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