When 10 mL of an aqueous solution of Fe^{2+} | vedclass

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(B) The balanced redox reaction is: $6Fe^{2+} + Cr_{2}O_{7}^{2-} + 14H^{+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_{2}O$. At the equivalence point,the mi

Vedclass · Accepted Answer

$18$

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(B) The balanced redox reaction is: $6Fe^{2+} + Cr_{2}O_{7}^{2-} + 14H^{+} ightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_{2}O$. At the equivalence point,the milli-equivalents of $Fe^{2+}$ equal the milli-equivalents of $K_{2}Cr_{2}O_{7}$. For $Fe^{2+}$,the n-factor is $1$ $(Fe^{2+} ightarrow Fe^{3+} + e^{-})$. For $K_{2}Cr_{2}O_{7}$,the n-factor is $6$ $(Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} ightarrow 2Cr^{3+} + 7H_{2}O)$. Let $M$ be the molarity of $Fe^{2+}$ solution. $M 	imes 10 	imes 1 = 0.02 	imes 15 	imes 6$. $10M = 1.8$. $M = 0.18 \ M = 18 	imes 10^{-2} \ M$. Thus,the value of $x$ is $18$.

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