Chemistry Flashcards

Question 1

 

In an electrolysis experiment, the same quantity of electricity deposited 27 g of silver (A_r: 108) and 14 g of cadmium (A_r: 112).

 

What is the charge on cadmium?

 

A)  2+

 

B)  3+

 

C)  4+

 

D)  5+

 

 

Answer: A

 

Note:
- same quantity of electricity = same no. of moles of electrons being supplied in both experiments

 

- Half-equations for the reduction of each metals ion at the cathode:

 

Ag⁺ + e^– → Ag
Cd^x+ + xe^– → Cd
x is unknown.

 

No. of moles of Ag = 27/108 = 0.25 mol
Moles of e^– = 0.25 mol
since 1 mol of e^– will produce 1 mol of Ag.

 

No. of moles of Cd = 14/112 = 0.125 mol
Moles of e^– = 0.25 mol(from above)

 

Since 0.25 mol of e^– produced 0.125 mol of Cd, it takes 2 mol of e^– to produce 1 mol of Cd i.e. Cd²⁺ + 2e^– → Cd

 

x = 2.

 

Question 2

 

In an electrolysis experiment, the same quantity of electricity deposited 27 g of silver (A_r: 108) and 14 g of cadmium (A_r: 112).

 

What is the charge on cadmium?

 

A)  2+

 

B)  3+

 

C)  4+

 

D)  5+

 

 

Answer: A

 

Note:
- same quantity of electricity = same no. of moles of electrons being supplied in both experiments

 

- Half-equations for the reduction of each metals ion at the cathode:

 

Ag⁺ + e^– → Ag
Cd^x+ + xe^– → Cd
x is unknown.

 

No. of moles of Ag = 27/108 = 0.25 mol
Moles of e^– = 0.25 mol
since 1 mol of e^– will produce 1 mol of Ag.

 

No. of moles of Cd = 14/112 = 0.125 mol
Moles of e^– = 0.25 mol(from above)

 

Since 0.25 mol of e^– produced 0.125 mol of Cd, it takes 2 mol of e^– to produce 1 mol of Cd i.e. Cd²⁺ + 2e^– → Cd

 

x = 2.