Q21.29 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 reduction occur? 

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

(d) At which electrode are electrons generated? 

(e) Which electrode is positively charged?

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

(g) Suggest a solution for the anode 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 cations 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. to the $\text{Ni}$  electrode from the ${\text{Fe/Fe}}^{\text{2 + }}$ electrode.
2.  ${\text{Fe/Fe}}^{\text{2 + }}$Half cell 
3.  ${\text{Fe/Fe}}^{\text{2 + }}$Half cell 
4.   $\text{Ni}$electrode
5.  $\text{Ni}$ electrode
6.  $\text{Fe}$ electrode 
7.  ${\text{1MNi}}^{\text{2 + }}$
8.  ${\text{K}}^{\text{ + }}$and  ${\text{Cl}}^{\text{ - }}$
9. The cathode can be made of inactive materials.
10. anions travel from the anode to the anode
11. oxidation:  $\text{Fe}\to {\text{Fe}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}$

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

       Overall:  ${\text{Fe + Ni}}^{\text{2 + }}\to {\text{Ni + Fe}}^{\text{2 + }}$

Step 1: voltmeter means,

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

Step 2: Determining which direction electrons 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. As a result, electrons travel from the anode to the cathode, where they are utilised to decrease the metals in the electrolyte solution, allowing it to flow from the  ${\text{Fe/Fe}}^{\text{2 + }}$to the Ni electrode.

Step 3: Determining in which half-cell does reduction occur

Oxidation happens in the  ${\text{Fe/Fe}}^{\text{2 + }}$half-cell, since electrons are liberated from the$\text{Fe}$  metal to create  ${\text{Fe}}^{\text{2 + }}$

Step 4: Determining in which half-cell do electrons leave the cell.

Electrons enter the  ${\text{Fe/Fe}}^{\text{2 + }}$half-cell as anodes.

Step 5: Determining at which electrode are electrons generated

The cathode - the   electrode - consumes electrons.

Step 6: Determining in which electrode is positively charged.

The cathode is negatively charged since it is where electrons go.  $\text{Ni}$ is the negatively charged electrode.

Step 7: Determining in which electrode increases in mass during cell operation

As the cell functions, the mass of the anode reduces as it oxidises. The bulk of the $\text{Fe}$  electrode drops.

Step 8: Determining a solution for the anode electrolyte.

A nickel cation is ${\text{Ni}}^{\text{2 + }}$ , thus it can be utilised in the electrolyte solution; however, because the concentration of  ${\text{Fe}}^{\text{2 + }}$in solution is  ${\text{1M,Ni}}^{\text{2 + }}$solutions like ${\text{1MNi}}^{\text{2 + }}$ cannot be employed.

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

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

Step 10: Determining for 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 can be utilised in a variety of situations.

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

As the metal oxidises, anions migrate to the positive electrode, the anode, to balance the rising positive charge in the solution.

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

oxidation:  $\text{Fe}\to {\text{Fe}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}$

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

Overall:  ${\text{Fe + Ni}}^{\text{2 + }}\to {\text{Ni + Fe}}^{\text{2 + }}$