# The thermodynamic potential of the cell should be determined. Concept Introduction: The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs. ### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213 ### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

#### Solutions

Chapter 24, Problem 24.1QAP

(a)

Interpretation Introduction

## Interpretation: The thermodynamic potential of the cell should be determined.Concept Introduction:The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs.

Expert Solution

The thermodynamic potential of the cell is 1.409V .

### Explanation of Solution

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

The expression for electrode potential is:

Ecell=Ecell00.0592nlog[Red][Ox]

Here, standard electrode potential is Ecell0 , 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=ErightEleft ...... (I)

Here, the electrode potential of cathode is Eright andelectrode 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++4e2H2O

The expression for the electrode potential of cathode is:

Eright=Ecell00.0592nlog1[ Pb 2+]2 ...... (II)

The expression for the electrode potential of anode is:

Eleft=Ecell00.0592nlog1[pO2][ H +]4 ...... (III)

Substitute 0.126 for Ecell0 , 2 for n and 0.150 for Pb2+ in equation (II).

Eright=0.1260.05924log1 [ 0.150]2=0.1504V

Substitute 1.299 for Ecell0 , 0.850 for pO2 , 4 for n and 0.215 for H+ in equation (III).

Eleft=1.2990.05924log1[0.850] [ 0.215]4=1.2584V

Substitute 0.1504V for Eright and 1.2584V for Eleft .

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

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

(b)

Interpretation Introduction

### Interpretation:The IR drop when the current is 0.220 A should be determined.Concept Introduction:The current is directly proportional to voltage.

Expert Solution

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

### Explanation of Solution

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

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

V=(0.220A)(0.900Ω)=0.198V

Therefore, the IR drop is 0.198V .

### Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

### Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts! 