15 text{ mL} of aqueous solution of Fe^{2+} i | vedclass

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(C) The balanced redox reaction in acidic medium is: $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$ According to the law of

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) The balanced redox reaction in acidic medium is: $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ ightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$ According to the law of equivalence: $n_{eq}(Fe^{2+}) = n_{eq}(Cr_2O_7^{2-})$ $(Molarity 	imes Volume 	imes n-factor)_{Fe^{2+}} = (Molarity 	imes Volume 	imes n-factor)_{Cr_2O_7^{2-}}$ For $Fe^{2+} ightarrow Fe^{3+}$,the $n-factor = 1$. For $Cr_2O_7^{2-} ightarrow 2Cr^{3+}$,the $n-factor = 6$. $\left(\frac{15 	imes M_{Fe^{2+}}}{1000}ight) 	imes 1 = \left(\frac{20 	imes 0.03}{1000}ight) 	imes 6$ $15 	imes M_{Fe^{2+}} = 20 	imes 0.03 	imes 6$ $15 	imes M_{Fe^{2+}} = 3.6$ $M_{Fe^{2+}} = \frac{3.6}{15} = 0.24 	ext{ M}$ $0.24 	ext{ M} = 24 	imes 10^{-2} 	ext{ M}$ Thus,the value is $24$.

$15 \text{ mL}$ of aqueous solution of $Fe^{2+}$ in acidic medium completely reacted with $20 \text{ mL}$ of $0.03 \text{ M}$ aqueous $Cr_2O_7^{2-}$. The molarity of the $Fe^{2+}$ solution is $\times 10^{-2} \text{ M}$ (Round off to the Nearest Integer).

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