For the balanced chemical reaction: IO_3^- + | vedclass

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(A) The given reaction is: $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$.Step-$(I)$: Identify oxidation and reduction half-reactions.Oxidation: $I^- 	o I_2

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$5, 6, 3, 3$

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(A) The given reaction is: $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$.Step-$(I)$: Identify oxidation and reduction half-reactions.Oxidation: $I^- 	o I_2$Reduction: $IO_3^- 	o I_2$Step-$(II)$: Balance the reduction half-reaction:$2IO_3^- + 12H^+ + 10e^- 	o I_2 + 6H_2O$Step-$(III)$: Balance the oxidation half-reaction:$2I^- 	o I_2 + 2e^-$Step-$(IV)$: Multiply the oxidation half-reaction by $5$ to balance the electrons:$10I^- 	o 5I_2 + 10e^-$Step-$(V)$: Add both half-reactions:$2IO_3^- + 10I^- + 12H^+ 	o 6I_2 + 6H_2O$Step-$(VI)$: Simplify by dividing by $2$:$IO_3^- + 5I^- + 6H^+ 	o 3H_2O + 3I_2$Comparing with $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$,we get $a = 5, b = 6, c = 3, d = 3$.

For the 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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