Chapter 14.5, Problem 10TY

Interpretation Introduction

Interpretation:

The formation constant and ${\text{E}}^{\text{0}}$ of the given reaction has to be calculated.

${\text{Cu}}^{\text{+}}\text{+2}\text{ethylenediamine}\stackrel{}{\rightleftharpoons }\text{Cu}{\left(\text{ethylenediamine}\right)}_{\text{2}}^{\text{+}}$

Concept Introduction:

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$

Nernst equation:

In electrochemistry, the standard reduction potentials are related with the temperature and gas constant and standard electrode potentials and concentrations.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{0}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{log}\frac{{\left[\text{cathode}\right]}^{\text{n}}}{{\left[\text{anode}\right]}^{\text{n}}}$

Solubility constant:

The general form of a problem where the problem has the relation between ${\text{E}}^{\text{0}}$ values for half reactions is,

${\text{K=10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}$

Where,

$\text{K}$= solubility constant

$\text{n}$= no of electrons.

${\text{E}}^{\text{0}}$=cell potential