Chapter 14, Problem 14.32P

Interpretation Introduction

Interpretation:

$\text{E}{}_{}^{\text{o}}$ value for the half-reaction has to be calculated.

Concept introduction:

Nernst equation can be used to determine the cell potential at any instant, the difference from the standard state.

${\text{E = E}}^{\text{o }}\text{- }\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\frac{{\left[\text{B}\right]}^{\text{b}}{\left[{\text{H}}^{\text{+}}\right]}^{\text{m}}}{{\left[\text{A}\right]}^{\text{a}}}$

Where:

$\text{E}$ = cell potential under nonstandard conditions

$\text{E}{}_{}^{\text{o}}$ = cell potential under standard conditions
n = number of moles of electrons exchanged in the electrochemical reaction

Solubility product: ${\text{K}}_{\text{sp}}$ It is defined as the product formed from mathematical product of its concentration of dissolved ion raised up to the power of their stoichiometric coefficient.

$\begin{array}{l}{\text{M}}_{\text{y}}{\text{X}}_{\text{z(s)}}\underset{}{\rightleftharpoons }{\text{yM}}_{{}_{\text{(aq)}}}^{\text{Z+}}\text{+}{\text{zX}}_{\text{(aq)}}^{\text{y-}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{M}}_{{}_{}}^{\text{Z+}}\right]}^{\text{y}}{\left[{\text{X}}_{{}_{}}^{\text{y-}}\right]}^{\text{z}}\end{array}$