Question

A rechargeable battery is constructed based on a concentration cell constructed of two Ag/Ag+half-cells. The volume of each half-cell is 2.2 Land the concentrations of Ag+ in the half-cells are 1.35 M and 1.0×10⁻³ M .

Part A: What is the concentration of Ag⁺ at the cathode after the cell is run at 3.8 A for 5.6 h ?   

Part B: What is the concentration of Ag⁺ at the cathode when the cell is "dead"?

Part C: For how long can this battery deliver 2.4 A of current before it goes dead?

Answer 1

n(Ag⁺ at cathode) = 1.35M * 2 L = 2.7 mol (reduction)

   n(Ag⁺ at anode) = 1.1 x10^-3 * 2 L = 2.2 x 10^-3 mol (oxidation)

In order for the cell goes dead, the concentration of bothmust be the same; therefore

x = 1.3489 mol, mol of electron passed through the circuitwill be 1.3489 mol.

the amount of charge needed to cause 1.3489 mol of electronspassed through the circuit will be 130381.15 C due to every mole ofelectron need 96500 C.

Q = It = 130381.15 C = 2.8 * t; t = 46564.70 seconds.

2.   Q = 3.5 * (5.5 * 60 * 60) = 69300 C

mol of electrons produced = 69300/96500 = 0.7181 mol ofelectrons = n(Ag)

m(Ag) = 0.7181 * 107.9 = 77.49 g

3.   n(Ag) = 120/107.9 = 1.112 mol = mol ofelectrons needed

The amount of charge needed to produce this amount ofelcetrons is 107321.59 C

t = Q/I = 107321.59 / 10 = 10732.159 seconds