Chapter 19, Problem 19.58QP

Interpretation Introduction

Interpretation:

Need to calculate the Faraday constant for the electrolysis of CuSO₄.

Concept introduction:

Electrolysis of CuSO₄solution resulted in the deposition of copper at the cathode.

The half-cell reaction was written below

  ${\text{cathode (reduction) Cu}}^{\text{2+}}{}_{\text{(aq)}}+2{\text{e}}^{\text{-}}\stackrel{}{\to }{\text{ Cu}}_{\text{(s)}}$

From the given time and current the coulombs of charges flowing through the circuit can be calculated by the given formula.

    $\text{charges}\text{(C)}\text{=}\text{current}\text{(A)}\times \text{time}\text{(s)}$

Since the quantity of copper deposited was given so the number of moles of copper produced can be calculated. On applying it into stoichiometry of the reaction, the number of moles of electrons can be determined.

    $\text{Number}\text{of}\text{moles of Cu}\text{=}\frac{\text{Weight}\text{of}\text{Cu}\text{deposited}}{\text{Molar mass}}$

Reduction of Cu²⁺ion need two electron then the number of moles of electrons can calculated as given below

$\text{Number of mole}\text{of}{\text{e}}^{\text{-}}\text{(}n\text{)}\text{=}\text{Number}\text{of}\text{moles}\text{of}\text{Cu}\text{×}\frac{\text{2 mole}\text{Cu}}{\text{1}\text{mole}{\text{e}}^{\text{-}}}$

Both charges and number of moles of electrons can be attained as stated above further Faraday constant can be calculated by the following formula. 

    $\text{Faraday constant =}\text{charges}\text{(C)/number}\text{of}\text{moles}\text{of}{\text{e}}^{\text{-}}n\text{(mole)}$

To determine: Faraday constant for the electrolysis of CuSO₄.