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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 28 January · Evening Shift · Chemistry · Electrochemistry

Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes...

Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12 . The current in Amperes used for the given electrolysis is ____. (Nearest integer).

Correct answer

0

Step-by-step solution

Electrolysis of NaCl is NaCl+H2O(aq)→NaOH(aq)+12Cl2(g)+12H2(g)\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{ g})+\frac{1}{2} \mathrm{H}_2(\mathrm{ g}) Since during electrolysis pH changes to 12 So [OH⊖]=10−2and\left[\mathrm{OH}^{\ominus}\right]=10^{-2} and\left[\mathrm{H}^{+}\right]=10^{-12}SobyFaradaylawGramamountofsubstancedeposited=AmountofelectricitypassedSo by Faraday law Gram amount of substance deposited = Amount of electricity passed10−2×6001000×96500=I×t10−2×6001000×96500=I×5×60I=10−2×600×965001000×5×60I=1.93 ampere  So ,I=2 ampere (nearest integer) \begin{aligned} & 10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t} \\ & \frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60 \\ & \mathrm{I}=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60} \\ & \mathrm{I}=1.93 \text{ ampere } \\ & \text{ So }, I=2 \text{ ampere (nearest integer) }\end{aligned}$
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