In the balanced chemical reaction IO_3^- + aI | vedclass

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Question

(A) The given unbalanced equation is: $IO_3^- + aI^- + bH^+ \longrightarrow cH_2O + dI_2$To balance the redox reaction:$1$. Oxidation half-reaction: $

Vedclass · Accepted Answer

$5, 6, 3, 3$

Vedclass · Answer

(A) The given unbalanced equation is: $IO_3^- + aI^- + bH^+ \longrightarrow cH_2O + dI_2$To balance the redox reaction:$1$. Oxidation half-reaction: $2I^- \longrightarrow I_2 + 2e^-$$2$. Reduction half-reaction: $IO_3^- + 6H^+ + 5e^- \longrightarrow \frac{1}{2}I_2 + 3H_2O$Multiplying the oxidation half-reaction by $5$ and the reduction half-reaction by $2$ to balance electrons:$10I^- \longrightarrow 5I_2 + 10e^-$$2IO_3^- + 12H^+ + 10e^- \longrightarrow I_2 + 6H_2O$Adding them together:$2IO_3^- + 10I^- + 12H^+ \longrightarrow 6I_2 + 6H_2O$Dividing by $2$ to simplify:$IO_3^- + 5I^- + 6H^+ \longrightarrow 3H_2O + 3I_2$Comparing with the given equation,we get $a=5, b=6, c=3, d=3$.Thus,the correct option is $A$.

In the balanced chemical reaction $IO_3^- + aI^- + bH^+ \longrightarrow cH_2O + dI_2$,the values of $a, b, c,$ and $d$ are respectively:

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