Question

A student does an experiment to determine the molar solubility of zinc cyanide. He constructs a voltaic cell at 298 K consisting of a 0.862 M zinc nitrate solution and a zinc electrode in the cathode compartment, and a saturated zinc cyanide solution and a zinc electrode in the anode compartment. If the cell potential is measured to be 0.113 V, what is the value of Ksp for zinc cyanide at 298 K based on this experiment?

Ksp for Zn(CN)2 =

Answer 1

Given data,

E_cell = 0.113 V

[Zn²⁺] = 0.862 M

Let us consider a reaction,

Zn(CN)₂ (s)-----> Zn²⁺ (aq) + 2 CN^- (aq)

K_sp = [Zn²⁺][CN^-]²

[CN^-] = √K_sp/[Zn²⁺]

According to the Nernst equation,

E_cell = E⁰_cell – (0.0591 V/n)*log ([Zn²⁺]_dil/[Zn²⁺]_conc)

0.113 V = -(0.0591 V/2)*log (√K_sp/[Zn²⁺])

log (√K_sp/[Zn²⁺]) = -(0.113 * 2) / (0.0591)

= -3.82

(√K_sp/[Zn²⁺]) = 10^-3.82

K_sp/[Zn²⁺] = (1.5 x 10^-4)²

K_sp/[Zn²⁺] = 2.25 x 10^-8

K_sp = 2.25 x 10^-8 x [Zn²⁺]

= 2.25 x 10^-8 x 0.862

  K_sp = 1.93 x 10^-8