° = - 0.13 V Br2 (l) + 2 e - → 2 Br - (aq) E° = + 1.07 V

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰ reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb (s) → Pb2 + (aq) + 2 e - ε⁰ oxidation = - ε⁰ reduction

= - ( - 0.13 V) = + 0.13 V

reduction

Br2 (l) + 2 e - → 2 Br - (aq) ε⁰ reduction = + 1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰ reduction = + 0.13 V + 1.07 V = 1.20 V