Redox Reactions And Electrochemistry Question 437
Question: How many moles of $ K_2Cr_2O_7 $ can be reduced by 1 mole of $ S{{n}^{2+}} $ [MP PMT 2003]
Options:
A) 1/3
B) 1/6
C) 2/3
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
$ Cr_2O_7^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7H_2O $ $ (S{{n}^{2+}}\to S{{n}^{4+}}+2{{e}^{-}})\times 3 $ $ \overline{Cr_2O_7^{2-}+14{{H}^{+}}+3S{{n}^{2+}}\to 3S{{n}^{4+}}+2C{{r}^{3+}}+7H_2O} $ It is clear from this equation that 3 moles of $ S{{n}^{2+}} $ reduce one mole of $ Cr_2O_7^{2-} $ , hence 1 mol. of $ S{{n}^{2+}} $ will reduce $ \frac{1}{3} $ moles of $ Cr_2O_7^{2-} $ .
BETA
