
$S{O_2}$ act as an oxidant while reacting with:
(A) Acidified $KMn{O_4}$
(B) Acidified ${K_2}C{r_2}{O_7}$
(C) ${H_2}S$
(D) Acidified ${C_2}{H_5}OH$
Answer
233.1k+ views
Hint: Try to recall that oxidants are those substances which decrease their oxidation number and increase the oxidation number of the other species in a redox reaction and act as oxidising agents. Now by using this you can easily find the correct option to the given question.
Complete step by step solution:
> It is known to you that oxidants are those species which increases the oxidation number of other species in a redox reaction. Now, to check on reacting with which species (given in options) $S{O_2}$act as an oxidant, you can easily find by inspecting each option:
> Acidified $KMn{O_4}$- In $KMn{O_4}$, the oxidation number of manganese(Mn) is +7 which is its maximum oxidation number and cannot be oxidized further. So, it will decrease its oxidation number in a redox reaction and thus,$S{O_2}$ do not act as an oxidant for Acidified $KMn{O_4}$.
> Acidified ${K_2}C{r_2}{O_7}$-In ${K_2}C{r_2}{O_7}$, the oxidation number of chromium(Cr) is +6 which is its maximum oxidation number and cannot be oxidized further. So, it will decrease its oxidation number in a redox reaction and thus,$S{O_2}$ do not act as an oxidant for Acidified ${K_2}C{r_2}{O_7}$.
${H_2}S$- In ${H_2}S$, sulphur has oxidation number of (-2) in ${H_2}S$ and this is minimum oxidation number of sulphur and cannot be reduced further. Therefore, sulphur will increase its oxidation number (from -2 to 0) in a redox reaction with $S{O_2}$ and thus, $S{O_2}$ act as an oxidant for ${H_2}S$.
> Acidified ${C_2}{H_5}OH$- In this case also, $S{O_2}$ do not act as an oxidant forAcidified ${C_2}{H_5}OH$.
Therefore, from above discussion we can easily conclude that option C is the correct option for the given question.
Note: It should be remembered to you that Acidified ${K_2}C{r_2}{O_7}$ is one of the most commonly used moderate oxidising agents in organic chemistry.
Also, you should remember that $S{O_2}$ is a colorless gas which acts as a bleaching agent to remove excess chlorine and also act as a disinfectant.
Complete step by step solution:
> It is known to you that oxidants are those species which increases the oxidation number of other species in a redox reaction. Now, to check on reacting with which species (given in options) $S{O_2}$act as an oxidant, you can easily find by inspecting each option:
> Acidified $KMn{O_4}$- In $KMn{O_4}$, the oxidation number of manganese(Mn) is +7 which is its maximum oxidation number and cannot be oxidized further. So, it will decrease its oxidation number in a redox reaction and thus,$S{O_2}$ do not act as an oxidant for Acidified $KMn{O_4}$.
> Acidified ${K_2}C{r_2}{O_7}$-In ${K_2}C{r_2}{O_7}$, the oxidation number of chromium(Cr) is +6 which is its maximum oxidation number and cannot be oxidized further. So, it will decrease its oxidation number in a redox reaction and thus,$S{O_2}$ do not act as an oxidant for Acidified ${K_2}C{r_2}{O_7}$.
${H_2}S$- In ${H_2}S$, sulphur has oxidation number of (-2) in ${H_2}S$ and this is minimum oxidation number of sulphur and cannot be reduced further. Therefore, sulphur will increase its oxidation number (from -2 to 0) in a redox reaction with $S{O_2}$ and thus, $S{O_2}$ act as an oxidant for ${H_2}S$.
> Acidified ${C_2}{H_5}OH$- In this case also, $S{O_2}$ do not act as an oxidant forAcidified ${C_2}{H_5}OH$.
Therefore, from above discussion we can easily conclude that option C is the correct option for the given question.
Note: It should be remembered to you that Acidified ${K_2}C{r_2}{O_7}$ is one of the most commonly used moderate oxidising agents in organic chemistry.
Also, you should remember that $S{O_2}$ is a colorless gas which acts as a bleaching agent to remove excess chlorine and also act as a disinfectant.
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