How many grams of aluminum can form by passing $\mathbf{305}\mathbf{C}$through an electrolytic cell containing a molten aluminum salt?

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 $305\mathrm{C}$is passed through an electrolytic cell containing molten aluminum salt.

- The charge of the aluminum ion is $3^{*}$, so, the half-reaction for ${\mathrm{Al}}^{3+}$ reduction is

${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

- Faraday constant: charge of 1 mole of electrons $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mole}}^{-}\right.)$

- Charge = $305\mathrm{C}$

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

Because one mole of aluminum requires three moles of electrons, the total number of moles of aluminum created is

role="math" localid="1664012563984" $\mathrm{Moles}\mathrm{Al}=3.161\times {10}^{-3}{\mathrm{mole}}^{-}\times \frac{1\mathrm{mol}\mathrm{Al}}{3{\mathrm{mole}}^{-}}=1.0537\times {10}^{-3}\mathrm{mol}\mathrm{Al}$

The molar mass of aluminum is $26.9815\mathrm{g}/\mathrm{mol}$, so, the mass of aluminum produced is

$\mathrm{MassAl}=1.0537\times {10}^{-3}\mathrm{molAl}\times \frac{26.9815\mathrm{g}}{1\mathrm{molAl}}=0.028\mathrm{g}$

Therefore, $\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{aluminum}\mathrm{produced}\mathrm{is}0.028\mathrm{g}$.