A) + 4 to 0
B) + 4 to + 2
C) + 4 to + 6
D) + 6 to + 4
Correct Answer: C
Solution :
Acidified \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]solution oxidises \[S{{O}_{2}}\]into \[C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\] \[\overset{+4}{\mathop{3S{{O}_{2}}}}\,+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{K}_{2}}S{{O}_{4}}\] \[+C{{r}_{2}}{{(\overset{+6}{\mathop{S{{O}_{4}}}}\,)}_{3}}+{{H}_{2}}O\] Hence, oxidation state of sulphur changes from +4 to +6.You need to login to perform this action.
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