Question 20.15: Calculating the Amount of Product from the Amount of Charge ...

Calculating the Amount of Product from the Amount of Charge in an Electrolysis

When an aqueous solution of potassium iodide is electrolyzed using platinum electrodes, the half-reactions are

\begin{aligned} 2 \mathrm{I}^{-}(a q) & \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{e}^{-} \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} & \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q) \end{aligned}

How many grams of iodine are produced when a current of 8.52 \mathrm{~mA} flows through the cell for 10.0 \mathrm{~min} ?

PROBLEM STRATEGY

The current in amperes multiplied by the time in seconds equals the charge in coulombs. Convert the charge in coulombs to moles of electrons. Then, noting that each mole of iodine produced requires 2 \mathrm{~mol}  \mathrm{e}^{-}, convert \mathrm{mol}  \mathrm{e}^{-}to \mathrm{mol}  I_{2}. Finally, convert \mathrm{mol}  I_{2} to \mathrm{g}  I_{2}.

\text { Coulombs } \longrightarrow \mathrm{mol}   \mathrm{e}^{-} \longrightarrow \mathrm{mol}  \mathrm{I}_{2} \longrightarrow \mathrm{g}  \mathrm{I}_{2}