aIO_3^- + bI^- + 6H^+ to cI_2 + 3H_2OIn the a | vedclass

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(A) To balance the redox reaction $aIO_3^- + bI^- + 6H^+ 	o cI_2 + 3H_2O$:$1$. Identify the oxidation states: $I$ in $IO_3^-$ is $+5$,$I$ in $I^-$ is

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$1, 5, 3$

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(A) To balance the redox reaction $aIO_3^- + bI^- + 6H^+ 	o cI_2 + 3H_2O$:$1$. Identify the oxidation states: $I$ in $IO_3^-$ is $+5$,$I$ in $I^-$ is $-1$,and $I$ in $I_2$ is $0$.$2$. Oxidation half-reaction: $2I^- 	o I_2 + 2e^-$.$3$. Reduction half-reaction: $IO_3^- + 6H^+ + 5e^- 	o \frac{1}{2}I_2 + 3H_2O$.$4$. To balance electrons,multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:$10I^- 	o 5I_2 + 10e^-$$2IO_3^- + 12H^+ + 10e^- 	o I_2 + 6H_2O$$5$. Adding these gives: $2IO_3^- + 10I^- + 12H^+ 	o 6I_2 + 6H_2O$.$6$. Dividing by $2$ to simplify: $1IO_3^- + 5I^- + 6H^+ 	o 3I_2 + 3H_2O$.Thus,$a = 1$,$b = 5$,and $c = 3$.

$aIO_3^- + bI^- + 6H^+ \to cI_2 + 3H_2O$
In the above reaction,the coefficients $a$,$b$,and $c$ are respectively:

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$aIO_3^- + bI^- + 6H^+ \to cI_2 + 3H_2O$In the above reaction,the coefficients $a$,$b$,and $c$ are respectively: