# The electrode potential of the following half-cell should be determined: Ag + ( 0.0436 M ) | Ag Concept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell. For a half-reaction, the standard electrode of hydrogen is taken as a reference electrode.

### 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 22, Problem 22.1QAP

(a)

Interpretation Introduction

## Interpretation: The electrode potential of the following half-cell should be determined:Ag+(0.0436 M)|AgConcept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell. For a half-reaction, the standard electrode of hydrogen is taken as a reference electrode.

Expert Solution

The electrode potential of the half-cell is 0.7185V .

### Explanation of Solution

The half-cell is Ag+(0.0436M)|Ag , the concentration [Ag+] is 0.0436M .

In the half reaction, reduction of Ag+ to Ag takes place. The other half reaction will be for reduction of hydrogen in the hydrogen electrode with electrode potential 0 V.

The expression of electrode potential by using the nearest equation for a cell, at 25°C is:

Ecell=Ecello(0.0592V)nlog(aP)(aR) ........ (I)

Here, the standard electrode potential of a reaction is E , the number of electron involved in the reaction is n , the concentration of the product of the reaction is aP and the concentration of the reactant is aR .

Here, the standard electrode potential of a reaction is 0.799V , because the used electrode is a silver (Ag) electrode.

The half cell reaction is as follows:

Ag+(aq)+eAg(s)

The number of electrons involved in the reaction is 1 .

Since, this is the half-cell reaction thus, only reactant is present in the aqueous form.

Ehalf-cell=Eo(0.0592V)nlog1(a Ag + )

Substitute 0.799V for E , 1 for n , and 0.0436M for aAg+ in equation (I).

Ehalf-cell=(0.799V)( 0.0592V)1log(1)( 0.0436M)=(0.799V)(0.0592V)log(22.94)=(0.799V)(0.0805V)=0.7185V

Thus, the electrode potential of the half-cell is 0.7185V .

(b)

Interpretation Introduction

### Interpretation: The electrode potential of the following half-cell should be determined:Fe3+(5.34×10−4 M),Fe2+(0.090 M)|PtConcept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell.

Expert Solution

The electrode potential of the half-cell is 0.6932 V.

### Explanation of Solution

The half-cell is Fe3+(5.34×104M),Fe2+(0.090M)|Pt , the concentration of the product [Fe2+] of the reaction is 0.090M and the concentration of the reactant [Fe3+] is 5.34×104M .

The reaction takes place is as follows:

Fe3++eFe2+

Thus, the standard electrode potential of reaction is 0.771V .

The number of electrons involved in the reaction is 1 .

Substitute 0.771V for E* , 1 for n , 0.090M for aP and 5.34×104M for aR in Equation (I).

Ehalf-cell=(0.771V)( 0.0592V)1log( 0.090M)( 5.34× 10 4 M)=(0.771V)(0.0592V)log(168.54)=(0.771V)(0.1318V)=0.6932V

Thus, the electrode potential of the half-cell is 0.6932 V.

(c)

Interpretation Introduction

### Interpretation: The electrode potential of the following half-cell should be determined:AgBr(sat′d),Br−(0.037 M)|AgConcept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell.

Expert Solution

The electrode potential of the half-cell is 0.158 V.

### Explanation of Solution

The half-cell is AgBr(satd),Br(0.037M)|Ag , the concentration of [Br] is 0.037M .

The half cell reaction is as follows:

AgBr(s)+eAg(s)+Br          Eo=0.073 V

Here, the standard electrode potential of reaction is 0.073V .

The number of electrons involved in the reaction is 1 .

The electrode potential of the half-cell is calculated as follows:

Ehalf-cell=Eo0.0592nlog[Br]

Substitute 0.073V for Eo , 1 for n and 0.037M for [Br] thus,

Ehalf-cell=(0.073V)( 0.0592V)1log(0.037M)=(0.073V)+(0.085V)=0.158V

Thus, the electrode potential of the half-cell is 0.158V .

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