A solution containing 0.14M Pb2+, 1.5×10−6M Pb4+, 1.5×10−6M Mn2+, 0.14M MnO−4, and 0.83M HNO3 was prepared. For this solution, the balanced reduction half‑reactions and overall net reaction shown can occur. 5[Pb4++2e−↽−−⇀Pb2+] ∘+=1.690 V
2[MnO−4+8H++5e−↽−−⇀Mn2++4H2O] ∘−=1.507 V
5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO−4+16 H+

A) Determine ∘cell , Δ∘ , and for this reaction. B) Calculate the value for the cell potential, cell , and the free energy, Δ , for the given conditions. C) Calculate the value of cell for this system at equilibrium. D) Determine the pH at which the given concentrations of Pb2+, Pb4+, Mn2+, and MnO−4 would be at equilibrium.

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

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.

Answer the same molecular formula;