Chapter 19, Problem 46E

Interpretation Introduction

Interpretation:

The value of $K_{sp}$ of Ag₃PO₄ should be calculated.

Concept introduction:

${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ dissolves in an aqueous solution as follows;

$\begin{array}{l}{\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}(s)\rightleftharpoons 3A{g}^{+}(aq)+{\text{PO}}_{\text{4}}{}^{\text{3-}}\text{(aq)}\\ \\ {\text{K}}_{sp}={\left[A{g}^{+}(aq)\right]}^3\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\text{(aq)}\right]\end{array}$

In an electrochemical cell, there are two electrodes that can be observed.

If oxidation takes place on an electrode, that electrode is called anode. The species in that electrode remove electrons and itself gets oxidized.

If reduction takes place on an electrode, that electrode is called cathode. The species in that electrode absorbs electrons and itself gets reduced.

The electrode potential of cell can be calculated as follows:

$E^0{}_{\text{cell}}=E^0{}_{\text{cathode}} -E^0{}_{\text{anode}}$

Nernst equation for a cell can be represented as follows:

${\text{E}}_{cell}={\text{E}}^0{}_{cell}-\frac{0.0592\text{ V}}{\text{z}}\text{ log Q}$

Here,

z = number of moles of electrons transferred in the cell and Q is ratio of concentration of products to reactant in a cell.