Chapter 19, Problem 88A

Interpretation Introduction

Interpretation:

The reaction of dichromate ion with iodide ions needs to be balanced in acid solution.

Concept introduction:

Half reaction method to balance ionic equation in acidic solution involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 88A

Balanced equation in ionic form

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

Explanation of Solution

Reaction in ionic form:

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

Assigning the oxidation number-

${\stackrel{+6}{Cr}}_2{O}_7{}^{2-}+\stackrel{-1}{I^{-}}\to \stackrel{0}{I_2}+\stackrel{+3}{C{r}^{+3}}$

The oxidation numbers of atoms shows that

$I$ is oxidized and $Cr$ is reduced.

Change in oxidation number of $I=+1$

Change in oxidation number of $Crperatom=-6$

$I^{-}\to I_2+e^{-}$

$C{r}_2{O}_7{}^{2-}+6e^{-}\to C{r}^{+3}$

Balancing the atoms and charges:

$2I^{-}\to I_2+2e^{-}$

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

Now adding both reactions after multiplying oxidation half cell reaction with 3

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

Conclusion

Balanced equation in ionic form

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