Consider a titration of potassium dichromate | vedclass

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Question

(D) The balanced redox reaction in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ From the stoichiome

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The balanced redox reaction in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ From the stoichiometry of the balanced equation,$1 	ext{ mole}$ of $Cr_2O_7^{2-}$ reacts with $6 	ext{ moles}$ of $Fe^{2+}$ (Mohr's salt). Alternatively,using the $n$-factor method: $n$-factor of $Cr_2O_7^{2-} = 6$ $n$-factor of $Fe^{2+} = 1$ By the law of equivalence,$	ext{moles of } Cr_2O_7^{2-} 	imes (n	ext{-factor}) = 	ext{moles of } Fe^{2+} 	imes (n	ext{-factor})$ $1 	imes 6 = 	ext{moles of } Fe^{2+} 	imes 1$ Therefore,$6 	ext{ moles}$ of Mohr's salt are required. Hence,$(D)$ is the correct option.

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is

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