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

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.