The equivalent weight of K_2Cr_2O_7 in an aci | vedclass

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Question

(C) In an acidic medium,the reduction half-reaction for potassium dichromate $(K_2Cr_2O_7)$ is:$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_

Vedclass · Accepted Answer

$M/6$

Vedclass · Answer

(C) In an acidic medium,the reduction half-reaction for potassium dichromate $(K_2Cr_2O_7)$ is:$Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$Here,the change in the oxidation state of Chromium $(Cr)$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation number is $2 	imes (6 - 3) = 6$.Therefore,the n-factor for $K_2Cr_2O_7$ in an acidic medium is $6$.The equivalent weight is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{6}$

The equivalent weight of $K_2Cr_2O_7$ in an acidic medium is...... (Molecular weight $= M$)

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