Chapter 19, Problem 19.118QP

Interpretation Introduction

Interpretation:

The ionic product of water has to be calculated.

Concept Introduction:

Ionic product of water$\left({\text{K}}_{\text{w}}\right)$ is the equilibrium constant for the self ionization of water.  The self ionisation of water is given below.

  ${\text{H}}_2{\text{O}}_{\text{(l)}}\to {\text{H}}^{+}{}_{\text{(aq)}}+{\text{OH}}^{-}{}_{\text{(aq)}}$

The equilibrium constant $\left({\text{K}}_{\text{w}}\right)$

  ${\text{K}}_{\text{w}}=\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

    At room temperature the value of ${\text{K}}_{\text{w}}$ is around ${\text{10}}^{-14}$

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

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

The relation between electrode potential and ionic product of water: standard electrode potential and ionic product of water are related by the given equation.

  ${\text{E}}^{°}{}_{\text{cell}}\text{=}\frac{\text{0}\text{.0592V}}{\text{n}}{\text{logK}}_{\text{w}}$

Where,

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

  $\text{n}$ is the number of electrons

  ${\text{K}}_{\text{w}}$ is the ionic product of water