Equivalent weights of KMnO_4 and K_2Cr_2O_7 i | vedclass

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(D) In acidic medium,the reduction half-reaction for $KMnO_4$ is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$ Since $5$ electrons ar

Vedclass · Accepted Answer

$\frac{M_A}{5}, \frac{M_B}{6}$

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(D) In acidic medium,the reduction half-reaction for $KMnO_4$ is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$ Since $5$ electrons are involved,the n-factor is $5$. Therefore,the equivalent weight of $KMnO_4 = \frac{M_A}{5}$. Similarly,the reduction half-reaction for $K_2Cr_2O_7$ in acidic medium is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_2O$ Since $6$ electrons are involved,the n-factor is $6$. Therefore,the equivalent weight of $K_2Cr_2O_7 = \frac{M_B}{6}$.

Equivalent weights of $KMnO_4$ and $K_2Cr_2O_7$ in acidic medium are respectively. (Molecular weight of $KMnO_4 = M_A$ and Molecular weight of $K_2Cr_2O_7 = M_B$).

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