Chapter 5, Problem 5J.2E

Interpretation Introduction

Interpretation:

The equilibrium constant for the given cell reaction has to be calculated.

Concept Introduction:

Cell potential:

Cell potential of electrochemical cell is given by,

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

  Where,

  ${\text{E}}_{{}_{\text{cell}}}^{\text{o}}\text{-}$ Overall cell potential

  ${\text{E}}_{\text{cathode}}^{\text{o}}\text{-}$ Cathode half-cell potential

  ${\text{E}}_{\text{anode}}^{\text{o}}\text{-}$ Anode half-cell potential.

Gibbs energy of cell:

Gibbs energy change of cell is given by

  ${\text{ΔG}}^{\text{o}}\text{=}{\text{-nFE}}_{\text{cell}}^{\text{o}}$

Where

${\text{ΔG}}^{\text{o}}-$ Gibb’s energy change

$\text{F-}$ Faraday constant

$\text{n-}$ Number of electrons.