JEE Main 2021 · Chemistry · Electrochemistry · medium

A current of 2 A is passed for 3860 s through 100 mL of 0.10 M CuSO4 solution using inert electrodes. After all Cu^2+ has been deposited, the…

Numerical answer

A current of

2\ \mathrm{A}

is passed for

3860\ \mathrm{s}

through

100\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{CuSO_4}

solution using inert electrodes. After all

\mathrm{Cu^{2+}}

has been deposited, the electrolysis of water continues. The total volume of gases (cathode plus anode) evolved at STP, in

\mathrm{mL}

, is (

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

, molar volume at STP

=22400\ \mathrm{mL}

)

Show answer and solution

Answer: 1120

Charge

=2\times3860=7720\ \mathrm{C}

, i.e.

7720/96500=0.08

mol of electrons. Cathode:

\mathrm{Cu^{2+}}

0.01

mol, needing

0.02

mol

e^-

. The remaining

0.06

mol

e^-

reduce water:

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

, giving

0.03

mol

\mathrm{H_2}

. Anode:

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

uses all

0.08

mol

e^-

, giving

0.02

mol

\mathrm{O_2}

. Total gas

=0.03+0.02=0.05

mol, volume

=0.05\times22400=1120\ \mathrm{mL}

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