The equivalent weight of K_2Cr_2O_7 in acidic | vedclass

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(C) In an acidic medium,the reduction half-reaction for $K_2Cr_2O_7$ is:$Cr_2O_7^{2-} + 14H^+ + 6e^- 	o 2Cr^{3+} + 7H_2O$The change in oxidation stat

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(C) In an acidic medium,the reduction half-reaction for $K_2Cr_2O_7$ is:$Cr_2O_7^{2-} + 14H^+ + 6e^- 	o 2Cr^{3+} + 7H_2O$The change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation state is $2 	imes (6 - 3) = 6$.Thus,the $n$-factor for $K_2Cr_2O_7$ is $6$.The molar mass of $K_2Cr_2O_7$ is $294 \ g/mol$.Equivalent weight $= \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{294}{6} = 49$.

The equivalent weight of $K_2Cr_2O_7$ in acidic medium is:

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