An industrial technician successfully electroplates a metal faucet with exactly of Chromium () from — Electrochemistry Chemistry Question
Question
An industrial technician successfully electroplates a metal faucet with exactly $1.04\text{ g}$ of Chromium ($Cr$) from an electrolytic bath containing aqueous Chromium(III) sulphate, $Cr_2(SO_4)_3$. If exactly $19.3\text{ minutes}$ is allowed for the steady plating process, what is the exact continuous current strictly needed in Amperes? (Atomic mass $Cr = 52\text{ g/mol}$, $1\text{ F} = 96500\text{ C}$)
💡 Solution & Explanation
In $Cr_2(SO_4)_3$, chromium exists as $Cr^{3+}$. The reduction reaction is $Cr^{3+} + 3e^- \rightarrow Cr$. Moles of $Cr$ deposited = $1.04\text{ g} / 52\text{ g/mol} = 0.02\text{ mol}$. Moles of electrons required = $0.02 \times 3 = 0.06\text{ mol } e^-$. Total charge $Q = 0.06\text{ F} \times 96500\text{ C/F} = 5790\text{ C}$. Time $t = 19.3\text{ min} \times 60\text{ s/min} = 1158\text{ s}$. Current $I = Q / t = 5790\text{ C} / 1158\text{ s} = 5\text{ A}$.