In the balanced chemical reactionIO_3^- + aI^ | vedclass

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(A) The given reaction is a redox reaction: $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$.Step $1$: Assign oxidation states.In $IO_3^-$,$I$ is $+5$. In $I^-

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$5, 6, 3, 3$

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(A) The given reaction is a redox reaction: $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$.Step $1$: Assign oxidation states.In $IO_3^-$,$I$ is $+5$. In $I^-$,$I$ is $-1$. In $I_2$,$I$ is $0$.Step $2$: Write half-reactions.Reduction: $IO_3^- + 6H^+ + 5e^- 	o \frac{1}{2}I_2 + 3H_2O$Oxidation: $I^- 	o \frac{1}{2}I_2 + e^-$Step $3$: Balance electrons by multiplying the oxidation half-reaction by $5$:$IO_3^- + 6H^+ + 5e^- 	o \frac{1}{2}I_2 + 3H_2O$$5I^- 	o \frac{5}{2}I_2 + 5e^-$Step $4$: Add the half-reactions:$IO_3^- + 5I^- + 6H^+ 	o 3I_2 + 3H_2O$Comparing this with the given equation $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$,we get $a = 5, b = 6, c = 3, d = 3$.

In the balanced chemical reaction
$IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$
$a, b, c, d$ respectively correspond to

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In the balanced chemical reaction$IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$$a, b, c, d$ respectively correspond to