For the following balanced equation,a, b, c a | vedclass

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

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$5, 6, 3, 3$

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(A) The given unbalanced equation is: $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$$1$. Assign oxidation states:In $IO_3^-$,$I$ is $+5$.In $I^-$,$I$ is $-1$.In $I_2$,$I$ is $0$.$2$. Identify oxidation and reduction half-reactions:Oxidation: $I^- 	o I_2$ (Change in oxidation state per $I$ atom is $1$,for $2$ atoms it is $2$)Reduction: $IO_3^- 	o I_2$ (Change in oxidation state per $I$ atom is $5$,for $2$ atoms it is $10$)$3$. Balance the electrons:Multiply the oxidation half-reaction by $5$ to balance the electron change $(5 	imes 2 = 10)$:$5I^- 	o 2.5I_2$ (or $10I^- 	o 5I_2$)$4$. Combine and balance atoms:$IO_3^- + 5I^- + 6H^+ 	o 3I_2 + 3H_2O$Comparing this with the given equation $IO_3^- + aI^- + bH^+ 	o cH_2O + dI_2$,we get:$a = 5$$b = 6$$c = 3$$d = 3$Thus,the correct option is $A$.

For the following balanced equation,$a, b, c$ and $d$ will be respectively:
$IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$

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For the following balanced equation,$a, b, c$ and $d$ will be respectively:$IO_3^- + aI^- + bH^+ \to cH_2O + dI_2$