Chapter 19, Problem 19.59QP

Interpretation Introduction

Interpretation:

Need to calculate the molar mass of the unknown metal deposited in an electrolytic cell connected in series with a cell containing AgNO₃ solution.

Concept introduction:

In the present case two electrolytic cell containing AgNO₃ and XCl₃ were connected in series, so the quantity of electricity flowing through both of the cell will be same.

    Half-cell reaction for cell-1:

    ${\text{Cathode (reduction) Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Ag}}_{\text{(s)}}$

The amount of the silver deposited was given, from which the amount of electricity flowing though the cell can be calculated in steps.  Number of moles of silver deposited was calculated first.  Since one mole of electron is needed to reduce one mole of Ag⁺_.  Therefore number of moles of electron needed is equal to number of moles of sliver deposited.  The coulombs of electron passing can be calculated by the equation given below.

  $\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight}\text{(g)}}{\text{Atomic}\text{mass}\text{(g}{\text{mole}}^{\text{-1}}\text{)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of Silver}\\ \text{then}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons}\text{×}\text{Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}\end{array}$

Half-cell reaction for cell-2 was given below:

Since the formula of the electrolyte was given as XCl_3. The electrolytic cell can written as follows.

    ${\text{Cathode (reduction) X}}^{\text{3+}}{}_{\text{(aq)}}{\text{+3e}}^{\text{-}}\stackrel{}{\to }{\text{X}}_{\text{(s)}}$

The above cell was connected in series with AgNO₃,  Therefore the amount of charges will be passing through both the cell. From the coulombs of charges obtained, the number of moles of electrons utilized can be calculated.

    $\text{Number}\text{of}\text{moles of electrons (}n\text{)}\text{=}\frac{\mathrm{Charges}C}{\text{Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}}$

From the half-cell reaction it was known that three mole of electron will be needed to produce one mole of X metal,

So,

  $\text{number}\text{of}\text{moles}\text{of}\text{X=}\text{number}\text{of}\text{mole}\text{of}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}\text{X}}{\text{3mole}{\text{e}}^{\text{-}}}$

On attaining the number of moles of X, we can get the molar mass of X by the given equation

    $\text{molar}\text{mass}\text{of}\text{X}\text{(g/mole) = }\frac{\text{Weight}\text{of}\text{X deposited(g)}}{\text{number}\text{of}\text{moles}\text{of}\text{X}\text{(mole)}}$