Chapter 19, Problem 19.57QP

Interpretation Introduction

Interpretation:

Need to calculate the molar mass of copper from the given quantity of current passing through the cell contains CuSO₄ solution.

Concept introduction:

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

The half-cell reaction was written as 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.

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

On dividing the number of charges with Faraday constant we can attain the number of moles of electrons involved in the reaction.

    $\text{Number}\text{of}\text{moles}\text{of}\text{electron (}n\text{) =}\text{charges/Faraday constant (96500 C/mole)}$

From knowing the number of mole of electrons and using the stoichiometry of the reaction, the number of moles of the Cu²⁺ reduced can be determined. In the present case two moles of electrons was accepted by Cu²⁺ ion to produce Cu metal.

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

Then

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

To find: the molar mass of copper obtained through the electrolysis of aqueous CuSO₄solution_, for a given period of time upon passing the given quantity of current.