When a steady current of 2 {\A} was passed through two electrolytic cells {\A} and {\B} containing electrolytes {\ZnSO}_4 and {\CuSO}_4 connected in series, 2 {\g} of {\Cu} were deposited at the cathode of cell {\B}. How long did the current flow? What mass of {\Zn} was deposited at cathode of cell A? [Atomic mass: {\Cu}=63.5 {\g} {\mol}^{-1}, {\Zn}=65 {\g} {\mol}^{-1} ; 1 F=96500 {\C} {\mol}^{-1}]

CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

Question : 29 of 37

Marks: +1, -0

When a steady current of

2 {\A}

was passed through two electrolytic cells

{\A}

and

{\B}

containing electrolytes

{\ZnSO}_4

and

{\CuSO}_4

connected in series,

2 {\g}

{\Cu}

were deposited at the cathode of cell

{\B}

. How long did the current flow? What mass of

{\Zn}

was deposited at cathode of cell A?
[Atomic mass:

{\Cu}=63.5 {\g} {\mol}^{-1}, {\Zn}=65 {\g} {\mol}^{-1}

;

1 F=96500 {\C} {\mol}^{-1}]

Solution:

\; {\Zn}^{2+}( {\aq})+\;{{2 {\e}^{-}}}↙{{2 {\mol}}} → \;{{ {\Zn}( {\ s}}↙{{1 {\mol}}})}

\; {\Cu}^{2+}+\;{{2 e^{-}}}↙{{2 {\mol}}} → \;{{ {\Cu}(s)}}↙{{1 {\mol}}}

( 2 gm given)
The charge

{\Q}

on a mole of electrons,

{\Q}=n {\F}

Calculation of time for the flow of current:

\;n=1 {\mol}

\;Q=1 × 96500 {\C} {\mol}^{-1}=96500 {\C}

Molar mass of

{\Cu}=63.5 {\gm} {\mol}^{-1}

\; ∵ \;\; 63.5 {\gm} \;\text " of "\; {\Cu} \;\text " is deposited by electric charge "\;

\;=96500 {\C}

\; ∴ \;\; 2 {\gm} \;\text " of "\; {\Cu} \;\text " is deposited by electric charge "\;

\;=\;{96500}/{63.5} × 2=3039.37 {\C}

Let

2 {\A}

of current be passed for time

t

, quantity of electricity used

=2 {\A} × t=3039.37 {\C}

\;\;\text " or, "\; t=\;{3039.37 {\C}}/{2}=1519.68 {\ s}

\;=25 {\min} .33 {\ s}

Calculation of mass of

{\Zn}

deposited:

\;{ {\W}_1}/{ {\W}_2}\;=\;{ {\E}_1}/{ {\E}_2}=\;{\;\text " Mass of "\; {\Zn}}/{\;\text " Mass of "\; {\Cu}}

\;=\;{\;\text " Molar mass of "\; {\Zn} \/ \;\text " Charge on "\; {\Cu}}/{\;\text " Molar mass of "\; {\Cu} \/ \;\text " Charge on "\; {\Cu}}

Amount of

{\Zn}

deposited:

=2 × \;{\;{65}/{2}}/{{635}/{2}}=2.0472 {\gm}

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