Match the following Column-I (Half Reaction) | vedclass

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

Question

(A) The $n$-factor is the change in oxidation state per molecule.$A$. $C_2O_4^{2-} \rightarrow 2CO_2$: Oxidation state of $C$ changes from $+3$ to $+4

Vedclass · Accepted Answer

$A-P, B-Q, C-S, D-R$

Vedclass · Answer

(A) The $n$-factor is the change in oxidation state per molecule.$A$. $C_2O_4^{2-} ightarrow 2CO_2$: Oxidation state of $C$ changes from $+3$ to $+4$. Total change $= 2 	imes (4-3) = 2$. So,$A-P$.$B$. $Cr_2O_7^{2-} ightarrow 2Cr^{3+}$: Oxidation state of $Cr$ changes from $+6$ to $+3$. Total change $= 2 	imes (6-3) = 6$. So,$B-Q$.$C$. $2S_2O_3^{2-} ightarrow S_4O_6^{2-} + 2e^-$: Oxidation state of $S$ changes from $+2$ to $+2.5$. Total change for $2$ moles of $S_2O_3^{2-}$ is $2 	imes [2 	imes (2.5-2)] = 2$. However,per mole of $Na_2S_2O_3$,$n$-factor is $1$. So,$C-S$.$D$. $MnO_4^{-} ightarrow MnO_2$: Oxidation state of $Mn$ changes from $+7$ to $+4$. Total change $= 7-4 = 3$. So,$D-R$.Thus,the correct match is $A-P, B-Q, C-S, D-R$.

Column-$I$	Column-$II$
$A$. $C_2O_4^{2-} \rightarrow 2CO_2$	$P$. $2$
$B$. $Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+}$	$Q$. $6$
$C$. $2Na_2S_2O_3 \rightarrow Na_2S_4O_6$	$R$. $3$
$D$. $MnO_4^{-} \rightarrow MnO_2$	$S$. $1$

Similar Questions

In acidic medium,$K_2Cr_2O_7$ shows oxidising action as represented in the half reaction:
$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$
$X, Y, Z$ and $A$ are respectively:

Which of the following is a redox reaction?

Assign $A, B, C, D$ from the given type of reaction.
$3Mg_{(s)} + N_{2(g)} \longrightarrow Mg_3N_{2(s)}$

Acidified potassium dichromate on reacting with a sulphite is reduced to

In the given chemical reaction,colors of the $Fe^{2+}$ and $Fe^{3+}$ ions,are respectively :
$5 Fe^{2+} + MnO_{4}^{-} + 8 H^{+} \rightarrow Mn^{2+} + 4 H_{2} O + 5 Fe^{3+}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module