Lead is to be deposited at a cathode from a solution that is 0.150 M in Pb2+ and 0.215 M in HClO4. Oxygen is evolved at a pressure of 0.850 atm at a 30-cm2platinum anode. The cell has a resistance of 0.900 V.

(a) Calculate the thermodynamic potential of the cell.

(b) Calculate the IR drop if a current of 0.220 A is to be used.

Short Answer

Expert verified

(a) The thermodynamic potential of the cell is -1.409 V.

(b) The IR drop when the current is 0.220 A is 0.198 V.

Step by step solution

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01

Part (a) Step 1: Given Information

The thermodynamic potential of the cell should be determined.

02

Part (a) Step 2: Explanation

The solution is 0.150 M in Pb2+ and 0.215 M in HClO4.

The expression for electrode potential is:

Ecell=E0cell-0.0592nlog[Red][Ox]

Here, standard electrode potential is E0cell, number of electrons is n, reduction is Red and oxidation is Ox.

The electrode potential of the cell is equal to the difference between electrode potential of cathode and electrode potential of anode.

Ecell=Eright-Eleft…… (I)

Here, the electrode potential of cathode is Erightand electrode potential of anode is Eleft.

The overall reaction is:

2Pb2++2H2O2Pb(s)+O2+4H+

Write the reaction at cathode.

2Pb2++4e2Pb(s)

The reaction at anode is:

O2+4H++4e-2H2O

The expression for the electrode potential of cathode is:

Eright=E0cell-0.0592nlog1[Pb2+]2…… (II)

The expression for the electrode potential of anode is:

Eleft=E0cell-0.0592nlog1[pO2][H+]4…… (III)

Substitute -0.126 for , 2 for n and 0.150 for Pb2+ in equation (II).

Eright=-0.126-0.05924log1[0.150]2=-0.1504V

Substitute 1.299 for E0cell, 0.850 for pO2, 4 for n and 0.215 for H+ in equation (III).

Eleft=1.299-0.05924log1[0.850][0.215]4=1.2584V

Substitute -0.1504 V for Erightand 1.2584 V for Eleft.

Ecell=(-0.1504V)-(1.2584V)=-1.409V

Therefore, the thermodynamic potential of the cell is -1.409 V..

03

Part (b) Step 1: Given Information

The IR drop when the current is 0.220 A should be determined.

04

Part (b) Step 2: Explanation

The oxygen is evolved at pressure of 0.850 atm at a 30 cm2 platinum anode. And resistance of cell is 0.900 ohm.

The expression for IR drop is:

V=IR …… (IV)

Here, voltage of cell is V, the current is I and resistance of the cell is R.

Substitute 0.220 A for I and 0.900 ohm for R in Equation (IV).

V=(0.220A)(0.900Ω)=0.198V

Therefore, the IR drop is 0.198 V.

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(f) Determine the detection limit (ng) and in parts per million using a k factor of 2 (see Equation 1-12).

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