Question

Balance the following below equation.

$I_2+C{r}_2{O}_7^{2-}+H^{+}\to C{r}^{3+}+{IO}_3^{-}+H_{2}O$

• A. $3I_2+5C{r}_2{O}_7^{2-}+34H^{+}\to 10C{r}^{3+}+6{IO}_3^{-}+17H_{2}O$
• B. $3I_2+3C{r}_2{O}_7^{2-}+34H^{+}\to 10C{r}^{3+}+5{IO}_3^{-}+12H_{2}O$
• C. $2I_2+5C{r}_2{O}_7^{2-}+17H^{+}\to 10C{r}^{3+}+4{IO}_3^{-}+15H_{2}O$
• D. None of these

Answer 1

The correct option is A $3I_2+5C{r}_2{O}_7^{2-}+34H^{+}\to 10C{r}^{3+}+6{IO}_3^{-}+17H_{2}O$

The unbalanced redox equation is as follows:

$I_2+C{r}_2{O}_7^{2-}+H^{+}\to C{r}^{3+}+{IO}_3^{-}+H_{2}O$

Balance all atoms other than H and O.

$I_2+C{r}_2{O}_7^{2-}+H^{+}\to 2C{r}^{3+}+2{IO}_3^{-}+H_{2}O$

The oxidation number of chromium changes from +6 to +3. The change in the oxidation number of one chromium atom is 3.
For 2 Cr atoms, the change in the oxidation number is 6.
The oxidation number of iodine changes from 0 to +5. The change in the oxidation number of one I atom is 5. For 2 I atoms, the change in the oxidation number is 10.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $C{r}_2{O}_7^{2-}$ and $C{r}^{3+}$ with 5 and by multiplying $I_2$ and $I{O}_3^{-}$ with 3.

$3I_2+5C{r}_2{O}_7^{2-}+H^{+}\to 10C{r}^{3+}+6{IO}_3^{-}+H_{2}O$

O atoms are balanced by adding 16 water molecules on RHS.

$3I_2+5C{r}_2{O}_7^{2-}+H^{+}\to 10C{r}^{3+}+6{IO}_3^{-}+17H_{2}O$

Hydrogen atoms are balanced by adding 33 $H^{+}$ on LHS.

$3I_2+5C{r}_2{O}_7^{2-}+34H^{+}\to 10C{r}^{3+}+6{IO}_3^{-}+17H_{2}O$

This is the balanced chemical equation.