Question

Which is the correct oxidation half reaction for the following reaction K2Cr2O7 + H2O + S à SO2 + KOH + Cr2O3

Answer 1

Answer:

$S^0 \rightarrow S^{+4}+4e^-$

Explanation:

Hello,

In this case, for the reaction:

$K_2Cr_2O_7 + H_2O + S \rightarrow SO_2 + KOH + Cr_2O_3$

We first must assign the oxidation state of each element:

$K^{+1}_2Cr^{+6}_2O_7^{+2} + H_2^{+1}O^{-2} + S^0 \rightarrow S^{+4}O_2^{-2} + K^{+1}O^{-2}H^{+1} + Cr_2^{+3}O_3^{-2}$

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:

$S^0 \rightarrow S^{+4}+4e^-$

It means, that four electrons were lost due to the effect of the strong oxidizing agent, potassium dichromate.

Best regards.