Answer:
[tex]S^0 \rightarrow S^{+4}+4e^-[/tex]
Explanation:
Hello,
In this case, for the reaction:
[tex]K_2Cr_2O_7 + H_2O + S \rightarrow SO_2 + KOH + Cr_2O_3[/tex]
We first must assign the oxidation state of each element:
[tex]K^{+1}_2Cr^{+6}_2O_7^{+2} + H_2^{+1}O^{-2} + S^0 \rightarrow S^{+4}O_2^{-2} + K^{+1}O^{-2}H^{+1} + Cr_2^{+3}O_3^{-2}[/tex]
Thus, we should remember that the oxidation half-reaction applies for the element undergoing an increase in its oxidation state, such case is sulfur, for which passes from 0 to +4 as shown below:
[tex]S^0 \rightarrow S^{+4}+4e^-[/tex]
It means, that four electrons were lost due to the effect of the strong oxidizing agent, potassium dichromate.
Best regards.
