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}
When the current flows for 6.00 \times 10^{2} \mathrm{~s}(10.0 \mathrm{~min}), the amount of charge is
8.52 \times 10^{-3} \mathrm{~A} \times 6.00 \times 10^{2} \mathrm{~s}=5.11 \mathrm{C}
Note that two moles of electrons are equivalent to one mole of \mathrm{I}_{2}. Hence,
5.11 \cancel{C} \times \frac{1 \cancel{\text { mol } e^-}}{9.65 \times 10^{4} \cancel{C}} \times \frac{1 \cancel{\mathrm{mol I}}_{2}}{2 \cancel{\text { mol e }}} \times \frac{254 \mathrm{~g} \mathrm{I}_{2}}{1 \cancel{\mathrm{mol I}}_{2}}=\mathbf{6 . 7 3} \times 10^{-3} \mathbf{g ~ I}_{2}