Four cells containing CuSO₄(aq), AgNO₃(aq), Pb(NO₃)₂(aq), and Au(NO₃)₃(aq) are connected to power supplies that operate at 12.0 A for 1.00 h. At the end of this time, which cell will deposit the greatest mass of metal?

 

The correct option is

C) AgNO3(aq)

Step-by-step explanation

Current = 12.0 A.....(given).

Time = 1.00 hr

or = 3600 seconds.

(Because, 1 hr = 3600 seconds)

 

Step 1) :

Quantity of electricity passed = current (in amperes) × Time ( in seconds )

So,

Quantity of electricity passed = 12.0 amperes × 3600 seconds

= 43200 coulombs or

= 43200 C.

 

Let's try to solve this question one by one.

 

Step 2) :

A) CuSO4(aq) :

The dissociation of CuSO4(aq) is take place as follows :

CuSO4 (aq) ---> Cu²⁺(aq) + SO4^2-(aq)

So,

The reaction occuring at the cathode in CuSO4

Cu2+(aq) + 2e- ----> Cu(s)

 

We know,

1 F = 96500 C and

Mass of 1 mole of Cu = 63.55 g

 

Note : Number of electrons = number of Faraday.

 

Because,

2F = 2× 96500 C deposit Cu = 1 mol & = 63.55 g.

So,

43200 C will deposit Cu= (63.55 g × 43200 C) /(2×96500 C)

= 2,745,360 g /193000

= 14.22 g

 

Hence, grams of Cu are deposited during this process are 14.22 grams.

 

B) AgNO3(aq) :

The dissociation of AgNO3(aq) is take place as follows :

AgNO3(aq) ---> Ag+(aq) + NO3-(aq)

So,

The reaction occuring at the cathode in AgNO3 :

Ag+(aq) + 1e- ---> Ag(s)

 

We know,

1 F = 96500 C and

Mass of 1 mole of Ag = 108 g

 

Because,

1F = 1× 96500 C deposit Ag = 1 mol & = 108 g.

So,

43200 C will deposit Ag = (108 g × 43200 C) /(1×96500 C)

= 4,665,600 g /96500

= 48.35 g

 

Hence, grams of Ag are deposited during this process are 48.35 grams.

 

C) Pb(NO3)2(aq) :

The dissociation of Pb(NO3)2(aq) is take place as follows :

Pb(NO3)2(aq) ---> Pb2+(aq) + NO3-(aq)

So,

The reaction occuring at the cathode in Pb(NO3)2 :

Pb2+(aq) + 2e- ---> Pb(s)

 

We know,

1 F = 96500 C and

Mass of 1 mole of Pb = 207.2 g

 

Because,

2F = 2× 96500 C deposit Pb = 1 mol & = 207.2 g

So,

43200 C will deposit Ag = (207.2 g × 43200 C) /(2×96500 C)

= 8,951,040 g /193000

= 46.38 g

 

Hence, grams of Pb are deposited during this process are 46.38 grams.

 

D) Au(NO₃)₃(aq) :

The dissociation of Au(NO₃)₃(aq) is take place as follows :

Au(NO₃)₃(aq) ---> Au³⁺?(aq) + NO₃^-(aq)

So,

The reaction occuring at the cathode in Au(NO₃)₃ :

Au³⁺(aq) + 3e- ---> Au(s)

 

We know,

1 F = 96500 C and

Mass of 1 mole of Au = 197 g

 

Because,

3F = 3× 96500 C deposit Au = 1 mol & = 197 g

So,

43200 C will deposit Au = (197 g × 43200 C) /(3×96500 C)

= 8,510,400 g /289500

= 29.40 g

 

Hence, grams of Au are deposited during this process are 29.40 grams.