How many moles of K_2Cr_2O_7 can be reduced b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The balanced half-reactions are:$Cr_2O_7^{2-} + 14H^{+} + 6e^- 	o 2Cr^{3+} + 7H_2O$$(Sn^{2+} 	o Sn^{4+} + 2e^-) 	imes 3$Adding these gives the

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) The balanced half-reactions are:$Cr_2O_7^{2-} + 14H^{+} + 6e^- 	o 2Cr^{3+} + 7H_2O$$(Sn^{2+} 	o Sn^{4+} + 2e^-) 	imes 3$Adding these gives the overall balanced equation:$Cr_2O_7^{2-} + 14H^{+} + 3Sn^{2+} 	o 3Sn^{4+} + 2Cr^{3+} + 7H_2O$From the stoichiometry of the balanced equation,$3 \ mol$ of $Sn^{2+}$ are required to reduce $1 \ mol$ of $Cr_2O_7^{2-}$.Therefore,$1 \ mol$ of $Sn^{2+}$ will reduce $\frac{1}{3} \ mol$ of $K_2Cr_2O_7$.

How many moles of $K_2Cr_2O_7$ can be reduced by $1 \ mol$ of $Sn^{2+}$?

Similar Questions

The oxidation state of chromium in the final product formed in the reaction between $KI$ and acidified $K_{2}Cr_{2}O_{7}$ solution is:

What is the oxidation state of chromium in the final product formed by the reaction between $KI$ and acidic $K_2Cr_2O_7$ solution?

In the reduction of dichromate by $Fe(II)$,the number of electrons involved per chromium atom is:

The equivalent weight of $Na_2S_2O_3$ (Gram molecular weight $= M$) in the given reaction is $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.

Which of the following reactions involves both oxidation and reduction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module