Chapter 18, Problem 18.73QP

Interpretation Introduction

Interpretation:

The solubility product of $\text{AgBr}$ has to be calculated

Concept Introduction:

Solubility Product $\left({\text{K}}_{\text{sp}}\right)$: Solubility is the property of a salute to dissolve in a solvent.  There are some solutes which are not completely soluble in the given solvent.  For example $\text{AgBr}$ is not completely soluble in water.  The dissolved phase of $\text{AgBr}$ is in equilibrium with the undissolved phase.  The solubility product is the product of the dissolved ion concentration in the solution.

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

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

Nernst equation is one of the important equation in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the  species that is involved in the chemical reaction and temperature.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{n}$ is the number of electrons involved in a reaction

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

$\left[\text{Red}\right]$ is the concentration of the reduced species

$\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({25}^{°}C)$, after substituting the values of all the constants the equation can be written as

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$