Question

there are half reaction AgI(s) + e -⇌ Ag(s) + I-(aq) E° = -0.150 V Ag+(aq)...

there are half reaction

AgI(s) + e -⇌ Ag(s) + I-(aq) E° = -0.150 V

Ag+(aq) + e-⇌Ag (s) E° =0,80 V

a) Calculate the standard potential of the reaction: AgI(s) + e - ⇌ Ag(s) + I-(aq)  

b) Calculate the solubility product K(s) du AgI (s)

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Answer #1

a) the standard potential of the reaction: AgI(s) + e - ⇌ Ag(s) + I-(aq)

Ksp = [Ag+] [I¯]

AgI(s) + e - ⇌ Ag(s) + I-(aq)    E° = -0.150 V

Ag+(aq) + e-⇌Ag (s)   E° = 0.80 V

Eo = 0.80-(-0.150) = 0.950V

Use the Nernst Equation:

Ecell = Eo - (0.0591 / n) log K

0 = -0.95 - (0.0591 / 1) log K

0.95 / -0.0591 = log K

log K = -16.07

K = 8.51 x 10¯17

b) The solubility product​ Ksp = 8.51 x 10¯17

Note that you never have to use the Ksp expression to calculate anything.

The Ksp is determined directly from the electrochemical data.

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