91Ó°ÊÓ

It's an electrochemical process that involves passing a current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

- The charge of $235\mathrm{C}$is passed through an electrolytic cell containing a molten radium salt.

- The charge of radium ion is $2^{+}$, so, the half-reaction for ${\mathrm{Ra}}^{2+}$reduction is

${\mathrm{Ra}}^{2+}+2{\mathrm{e}}^{-}→\mathrm{Ra}$

- Faraday constant: charge of a mole of electrons $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mol}}^{-}\right)$

- Charge $=235\mathrm{C}$

We can compute the number of moles of electrons using the Faraday constant since we know the charge.

$\begin{array}{c}\mathrm{Charge}={\mathrm{Molese}}^{-}×\mathrm{FMolese}=\frac{\mathrm{Charge}}{\mathrm{F}}\\ =\frac{235\mathrm{C}}{96485\mathrm{C}/{\mathrm{mole}}^{-}}\\ =2.4356×{10}^{-3}{\mathrm{mole}}^{-}\end{array}$

Because one mole of radium is created by two moles of electrons, the total number of moles of radium produced is

$\mathrm{Moles}\mathrm{Ra}=2.4356×{10}^{-3}{\mathrm{mole}}^{-}×\frac{1\mathrm{molRa}}{2{\mathrm{mole}}^{-}}=1.2178×{10}^{-3}\mathrm{mol}\mathrm{Ra}$

The molar mass of radium is $226\mathrm{g}/\mathrm{mol}$, so, the mass of radium produced is

$\mathrm{Mass}\mathrm{Ra}=1.2178×{10}^{-3}\mathrm{molRa}×\frac{226\mathrm{g}}{1\mathrm{mol}\mathrm{Ra}}=0.275\mathrm{g}$

Therefore,$\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{radium}\mathrm{produced}\mathrm{is}0.275\mathrm{g}$.