During the electrolysis of aqueous solutions, the preferential discharge of competing ions depends s — Electrochemistry Chemistry Question

💡 Solution & Explanation

$Na^+$ ($-2.71\text{ V}$) is much harder to reduce than water ($-0.83\text{ V}$), so $H_2$ forms. At the anode, $O_2$ evolution is thermodynamically favored but kinetically slow, creating an 'overvoltage' that allows $Cl^-$ to oxidize in concentrated brine. $Cu^{2+}$ ($+0.34\text{ V}$) is easily reduced over water. If a reactive copper anode is used, the anode itself oxidizes ($Cu \rightarrow Cu^{2+} + 2e^-$) rather than water oxidizing to $O_2$.