In a titration experiment, 10 ,mL of an FeCl | vedclass

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Question

(a) In titration at equivalence point, number of equivalents of both reactants must be same.
$	herefore N_{1} V_{1}=N_{2} V_{2}$
$=\frac{(N 	imes

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(a) In titration at equivalence point, number of equivalents of both reactants must be same.
$	herefore N_{1} V_{1}=N_{2} V_{2}$
$=\frac{(N 	imes V) 	ext { of } K _{2} Cr _{2} O _{7}}{V 	ext { of } FeCl _{2}}$
$N$ (normality) of $K _{2} Cr _{2} O _{7}$
$=\frac{	ext { Mass } 	imes 1000}{	ext { Equivalent mass } 	imes 	ext { Volume of }}$
$	ext { Equivalent mass }=\frac{	ext { Molar mass }}{n\, 	ext { factor }}$
$=\frac{294}{6}$
$=49\, 	ext { g/equi. }$
$\left( K _{2} Cr _{2} O _{7}ight.$ as oxidising agent is reduced to $\left.2 Cr ^{3+}ight)$
Oxidation state of $2\, Cr$ atoms change from $+\,6$ to $+\,3$
Total change is $3 	imes 2=6,(	herefore n$-factor $=6)$
$	herefore N\left(ight.$ of $\left.K _{2} Cr _{2} O _{7}ight)=\frac{1225 	imes 1000}{49 	imes 250}=0.1 \,N$
$	herefore N\left(ight.$ of $\left.FeCl _{2}ight)=\frac{0.1 	imes 25}{10}$
$=0.25 \,N$

${E^\Theta }$	$V$	$Cr$	$Mn$	$Fe$	$Co$	$Ni$	$Cu$
$\left( {{M^{2 + }}/M} \right)$	$-1.18$	$-0.91$	$-1.18$	$-0.44$	$-0.28$	$-0.25$	$+0.35$

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