JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Redox Reactions and Electrochemistry

Electrolysis of aqueous NaCl liberates $C{l}_2$ at the anode and $H_2$ with $O{H}^{-}$ at the cathode: $NaCl+H_{2}O\to NaOH+\frac{1}{2}C{l}_2+\frac{1}{2}{H}_2$. A final pH of 12 fixes $[O{H}^{-}]=10^{-2}M$, so the moles of $O{H}^{-}$ produced in 600 mL are $10^{-2}\times 0.6=6\times 10^{-3}mol$. One mole of electrons deposits one mole of $O{H}^{-}$ at the cathode, so the charge passed is $Q=nF=6\times 10^{-3}\times 96500=579C$. Over 5 min (300 s), the current is $I=Q/t=579/300=1.93A$, which rounds to 2 A.