In the following balanced chemical reaction,I | vedclass

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Question

(A) Let us balance the reaction using the oxidation number method.$(i) IO_3^- + I^- + H^+ 	o H_2O + I_2$$(ii)$ Oxidation number change:$\mathop{I}\li

Vedclass · Accepted Answer

$5, 6, 3, 3$

Vedclass · Answer

(A) Let us balance the reaction using the oxidation number method.$(i) IO_3^- + I^- + H^+ 	o H_2O + I_2$$(ii)$ Oxidation number change:$\mathop{I}\limits^{+5}O_3^- + \mathop{I}\limits^{-1} 	o \mathop{I}\limits^{0}_2$Decrease in oxidation number = $5$ units (per $I$ atom in $IO_3^-$)Increase in oxidation number = $1$ unit (per $I$ atom in $I^-$)$(iii)$ To balance the change,multiply $I^-$ by $5$:$IO_3^- + 5I^- + H^+ 	o H_2O + I_2$$(iv)$ Total $I$ atoms on the left is $6$,so $I_2$ on the right must be $3$:$IO_3^- + 5I^- + H^+ 	o H_2O + 3I_2$$(v)$ Balance $O$ atoms by adding $3H_2O$:$IO_3^- + 5I^- + H^+ 	o 3H_2O + 3I_2$$(vi)$ Balance $H$ atoms by adding $6H^+$:$IO_3^- + 5I^- + 6H^+ 	o 3H_2O + 3I_2$Comparing with $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$,we get $a=5, b=6, c=3, d=3$.

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

In the following balanced chemical reaction,$IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$,what are the values of $a, b, c$ and $d$ respectively?

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