Calculate the value of the solubility product constant for PbSO4 from the half-cell potentials. PbSO4(s) + 2e- → Pb(s) + SO42-(aq) Eº = -0.351 V Pb2+ + 2e- → Pb(s) Eº = -0.130 V

K s p  =  3.37  ×  10 ⁻⁸

Explanation:

Anode (oxidation) half reaction:

P b (s) → Pb²⁺(aq) + 2e⁻      E° = 0.130 V

Cathode (reduction) half reaction:

PbSO₄(s) + 2e- → Pb(s) + SO₄²⁻(aq)     Eº = -0.351 V

Adding the two half reactions together gives a net reaction equivalent to the solubility equilibrium of lead (II) sulfate salt:

 PbSO₄(s) → Pb²⁺ + SO₄²⁻(aq)

Ksp = [Pb²⁺] [SO₄²⁻]

E ° c e l l  =  0.130 -  0.351  =  − 0.221  V

using the Nernst equation;  -RTlnK =  − n F E ° c el l

where R is molar gas constant, T is temperature, n is number of moles of electron, F is Faraday's constant,

-RTlnK =  − ( 2  m o l ) ( 96485  C / m o l ) ( − 0.221  V )

-RTlnK  =  -42646 J / m o l

ln K  =  lnK s p  =  -(  42646 / R T)

lnK s p  =  -(  426466/ 8.314 × 298.15 )

Ksp = e∧(-17.204)

K s p  =  3.37  ×  10 ⁻⁸

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.