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

both adhesion and cohesion are attractions.

they are different in that adhesion is the force of attraction between the molecules of two different substances, whereas cohesion is the force of attraction between molecules of the same substance.

possible examples include capillary action, in which the adhesion between water and glass causes the water to rise in a narrow tube against the force of gravity.

because of hydrogen bonding, water is extremely cohesive.

this causes molecules on the surface of water to be drawn inward, which is why drops of water form beads on smooth surfaces.

hope this .

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.