A) \[S{{n}^{4+}}\]is the oxidizing agent because it undergoes oxidation
B) \[S{{n}^{4+}}\]is the reducing agent because it undergoes oxidation
C) \[{{H}_{2}}S{{O}_{3}}\]is the reducing agent because it undergoes oxidation
D) \[{{H}_{2}}S{{O}_{3}}\]is the reducing agent because it undergoes reduction
Correct Answer: C
Solution :
\[{{H}_{2}}\overset{+4}{\mathop{S{{O}_{3}}}}\,\left( aq \right)+\overset{4+}{\mathop{S{{n}_{3}}}}\,\left( aq \right)+{{H}_{2}}O\left( l \right)\xrightarrow[{}]{{}}\] \[S{{n}^{2+}}\left( aq \right)+HSO_{4}^{-}\left( aq \right)+3{{H}^{+}}\] Hence \[{{H}_{2}}S{{O}_{3}}\]is the reducing agent because it undergoes oxidation.You need to login to perform this action.
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