Given the balanced ionic equation representing a reaction:
2a1(s) + 3cu2+(aq) ==> 2a13+(aq) + 3cu(s)
which half-reaction represents the reduction that occurs? (i ) al ==> al3+ + 3e-
(2) al3+ + 3e- ==> al
(3) cu ==> cu2+ + 2e-
(4) cu2+ + 2e- ==> cu

It's not the same dearthe first one is in grams per mole while the other is not

Iimpact me by the one the yet but him.

Ineed with this also i hope you understand it

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.