Question

Balance the following equations:

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

• A. $C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$
• B. $C{r}_2{O}_7^{2-}+I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$
• C. $C{r}_2{O}_7^{2-}+6I^{-}+7H^{+}\to 2C{r}^{3+}+I_2+7H_{2}O$
• D. None of these

Answer 1

The correct option is A $C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$

The unbalanced redox equation is as follows:

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

Balance all atoms other than $H$ and $O.$

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

The oxidation number of $Cr$ changes from +6 to 3. The change in the oxidation number of 1 $Cr$ atom is 3. The total change in the oxidation number for 2 Cr atoms is 6.

The oxidation number of $I$ changes from -1 to 0. The change in the oxidation number of 1 atom of $I$ is 1. The total change in the oxidation number for 2 $I$ atoms is 2.

The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying $I^{-}$ and $I$ with 3.

$C{r}_2{O}_7^{2-}+6I^{-}+H^{+}\to 2C{r}^{3+}+3I_2+H_{2}O$

$O$ atoms are balanced by adding 6 water molecules on RHS.

$C{r}_2{O}_7^{2-}+6I^{-}+H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$

Hydrogen atoms are balanced by adding 13 $H^{+}$ atoms on the RHS.

$C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$

This is the balanced equation.