Question

In the following reaction,
$2I-+C{r}_2{O}_7^{2-}+14H^{+}\to I_2+2C{I}_{3+}+7H_{2}0$
Unbalanced parts are

• A. $H+,H_{2}O$
• B. $C{r}_2{O}_7^{2-},C{r}^{3+}$
• C. $I^{-},I_2$
• D. None of them are balanced

Answer 1

Answer: (C)

$(I)C{r}_2{O}_7^{2-}+6e^{-}\to 2C{r}^{3+}$
$(II)2I^{-}\to I_2+2e^{-}$
To balance electron (II) is to be multiplied by (3). Thus,
$6I^{-}+C{r}_2{O}_7^{2-}+14H^{+}\to 3I_2+2C{r}^{3+}+7H_2,O$
Thus, $I^{-}$ and $I_2$ are not balanced