Which of the following statements is true concerning the electrolysis of a 1.0 m aqueous solution of nai? reduction half-reaction e° (v) na+(aq) + e– na(s) –2.71 2h2o(l) + 2e– h2(g) + 2oh–(aq) –0.83 2h+(aq) + 2e– h2(g) 0.00 i2(s) + 2e– 2i–(aq) 0.54 o2(g) + 4h+(aq) + 4e– 2h2o(l) 1.23

Compound with sulfer  is bright red 

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.

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