In a balanced equation H_2SO_4 + x HI to H_2 | vedclass

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(C) To balance the redox reaction $H_2SO_4 + x \ HI 	o H_2S + y \ I_2 + z \ H_2O$,we analyze the oxidation states: $1$. Sulfur in $H_2SO_4$ goes from

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$x = 8, y = 4, z = 4$

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(C) To balance the redox reaction $H_2SO_4 + x \ HI 	o H_2S + y \ I_2 + z \ H_2O$,we analyze the oxidation states: $1$. Sulfur in $H_2SO_4$ goes from $+6$ to $-2$ in $H_2S$ (reduction,gain of $8 \ e^-$). $2$. Iodine in $HI$ goes from $-1$ to $0$ in $I_2$ (oxidation,loss of $1 \ e^-$ per atom). $3$. To balance the electrons,we multiply the iodine oxidation by $8$: $8 \ HI 	o 4 \ I_2 + 8 \ e^-$. $4$. The balanced equation is $H_2SO_4 + 8 \ HI 	o H_2S + 4 \ I_2 + 4 \ H_2O$. $5$. Comparing coefficients,we get $x = 8, y = 4, z = 4$.

In a balanced equation $H_2SO_4 + x \ HI \to H_2S + y \ I_2 + z \ H_2O$,the values of $x, y, z$ are

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