Chapter 18, Problem 51A

Interpretation Introduction

Interpretation: The electrode in the cathode portion of the cell for the given oxidation-reduction reaction needs to be identified from the given options.

  ${\text{2 Al}}_{\text{(aq)}}^{\text{+3}}{\text{ + 3 Mg}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Al(s) + 3 Mg}}_{\text{(aq)}}^{\text{+2}}$

1. Air
2. HCl
3. Mg
4. Al
5. H₂SO₄

Concept Introduction: In the electrochemical cell, the reactions at cathode and anode occur due to the difference in their reduction electrode potential value. The EMF of the cell can be calculated with help of electrode reduction potential values.

The reaction at each electrode is called as half-reaction and combination of both half-reaction gives the cell reaction of given electrochemical cell.

Answer to Problem 51A

Correct answer: The solution of HCl must act as cathode portion because ${\text{Al}}_{\text{(aq)}}^{\text{+3}}$ ions are present in it.

Explanation of Solution

Reason for correct option:

Given:

  ${\text{2 Al}}_{\text{(aq)}}^{\text{+3}}{\text{ + 3 Mg}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Al(s) + 3 Mg}}_{\text{(aq)}}^{\text{+2}}$

Half-reaction of the given galvanic cell can be written as:

  $\begin{array}{l}{\text{2 Al}}_{\text{(aq)}}^{\text{+3}}{\text{ + 3 Mg}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2Al(s) + 3 Mg}}_{\text{(aq)}}^{\text{+2}}\\ {\text{2 Al}}_{\text{(aq)}}^{\text{+3}}\stackrel{}{\to }\text{ 2Al(s)(reduction) at cathode}\\ {\text{3 Mg}}_{\text{(s)}}\stackrel{}{\to }{\text{ 3 Mg}}_{\text{(aq)}}^{\text{+2}}\text{ (Oxidation) at anode }\end{array}$

Thus, solution of HCl must act as cathode portion because ${\text{Al}}_{\text{(aq)}}^{\text{+3}}$ ions are present in it.

Conclusion

Reasons for incorrect options: Solid Mg acts as anode and air is non-electrolyte. H₂SO₄ cannot use as it will react with metals.