What will be the coefficients x, y, z in the | vedclass

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(C) The given unbalanced equation is: $IO_3^- + HSO_3^- 	o I^- + SO_4^{2-} + H^+$. Step $1$: Assign oxidation states. $I$ in $IO_3^-$ is $+5$,$I$ in

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$1, 3, 1$

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(C) The given unbalanced equation is: $IO_3^- + HSO_3^- 	o I^- + SO_4^{2-} + H^+$. Step $1$: Assign oxidation states. $I$ in $IO_3^-$ is $+5$,$I$ in $I^-$ is $-1$. $S$ in $HSO_3^-$ is $+4$,$S$ in $SO_4^{2-}$ is $+6$. Step $2$: Write half-reactions. Reduction: $IO_3^- + 6H^+ + 6e^- 	o I^- + 3H_2O$. Oxidation: $HSO_3^- + H_2O 	o SO_4^{2-} + 3H^+ + 2e^-$. Step $3$: Multiply oxidation half-reaction by $3$ to balance electrons: $3HSO_3^- + 3H_2O 	o 3SO_4^{2-} + 9H^+ + 6e^-$. Step $4$: Add the two half-reactions: $IO_3^- + 3HSO_3^- + 6H^+ + 3H_2O 	o I^- + 3SO_4^{2-} + 9H^+ + 3H_2O$. Step $5$: Simplify: $IO_3^- + 3HSO_3^- 	o I^- + 3SO_4^{2-} + 3H^+$. Comparing with $xIO_3^- + yHSO_3^- 	o zI^- + pH^+ + qSO_4^{2-}$,we get $x=1, y=3, z=1$.

What will be the coefficients $x, y, z$ in the balanced form of the following redox equation: $xIO_3^- + yHSO_3^- \to zI^{-} + pH^{+} + qSO_4^{2-}$?

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