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14. From the half-reactions below, calculate the solubility product of Mg(OH)2. Mg2+ + 2e º...

14. From the half-reactions below, calculate the solubility product of Mg(OH)₂.

Mg²⁺ + 2e º Mg(s) E° = -2.360 V

Mg(OH)₂(s) + 2e º Mg(s) + 2OH^- E° = -2.690 V

Expert Solution

oxidation :anode

Mg(s) --------------------> Mg²⁺ + 2e º E° = -2.360 V

reduction : cathode

Mg(OH)₂(s) + 2e º -------------------->Mg(s) + 2OH^- E° = -2.690 V

E° = E cathode - E anode

= -2.690 + 2.360

= -0.33

over all reaction

Mg(OH)₂(s) -------------------> Mg²⁺ + 2 OH-

Ecell = E^o cell - 0.0591 / n * log [Mg+2] [OH-]^2

0 = E^o cell - 0.0591 / n * log [Mg+2] [OH-]^2

E^o cell = 0.0591 / n * log [Mg+2] [OH-]^2

E^o cell = 0.0591 / 2 * log Ksp

-0.33 = 0.0591 / 2 * log Ksp

log Ksp = -11.17

Ksp = 6.76 x 10^-12

solubility product = Ksp = 6.76 x 10^-12

queen_honey_blossom answered 1 year ago

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