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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Find the time through which an average current 1.93Amp passed to reduce 0.2mol of $MgCl_2$ into Mg .The current efficiency is 60%

$\begin{array}{1 1}(a)\;\large\frac{10^5}{6}\normalsize sec\\(b)\;\large\frac{10^5}{3}\normalsize sec\\(c)\;10^5sec\\(d)\;\large\frac{5}{3}\normalsize\times 10^4sec\end{array}$

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1 Answer

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$MgCl_2+2e^-\rightarrow Mg+2Cl^-$
$MgCl_2\Rightarrow 0.2$
$2e^-\Rightarrow 0.4$
$Mg \Rightarrow 0.2$
No of mole of electron takes part=$\large\frac{1.96\times t}{96500}$
$\Rightarrow \large\frac{t}{5000}$
Consume mole of $e^-=\large\frac{t}{5000}$$\times 0.6$
Produce mole of Mg=$\large\frac{t\times 0.6}{5000\times 2}$
$\Rightarrow 0.2$
$t=\large\frac{0.2\times 2\times 5000}{0.6}$
$t=\large\frac{10^5}{3}$ sec
Hence (b) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

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