Based on the following data, what is the standard Gibbs free energy of formation of the sulfate ion at 298 K? ( R = 8.31 J/(K ∙ mol)) PbSO4(s) Pb2+(aq) + SO42–(aq); Ksp = 1.7 × 10–8

Substance ΔG°f (kJ/mol) at 298 K
Pb2+(aq)
–24.4

PbSO4(s)
–813.0
a. –788.6 kJ/mol

b. –793.1 kJ/mol

c. –837.4 kJ/mol

d. –744.3 kJ/mol

e. 832.9 kJ/mol

What do you not know about it

Answer : option a) 84explanation :

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

The answer to your question would be  h+ + cn– + li+ + oh– mc001-4.jpg li+ + cn– + h2o. the total ionic equation for the following reaction is  h+ + cn– + li+ + oh– mc001-4.jpg li+ + cn– + h2o.