A) 1/3
B) 1/6
C) 2/3
D) 1
Correct Answer: A
Solution :
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\] \[(S{{n}^{2+}}\to S{{n}^{4+}}+2{{e}^{-}})\times 3\] \[\overline{C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+3S{{n}^{2+}}\to 3S{{n}^{4+}}+2C{{r}^{3+}}+7{{H}_{2}}O}\] It is clear from this equation that 3 moles of \[S{{n}^{2+}}\] reduce one mole of \[C{{r}_{2}}O_{7}^{2-}\], hence 1 mol. of \[S{{n}^{2+}}\] will reduce \[\frac{1}{3}\] moles of \[C{{r}_{2}}O_{7}^{2-}\].
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