How to Balance this Equation by Oxidation number method? $H_2SO_4+HI->H_2O+H_2S+I_2$

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Answer 1 · 2017-06-17T12:34:41

Well, $"sulfur(VI+)"$ is REDUCED to $S"(-II)"$ ......

Well, $"sulfur(VI+)"$ is REDUCED to $S"(-II)"$ ......a formal 8 electron transfer.......

$"SO"_4^(2-)+8H^(+) +8e^(-) rarrS^(2-)+4H_2O(l)$ $(i)$

And of course sulfide anion has the same oxidation state as the charge on the ion, i.e. $S(-II)$ ; a formal eight electron reduction.

And for every reduction there is a corresponding oxidation......

$I^(-) rarr 1/2stackrel(0)I_2(s)+e^(-)$ $(ii)$

And we takes $(i)+8xx(ii)$ :

$8I^(-) +SO_4^(2-)+8H^(+) rarr 4stackrel(0)I_2(s) +S^(2-) + 4H_2O(l)$

The which, I think, is balanced with respect to mass and charge; as indeed it must be if it is to represent physical reality.

How to Balance this Equation by Oxidation number method? H_2SO_4+HI->H_2O+H_2S+I_2H_2SO_4+HI->H_2O+H_2S+I_2