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 1.9 L and the concentrations of Ag+ in the half-cells are 1.15 M and 1.3×10⁻³ M .

1)What is the concentration of Ag⁺ at the cathode after the cell is run at 3.6 A for 5.7 h ? Express your answer using two significant figures.

2)What is the concentration of Ag⁺ at the cathode when the cell is "dead"?

3)For how long can this battery deliver 2.3 A of current before it goes dead? Express your answer using two significant figures.

Answer 1

1) the current passed = 3.6 A

time = 5.7 hours

So charge passed = Q = IXt = 3.6 X 5.7 X 3600 seconds = 73872 Coulumbs

According to Faraday's law :

The amount of Ag deposited at cathode will be : 1 gram equivalent will be deposited if 1Faraday of charge is passed

Or 108 grams of Ag will be reduced if 96485 C of charge is passed

so amount of Ag reduced by passing 73872 = 108 /x 73872 / 96485 = 82.68 grams

Or 0.765 gram equivalents

Initially equivalent of Ag+ present = moles of Ag+ = Molarity X volume = 1.15 X 1.9 = 2.185

After passing current Ag+ equivalents left = 2.185 - 0.765 = 1.42 equivalents

concentration = 1.42 / 1.9 = 0.747 molar

2) The cell will be dead when the E = 0

Ecell = -0.0592 log [Q] = 0

concentration of two cells become = equal

1.15 + 0.0013 /2 = 0.575

The concentration of two cell will become = 0.576

3) To reach this concentration means we need to attain the equivalents = 0.576 X 1.9 = 1.094 equivalents

Initial equivalents = 2.185

so the equivalents deposited = 2.185 - 1.094 = 1.091 equivalents

For 1 equivalent the charge required = 1 Faraday = 96485 C

So for 1.091 = 96485 X 1.091 C = 105265.14 C

Charge = Current x time in seconds

So time = Charge / current = 105265.14 C / 2.3 = 45767.45 seconds

or 12.71 hours