Chapter 22, Problem 22.1QAP

Calculate the electrode potentials of the following half-cells.
(a) Ag⁺(0.0436 M)|Ag
(b) Fe³⁺ (5.34 × 10^-4M),Fe²⁺ (0.090 M)|Pt
(c) AgBr(sat’d),Br^- (0.037 M)|Ag

Interpretation Introduction

Interpretation: The electrode potential of the following half-cell should be determined:${\text{Ag}}^{+}\left(0.0436\text{M}\right)|\text{Ag}$

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.

Answer to Problem 22.1QAP

The electrode potential of the half-cell is $0.7185\text{V}$ .

Explanation of Solution

The half-cell is ${\text{Ag}}^{+}\left(0.0436\text{M}\right)|\text{Ag}$ , the concentration $\left[{\text{Ag}}^{+}\right]$ is $0.0436\text{M}$ .

In the half reaction, reduction of ${\text{Ag}}^{+}$ 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 $25°\text{C}$ is:

  $E_{\text{cell}}=E_{cell}^o-\frac{\left(0.0592\text{V}\right)}{n}\log \frac{\left(a_{\text{P}}\right)}{\left(a_{\text{R}}\right)}$ ........ (I)

Here, the standard electrode potential of a reaction is $E^{\circ }$ , the number of electron involved in the reaction is $n$ , the concentration of the product of the reaction is $a_{\text{P}}$ and the concentration of the reactant is $a_{\text{R}}$ .

Here, the standard electrode potential of a reaction is $0.799\text{V}$ , because the used electrode is a silver $\left(\text{Ag}\right)$ electrode.

The half cell reaction is as follows:

  ${\text{Ag}}^{+}\left(aq\right)+e^{-}\to \text{Ag}\left(s\right)$

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.

  $E_{\text{half-cell}}=E^{{}^o}-\frac{\left(0.0592\text{V}\right)}{n}\log \frac{1}{\left(a_{{\text{Ag}}^{+}}\right)}$

Substitute $0.799\text{V}$ for $E^{\circ }$ , $1$ for $n$ , and $0.0436\text{M}$ for $a_{{\text{Ag}}^{+}}$ in equation (I).

  $\begin{array}{c}{E}_{\text{half-cell}}=\left(0.799\text{V}\right)-\frac{\left(0.0592\text{V}\right)}{1}\log \frac{\left(1\right)}{\left(0.0436\text{M}\right)}\\ =\left(0.799\text{V}\right)-\left(0.0592\text{V}\right)\log \left(22.94\right)\\ =\left(0.799\text{V}\right)-\left(0.0805\text{V}\right)\\ =0.7185\text{V}\end{array}$

Thus, the electrode potential of the half-cell is $0.7185\text{V}$ .

Interpretation Introduction

Interpretation: The electrode potential of the following half-cell should be determined:${\text{Fe}}^{3+}\left(5.34\times {10}^{-4}\text{M}\right),{\text{Fe}}^{2+}\left(0.090\text{M}\right)|\text{Pt}$

Concept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell.

Answer to Problem 22.1QAP

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

Explanation of Solution

The half-cell is ${\text{Fe}}^{3+}\left(5.34\times {10}^{-4}\text{M}\right),{\text{Fe}}^{2+}\left(0.090\text{M}\right)|\text{Pt}$ , the concentration of the product $\left[{\text{Fe}}^{2+}\right]$ of the reaction is $0.090\text{M}$ and the concentration of the reactant $\left[{\text{Fe}}^{3+}\right]$ is $5.34\times {10}^{-4}\text{M}$ .

The reaction takes place is as follows:

  ${\text{Fe}}^{3+}+e^{-}\to {\text{Fe}}^{2+}$

Thus, the standard electrode potential of reaction is $0.771\text{V}$ .

The number of electrons involved in the reaction is $1$ .

Substitute $0.771\text{V}$ for $E*$ , $1$ for $n$ , $0.090\text{M}$ for $a_{\text{P}}$ and $5.34\times {10}^{-4}\text{M}$ for $a_{\text{R}}$ in Equation (I).

  $\begin{array}{c}{E}_{\text{half-cell}}=\left(0.771\text{V}\right)-\frac{\left(0.0592\text{V}\right)}{1}\log \frac{\left(0.090\text{M}\right)}{\left(5.34\times {10}^{-4}\text{M}\right)}\\ =\left(0.771\text{V}\right)-\left(0.0592\text{V}\right)\log \left(168.54\right)\\ =\left(0.771\text{V}\right)-\left(0.1318\text{V}\right)\\ =0.6932\text{V}\end{array}$

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

Interpretation Introduction

Interpretation: The electrode potential of the following half-cell should be determined:$\text{AgBr}\left(\text{sa<msup><mo>t</mo><mo>′</mo></msup><mo>d</mo>}\right),{\text{Br}}^{-}\left(0.037\text{M}\right)|\text{Ag}$

Concept introduction: The electrode potential is defined as the difference between the potential of two electrodes in a galvanic cell.

Answer to Problem 22.1QAP

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

Explanation of Solution

The half-cell is $\text{AgBr}\left(\text{sa<msup><mo>t</mo><mo>′</mo></msup><mo>d</mo>}\right),{\text{Br}}^{-}\left(0.037\text{M}\right)|\text{Ag}$ , the concentration of $\left[{\text{Br}}^{-}\right]$ is $0.037\text{M}$ .

The half cell reaction is as follows:

  $\text{AgBr}\left(s\right)+e^{-}\rightleftharpoons \text{Ag}\left(s\right)+{\text{Br}}^{-}{\text{ E}}^o=0.\text{073 V}$

Here, the standard electrode potential of reaction is $0.073\text{V}$ .

The number of electrons involved in the reaction is $1$ .

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

  $E_{\text{half-cell}}=E_{}^o-\frac{0.0592}{n}\log \left[{\text{Br}}^{-}\right]$

Substitute $0.073\text{V}$ for $E^o$ , $1$ for $n$ and $0.037\text{M}$ for $\left[{\text{Br}}^{-}\right]$ thus,

  $\begin{array}{c}{E}_{\text{half-cell}}=\left(0.073\text{V}\right)-\frac{\left(0.0592\text{V}\right)}{1}\log \left(0.037\text{M}\right)\\ =\left(0.073\text{V}\right)+\left(0.085\text{V}\right)\\ =0.158\text{V}\end{array}$

Thus, the electrode potential of the half-cell is $0.158\text{V}$ .