Chapter 21, Problem 57A

(a)

Interpretation Introduction

Interpretation: The value of the standard electrode potential of the cell and overall cell reaction needs to be determined for the given cell.

  $\text{Sn}\left|{\text{Sn}}^{2+}\right|\left|{\text{Pb}}^{2+}\right|\text{Pb}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and decreases during reduction.

Explanation of Solution

The given cell is as follows:

  $\text{Sn}\left|{\text{Sn}}^{2+}\right|\left|{\text{Pb}}^{2+}\right|\text{Pb}$

The two-half reactions can be represented as follows:

  $\begin{array}{l}\text{Sn}\to {\text{Sn}}^{2+}+2e\\ {\text{Pb}}^{2+}+2e\to \text{Pb}\end{array}$

The standard cell potential for the above two reactions are as follows:

  $\begin{array}{l}\text{Sn}\to {\text{Sn}}^{2+}+2e{E}^{\text{o}}=-\text{0}\text{.136V}\\ {\text{Pb}}^{2+}+2e\to \text{Pb }{E}^{\text{o}}=\text{0}\text{.13V}\end{array}$

The standard cell potential can be calculated as follows:

  $E_{\text{cell}}^{\text{o}}=E_{\text{cathode}}-E_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=0.13\text{ V}-\left(-0.136\text{ V}\right)\\ =\left(0.13+0.136\right)\text{ V}\\ =\text{0}\text{.266 V}\end{array}$

The standard cell potential is $\text{0}\text{.266 V}$ .

(b)

Interpretation Introduction

Interpretation: The value of the standard electrode potential of the cell and overall cell reaction needs to be determined for the given cell.

  ${\text{H}}_2\left|{\text{H}}^{+}\right|\left|{\text{Br}}_2\right|{\text{Br}}^{-}$

Concept Introduction: In a voltaic cell, oxidation takes place at the anode and reduction takes place at the cathode. In oxidation, electrons are released and in reduction, electrons are added to the reactant. The oxidation state of the reactant increases during oxidation and decreases during reduction.

Explanation of Solution

The given cell is as follows:

  ${\text{H}}_2\left|{\text{H}}^{+}\right|\left|{\text{Br}}_2\right|{\text{Br}}^{-}$

The two-half reactions can be represented as follows:

  $\begin{array}{l}{\text{H}}_2\to 2{\text{H}}^{2+}+2e\\ {\text{Br}}_2+2e\to 2{\text{Br}}^{-}\end{array}$

The standard cell potential for the above two reactions are as follows:

  $\begin{array}{l}{\text{H}}_2\to 2{\text{H}}^{2+}+2e{E}^{\text{o}}=0\text{ V}\\ {\text{Br}}_2+2e\to 2{\text{Br}}^{-}{E}^{\text{o}}=\text{1}\text{.09 V}\end{array}$

The standard cell potential can be calculated as follows:

  $E_{\text{cell}}^{\text{o}}=E_{\text{cathode}}-E_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=1.09\text{ V}-\left(0\text{ V}\right)\\ =1.09\text{ V}\end{array}$

The standard cell potential is $1.09\text{ V}$ .