Chapter 16, Problem 16.30P

(a)

Interpretation Introduction

Interpretation:

From given reduction potential equilibrium constant for ${\text{I}}_{\text{2}\left(\text{aq}\right)}{\text{+I}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{I}}_{\text{3}}^{\text{-}}$ has to be calculated.

Concept introduction:

Equilibrium constant (K): In reversible chemical reaction, the relationship between the quantity of reactant and product present at equilibrium at given temperature called equilibrium constant which is represented by number.

$K=10^{\raisebox{1ex}{$n{E}^o$}\!\left/ \!\raisebox{-1ex}{$0.05916$}\right.}$

Where,

$n$ is number of electrons

$E^0$ is reduction potential

(b)

Interpretation Introduction

Interpretation:

From given reduction potential equilibrium constant for ${\text{I}}_{\text{2}\left(\text{s}\right)}{\text{+I}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{I}}_{\text{3}}^{\text{-}}$ has to be calculated.

Concept introduction:

Equilibrium constant (K): In reversible chemical reaction, the relationship between the quantity of reactant and product present at equilibrium at given temperature called equilibrium constant which is represented by number.

$K=10^{\raisebox{1ex}{$n{E}^o$}\!\left/ \!\raisebox{-1ex}{$0.05916$}\right.}$

Where,

$n$ is number of electrons

$E^0$ is reduction potential

(c)

Interpretation Introduction

Interpretation:

From given reduction potential, solubility of solid Iodine in water has to be calculated.

Concept introduction:

Solubility: Solubility is defined as maximum quantity of solute dissolved in given quantity of solvent at given temperature.

The solubility equilibrium law equation is,

$\text{K=}\frac{\left[\text{Product}\right]}{\left[\text{reactant}\right]}$

Note: If substance is pure solid or liquid or gas, its concentration is equal to unity.