Redox Reactions And Electrochemistry Question 214
Question: If 10 g of $ V_2O_5 $ , is dissolved in acid and is reduced to $ {{V}^{2+}} $ by zinc metal, how many mole of I, could be reduced by the resulting solution if it is further oxidised to $ V{{O}^{2+}} $ ions- [Assume no change in state of $ Z{{n}^{2+}} $ ions] (V=51,O=16, I=127):
Options:
A) 0.11 mole of $ I_2 $
B) 0.22 mole of $ I_2 $
C) 0.055 mole of $ I_2 $
D) 0.44 mole of $ I_2 $
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Answer:
Correct Answer: A
Solution:
[a] $ \underset{+5}{\mathop{V_2O_5}},+6{{e}^{-}}\to \underset{+2}{\mathop{2{{V}^{2+}}}}, $ Molecular mass of $ V_2O_5=51\times 2+5\times 16=182 $ 1 mol of $ V_2O_5 $ produces 2 moles of $ {{V}^{2+}} $
$ \therefore \frac{10}{182} $ moles of $ V_2O_5 $ produce $ 2\times \frac{10}{182}=0.11 $ moles of $ {{V}^{2+}} $ $ \underset{+2}{\mathop{{{V}^{2+}}}},\to V{{\underset{0}{\mathop{O}},}^{2+}}+2{{e}^{-}} $ (1)
$ \underset{0}{\mathop{I_2}},+2{{e}^{-}}\to \underset{-1}{\mathop{2{{I}^{-}}}}, $ (2)
On adding (1) and (2) $ {{V}^{2+}}+I_2\to 2{{I}^{-}}+V{{O}^{2+}} $ 1 mole of $ {{V}^{2+}} $ reduces 1 mole of $ I_2 $
$ \therefore 0.11 $ moles of $ {{V}^{2+}} $ reduce 0.11 moles of $ I_2 $
BETA
