Q21.30 P - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the following voltaic cell

(a) In which direction do electrons flow in the external circuit? 

(b) In which half-cell does oxidation occur?

(c) In which half-cell do electrons enter the cell?

(d) At which electrode are electrons consumed?

(e) Which electrode is negatively charged? 

(f) Which electrode decreases in mass during cell operation?

(g) Suggest a solution for the cathode electrolyte.

(h) Suggest a pair of ions for the salt bridge.

(i) For which electrode could you use an inactive material?

(j) In which direction do anions within the salt bridge move to maintain charge neutrality? 

(k) Write balanced half-reactions and an overall cell reaction.

Step-by-Step Solution

Answer

1. From  $\text{Zn}$ electrode to ${\text{Co/Co}}^{\text{2 + }}$ electrode
2.  ${\text{Co/Co}}^{\text{2 + }}$half cell
3.  $\text{Co}$ electrode
4.  $\text{Zn}$ electrode
5.  $\text{Zn}$ electrode
6. $\text{Co}$  electrode
7.  ${\text{1MZnn}}^{\text{2 + }}$
8. ${\text{K}}^{\text{ + }}$ and  ${\text{Cl}}^{\text{ - }}$
9. The cathode can be made of inactive materials.
10. Cations travel from the anode to the cathode.
11. Oxidation:  $\text{Zn}\to {\text{Zn}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}$

      Reduction: ${\text{Co}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to \text{Co}$ 

      Overall:  ${\text{Zn + Co}}^{\text{2 + }}\to {\text{Co + Zn}}^{\text{2 + }}$

Step 1: voltmeter means,

A voltmeter is a device that measures electric potential in volts.

Step 2: Determining the direction in which electrons do flow in the external circuit

Electrons flow from a high-concentration region to a low-concentration zone. Because the anode is the electrode that releases electrons, it is the location with the highest concentration. The electrons flow from the  $\text{Zn}$electrode to the  ${\text{Co/Co}}^{\text{2 + }}$ electrode as they migrate from the anode to the cathode, where they are used to charge the metals in the electrolyte solution.

Step 3: determining in which half-cell does oxidation occur

Half-cell  ${\text{Co/Co}}^{\text{2 + }}$

Step 4: determining in which half-cell do electrons enter the cell

 Electrons are ejected from the cathode, which is the  $\text{Co}$ electrode.

Step 5: determining at which electrode are electrons consumed

Electrons are produced in the  $\text{Zn}$ electrode, which is the anode.

Step 6: determining in which electrode is negatively charged

Because electrons travel away from the cathode, it is positively charged. The positively charged electrode is the $\text{Zn}$  electrode

Step 7: determining at which electrode decreases in mass during cell operation

As the cell functions, the mass of the cathode increases as the ions are reduced. The bulk of the  $\text{Co}$ electrode grows.

Step 8: determining a solution for the cathode electrolyte

Because a zinc cation is ${\text{Zn}}^{\text{2 + }}$ , it can be employed in electrolyte solutions like  ${\text{1MZn}}^{\text{2 + }}$

Step 9: determining a pair of ions for the salt bridge.

Ions in the salt bridge must be able to migrate at the same pace to each electrode. Salt bridges commonly use alkali cations and halide anions, which have virtually identical ionic sizes. In the solution, ${\text{K}}^{\text{ + }}{\text{K}}^{\text{ + }}$ and  ${\text{Cl}}^{\text{ - }}$can be employed

Step 10: determining at which electrode could you use an inactive material

Instead of interacting with ions in the solution, inactive materials can be employed to transfer electrons. These are suitable for usage in the cathode.

Step 11: Determining at which direction do anions within the salt bridge move to maintain charge neutrality

 As the number of ions in the solution decreases, cations migrate to the negative electrode, the cathode, to balance the decreasing positive charge in the solution.

Step 12: Determining balanced half-reactions and an overall cell reaction

Zinc oxidation is depicted as a half reaction:

 $\text{Zn}\to {\text{Zn}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}$

Cobalt reduction is represented by a half reaction:

 ${\text{Co}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to \text{Co}$

As a result, the net reaction is:

 ${\text{Zn + Co}}^{\text{2 + }}\to {\text{Co + Zn}}^{\text{2 + }}$