question_answer

In the following reaction, \[2{{I}^{-}}+C{{r}_{2}}{{O}^{2-}}_{7}+14{{H}^{+}}\to {{I}_{2}}+2C{{l}^{3+}}+7{{H}_{2}}O\] Unbalanced parts are

A)  \[{{H}^{+}},\,{{H}_{2}}O\]

B)  \[C{{r}_{2}}{{O}^{2-}}_{7},C{{r}^{3+}}\]

C)  \[{{I}^{-}},{{I}_{2}}\]

D)  None of them are balanced

Correct Answer: C

Solution :

[c] (I) \[C{{r}_{2}}{{O}_{7}}^{2-}+6{{e}^{-}}\to 2C{{r}^{3+}}\] (II) \[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}\] To balance electron (II) is to be multiplied by (3). Thus, \[\underline{6{{I}^{-}}}+C{{r}_{2}}{{O}^{2-}}_{7}+14{{H}^{+}}\to \underline{{{3I}_{2}}}+2C{{r}^{3+}}+7{{H}_{2}}O\] Thus, \[{{I}^{-}}\] and \[{{I}_{2}}\] are not balanced.