Question

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

Answer 1

`S^0 \rightarrow S^(+4)+4e^-`

Step-by-step 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.