When a steady current of 2A was passed through two electrolytic c… - Vidyadip

Question

When a steady current of $2A$ was passed through two electrolytic cells $A$ and $B$ containing electrolytes $ZnSO_4$ and $CuSO_4$ connected in series, $2g$ 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: $Cu = 63.5 \ g \ mol^{-1}, Zn = 65 \ g \ mol^{-1}; 1F = 96500 \ C \ mol^{-1}]$

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Answer

Steady current $= 2A$
copper deposited at cathode of cell $B = 2g$
Cell $B$ contains $CuSO_4$ and reaction may be represented as,
$CuSO _4 \rightarrow Cu ^{2+}+ SO _4^{2-}$
The reaction happens at cathode as,
$Cu^{2+} + 2e^- \rightarrow Cu$
So, $1 \ mol \ Cu$ deposited by $2F$ charge
$63.5 g \ Cu\rightarrow 2 \times 96500$
$2g \ Cu \rightarrow x$
$x =\frac{2 \times 2 \times 96500}{63.5}=6078.74 C$
we know formula
$Q = It$
$6078.74=2 \times t$
$t =\frac{6078.74}{2}=3039.37 \ Sec$
From faraday's second law of electrolysis,
$\frac{\omega t \text { of } C u}{\omega t \text { of } Z n}=\frac{E q \omega t C u}{E q \cdot \omega t Z n}$
$\frac{2}{x}=\frac{63.5 / 2}{65 / 2}$
$x=\frac{2}{0.9769}=2.047 g$
weight of $Zn$ deposited $= 2.047 g$

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