Question

In the following reaction,
$2I - + Cr_{2}O^{2-}_{7} + 14H^{+} \to I_{2} + 2CI_{3+} + 7H_{2}0$
Unbalanced parts are

• A. $H+, H_{2}O$
• B. $Cr_{2}O^{2-}_{7}, Cr^{3+}$
• C. $I^{-},I_{2}$
• D. None of them are balanced

Answer 1

Answer: (C)

$(I) Cr_{2}O^{2-}_{7} + 6e^{-} \to 2Cr^{3+}$
$(II) 2I^{-} \to I_{2} + 2e^{-}$
To balance electron (II) is to be multiplied by (3). Thus,
$6I^{-} + Cr_{2}O^{2-}_{7} + 14H^{+} \to 3I_{2} + 2Cr^{3+} + 7H_{2},O$
Thus, $I^{-}$ and $I_{2}$ are not balanced