Holooly Plus Logo

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} ?


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}

The "Step-by-Step Explanation" refers to a detailed and sequential breakdown of the solution or reasoning behind the answer. This comprehensive explanation walks through each step of the answer, offering you clarity and understanding.
Our explanations are based on the best information we have, but they may not always be right or fit every situation.
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Already have an account?

Related Answered Questions