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

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

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Metallic tin in the presence of $HCl$ is oxidized by $K_2Cr_2O_7$ to stannic chloride. What volume of decinormal dichromate solution would be reduced by $1 \ g$ of tin? (Atomic weight of $Sn = 118.7 \ g/mol$)

$KMnO_4$ reacts with ferrous sulphate according to the equation:
$MnO_4^- + 5Fe^{2+} + 8H^{+} \to Mn^{2+} + 5Fe^{3+} + 4H_2O$
Here,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

The number of moles of electrons required to reduce $0.2 \ mol$ of $Cr_2O_7^{2-}$ to $Cr^{3+}$ is:

In alkaline condition $KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:

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