20 mL of Fe^{2+} solution of certain concent | vedclass

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

Vedclass · Accepted Answer

$0.012$

Vedclass · Answer

(D) The reaction between $Fe^{2+}$ and $Cr_2O_7^{2-}$ in acidic medium is: $Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}$.Here,the $n$-factor for $K_2Cr_2O_7$ is $6$.The reaction between $Fe^{2+}$ and $MnO_4^-$ in acidic medium is: $MnO_4^- + 8H^{+} + 5Fe^{2+} \longrightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$.Here,the $n$-factor for $KMnO_4$ is $5$.Since the same amount of $Fe^{2+}$ reacts in both cases,the number of equivalents of $K_2Cr_2O_7$ must equal the number of equivalents of $KMnO_4$:$M_1 	imes V_1 	imes n_1 = M_2 	imes V_2 	imes n_2$Given: $M_1 = 0.01 \ M$,$V_1 = 20 \ mL$,$n_1 = 6$,$V_2 = 20 \ mL$,$n_2 = 5$.$0.01 	imes 20 	imes 6 = M_2 	imes 20 	imes 5$$M_2 = \frac{0.01 	imes 6}{5} = 0.012 \ M$.Thus,the molarity of $KMnO_4$ is $0.012 \ M$.

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