In the balanced chemical reaction,IO_3^- + a; | vedclass

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Question

(A) The given reaction is $IO_3^- + aI^{-} + bH^{+} 	o cH_2O + dI_2$. Step $1$: Identify oxidation and reduction half-reactions. Oxidation: $I^{-}

Vedclass · Accepted Answer

$5, 6, 3, 3$

Vedclass · Answer

(A) The given reaction is $IO_3^- + aI^{-} + bH^{+} 	o cH_2O + dI_2$. Step $1$: Identify oxidation and reduction half-reactions. Oxidation: $I^{-} 	o I_2$ Reduction: $IO_3^- 	o I_2$ Step $2$: Balance the reduction half-reaction: $2IO_3^- + 12H^{+} + 10e^{-} 	o I_2 + 6H_2O$. Step $3$: Balance the oxidation half-reaction: $2I^{-} 	o I_2 + 2e^{-}$. Step $4$: Multiply the oxidation half-reaction by $5$ to balance electrons: $10I^{-} 	o 5I_2 + 10e^{-}$. Step $5$: Add the two half-reactions: $2IO_3^- + 10I^{-} + 12H^{+} 	o 6I_2 + 6H_2O$. Step $6$: Simplify by dividing by $2$: $IO_3^- + 5I^{-} + 6H^{+} 	o 3H_2O + 3I_2$. 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,$IO_3^- + a\;I^{-} + b\;H^{+} \to c\;H_2O + d\;I_2$,$a$,$b$,$c$,and $d$ respectively correspond to

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