Chapter 20, Problem 34A

Interpretation Introduction

Interpretation:

With the help of cell representation, which one is oxidized and which one is reduced that will be explained.

Concept introduction:

A electrochemical cell is represented by writing the anode where oxidation occurs on the left hand side and cathode where reduction occurs on the right hand side.

Answer to Problem 34A

The given cell representation is

$\text{Al}|{\text{Al}}^{\text{+3}}\parallel {\text{Cu}}^{\text{+2}}|\text{Cu}$

The oxidation half reaction

$\text{Al}\left(\text{s}\right)\to {\text{Al}}^{\text{+3}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}$

The reduction half reaction

${\text{Cu}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Cu}$

Explanation of Solution

A electrochemical cell is represented by writing the anode where oxidation occurs on the left hand side and cathode where reduction occurs on the right hand side.

The given cell representation is

$\text{Al}|{\text{Al}}^{\text{+3}}\parallel {\text{Cu}}^{\text{+2}}|\text{Cu}$

The oxidation half reaction

$\text{Al}\left(\text{s}\right)\to {\text{Al}}^{\text{+3}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}$

The reduction half reaction

${\text{Cu}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Cu}$

Conclusion

The given cell representation is

$\text{Al}|{\text{Al}}^{\text{+3}}\parallel {\text{Cu}}^{\text{+2}}|\text{Cu}$

The oxidation half reaction

$\text{Al}\left(\text{s}\right)\to {\text{Al}}^{\text{+3}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}$

The reduction half reaction

${\text{Cu}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Cu}$