$ Cr_2O_7^{2-}+14{{H}^{+}}+{{6I}^{-}}\xrightarrow{{}}2C{{r}^{3+}}+7H_2O+3I_2 $ Which element is reduced-
A) I
B) O
C) H
D) Cr
Correct Answer: D
[d] $ 2{{I}^{-}}\to I_2 $ is oxidation (loss of electrons); $ Cr( +6 ) $ changes to $ Cr( +3 ) $ by gain of electrons. Hence Cr is reduced.
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