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\[C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]. \[{{\overset{+\,4}{\mathop{3SO}}\,}_{2}}+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+\,C{{r}_{2}}{{(\overset{+6}{\mathop{S}}\,{{O}_{4}})}_{3}}+{{H}_{2}}O\] Hence, oxidation state of sulphur changes from +4 to +6.
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