Redox Reactions And Electrochemistry Question 205
In the following reaction, $ 2{{I}^{-}}+Cr_2{{O}^{2-}}_7+14{{H}^{+}}\to I_2+2Cr^{3+}+7H_2O $ Unbalanced parts are
Options:
A) $ {{H}^{+}},,H_2O $
B) $ Cr_2{{O}^{2-}}_7, Cr^{3+}$
C) $ {{I}^{-}},I_2 $
D) None of them are balanced
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Answer:
Correct Answer: C
Solution:
[c] (I) $ Cr_2O_7^{2-}+6{{e}^{-}}\to 2Cr^{3+} $ (II) $ 2{{I}^{-}}\to I_2+2{{e}^{-}} $ To balance electron (II) is to be multiplied by (3). Thus, $ \underline{6{{I}^{-}}}+Cr_2O_7^{2-}+14{{H}^{+}}\to \underline{3I_2}+2Cr^{3+}+7H_2O $ Thus, $ {{I}^{-}} $ and $ I_2 $ are balanced.
BETA
