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The EMF of cell Hg/mercurous nitrate (0.01M)$\parallel$ mercurous nitrate (0.1M)/Hg was found to be 0.0295V at $25^{\large\circ}C$ .Then molecular formula of mercurous nitrate is

$\begin{array}{1 1}(a)\;Hg(NO_3)_2\\(b)\;Hg(NO_3)\\(c)\;Hg_2(NO_3)_2\\(d)\;None\end{array}$

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For given cell the reaction occuring at two electrodes are
At anode $nHg \rightarrow (Hg_n^{+n})_A+me^-$
At cathode $(Hg_n^{+n})_C+ne^-\rightarrow nHg$
Net reaction $(Hg_n^{+n})_C\rightarrow (Hg_n^{+n})_A$
So this is a concentration cell for which $E^0_{cell}=0$.The $E_{cell}$ will be given as
$E_{cell}=-\large\frac{0.059}{n}$$\log \big[\large\frac{(Hg_n^{+n})_A}{(Hg_n^{+n})_C}\big]$
$0.0295=\large\frac{0.059}{n}$$\log \big[\large\frac{(Hg_n^{+n})_C}{(Hg_n^{+n})_A}\big]$
$\Rightarrow \large\frac{0.059}{n}$$\log \large\frac{0.1}{0.01}$
$n=2$ and following of mercurous nitrate is $Hg_2(NO_3)_2$
Hence (c) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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