Question

Predict the approximate Ksp of Cuz(AsO4)2 based on the measured potential of Cell 7. Use the equation given in the Background.

Answer 1

Answer:

a. 1 x 10^-35

Explanation:

The correct compound given is: $\mathsf{Cu_3(AsO_4)}_2$

To predict the approximate Ksp value of the given compound, we will need to express the oxidation-reduction half-reaction of the compound and its dissociation, then, we will use the Nernst equation to determine the approximate Ksp value.

To start with the reduction half-reaction:

$\mathsf{Cu_3(AsO_4)_{2}(s) + 6e^- \to 2As O_{4}^{3-}_{(aq)}+3Cu(s) }$

The oxidation half-reaction is:

$\mathsf{3Cu(s) \to 3CU^{2+}_{(aq)} + 6e^-}$

The overall cell reaction now is:

$\mathsf{Cu_3(AsO_4)_{2}(s) \to 3Cu^+ (aq) + 2As O_{4}^{3-}_{(aq)} }$

From the reduction half-reduction, the number of moles of electrons (n) transferred is 6 moles.

By applying the Nernst equation:

$\mathsf{E_{cell} = E^0_{cell} -\dfrac{0.0591V}{n}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }$

At standard conditions;

The standard cell potential $\mathsf{E^0_{cell} = -0.342 \ V}$

and $\mathsf{E_{cell} = 0 \ V}$ since it is at equilibrium.

∴

$\mathsf{0 = -0.342 -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 } \\ \\ \\ \mathsf{0.342 = -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }$

$\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = \dfrac{-(0.342)*6}{0.0591 }}$

$\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = -34.7}$

$\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 \simeq -35}$

$\mathsf{[Cu^{2+}]^3[AsO_4^{3-}]^2 = 10^{-35}}$

$\mathbf{K_{sp} = [Cu^{2+}]^3[AsO_4^{3-}]^2 = 1\times 10^{-35}}$