JEE Main 2020 · Chemistry · Electrochemistry · hard

A current of 1.93 A is passed for 50 min through 100 mL of 0.10 M CuSO4 solution with inert electrodes. Using F=96500 C mol^-1 and molar volume 22400…

Numerical answer

A current of

1.93\ \mathrm{A}

is passed for

50\ \mathrm{min}

through

100\ \mathrm{mL}

0.10\ \mathrm{M\ CuSO_4}

solution with inert electrodes. Using

F=96500\ \mathrm{C\,mol^{-1}}

and molar volume

22400\ \mathrm{mL}

at STP, the total volume of gases (

\mathrm{O_2}

at anode plus

\mathrm{H_2}

at cathode) liberated, in mL at STP, is

Show answer and solution

Answer: 784

Q=1.93\times3000=5790

C, i.e.

0.060

mol e

^-

. Anode:

\mathrm{2H_2O\to O_2+4H^++4e^-}

gives

0.015

mol

\mathrm{O_2}=336

mL. Cathode:

\mathrm{Cu^{2+}}

(

0.010

mol) takes

0.020

mol e

^-

first; the remaining

0.040

mol e

^-

reduce water:

\mathrm{2H_2O+2e^-\to H_2+2OH^-}

, giving

0.020

mol

\mathrm{H_2}=448

mL. Total

=336+448=784

mL.

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