Chapter 19.2, Problem 23PP

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using half reaction method.

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 23PP

$C{r}_2{O}_7{}^{-2}+6I^{-1}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$

Explanation of Solution

The given redox reaction is:

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

Assigning the oxidation number-

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

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 $Crperatom=-6$

$I^{-1}\to I_2+e^{-}(Oxidation)$

$C{r}_2{O}_7{}^{-2}+6e^{-}\to C{r}^{+3}(\mathrm{Re}duction)$

Balancing the atoms and charges:

$2I^{-1}\to I_2+2e^{-}(Oxidation)$

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

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

$C{r}_2{O}_7{}^{-2}+6I^{-1}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$

Conclusion

Thus the balanced redox equation is:

$C{r}_2{O}_7{}^{-2}+6I^{-1}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$