Chapter 25, Problem 25.35P

(a)

Interpretation Introduction

Interpretation:

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ has to be calculated for the given reaction $\text{Cu}\left(s\right)\text{ + }{\left[Fe{\left(CN\right)}_6\right]}^{3-}\left(aq\right)\text{}\underset{}{\rightleftharpoons }{\text{ 2 Cu}}^{\text{+}}\left(aq\right)\text{ + }{\left[Fe{\left(CN\right)}_6\right]}^{4-}\left(aq\right)$.

Concept Introduction:

Standard potential $\left({\text{E}}_{\text{cell}}^{\text{o}}\right)$ can be calculated by the following formula,

  ${\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}$

Answer to Problem 25.35P

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ is $-0.163\text{ V}$.

Explanation of Solution

The given reaction is shown below,

  $\text{Cu}\left(s\right)\text{ + }{\left[Fe{\left(CN\right)}_6\right]}^{3-}\left(aq\right)\text{}\underset{}{\rightleftharpoons }{\text{ 2 Cu}}^{\text{+}}\left(aq\right)\text{ + }{\left[Fe{\left(CN\right)}_6\right]}^{4-}\left(aq\right)$

Reduction half-cell is given below,

  ${\left[Fe{\left(CN\right)}_6\right]}^{3-}\left(aq\right)\text{}{\text{ + e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\left[Fe{\left(CN\right)}_6\right]}^{4-}\left(aq\right)E_{red}^o=\text{0}\text{.358 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{Cu}\left(s\right)\underset{}{\to }{\text{ Cu}}^{\text{+}}\left(aq\right)+{\text{ 1e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[\text{Cu}|{\text{Cu}}^{+}\right]\text{=}-{\text{E}}_{red}^o\left[{\text{Cu}}^{+}|\text{Cu}\right]\text{=}-\text{0}\text{.521 V}\end{array}$

Calculate $E_{cell}^o$ for the following reaction as follows,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{0}\text{.358 V}\right)+\left(-\text{0}\text{.521 V}\right)\\ \\ =-0.163\text{ V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ has to be calculated for the given reaction ${\text{Fe}}^{\text{3+}}\left(aq\right)\text{ + Ag}\left(s\right)\underset{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}\left(aq\right){\text{ + Ag}}^{+}\left(aq\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 25.35P

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ is $-\text{0}\text{.951 V}$.

Explanation of Solution

The given reaction is shown below,

  ${\text{Fe}}^{\text{3+}}\left(aq\right)\text{ + Ag}\left(s\right)\underset{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}\left(aq\right){\text{ + Ag}}^{+}\left(aq\right)$

Reduction half-cell is given below,

  ${\text{Fe}}^{\text{3+}}\left(aq\right){\text{ + e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}\left(aq\right)E_{red}^o=\text{0}\text{.771 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{Ag}\left(s\right)\text{}\underset{}{\rightleftharpoons }{\text{ Ag}}^{+}\left(aq\right)+{\text{ e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[\text{Ag}|{\text{Ag}}^{+}\right]\text{=}-{\text{E}}_{red}^o\left[{\text{Ag}}^{+}|\text{Ag}\right]\text{=}-\text{0}\text{.7996 V}\end{array}$

Calculate $E_{cell}^o$ for the following reaction as follows,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{0}\text{.771 V}\right)+\left(-\text{0}\text{.7996 V}\right)\\ \\ =-\text{0}\text{.951 V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ has to be calculated for the given reaction $Zn\left(s\right){\text{ + F}}_2\left(g\right)\underset{}{\rightleftharpoons }{\text{ Zn}}^{\text{2+}}\left(aq\right){\text{ + 2 F}}^{-}\left(aq\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 25.35P

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ is $\text{3}\text{.63 V}$.

Explanation of Solution

The given reaction is shown below,

  $Zn\left(s\right){\text{ + F}}_2\left(g\right)\underset{}{\rightleftharpoons }{\text{ Zn}}^{\text{2+}}\left(aq\right){\text{ + 2 F}}^{-}\left(aq\right)$

Reduction half-cell is given below,

  ${\text{F}}_2\left(g\right){\text{+ 2e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ 2 F}}^{-}\left(aq\right)E_{red}^o=\text{2}\text{.866 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}Zn\left(s\right)\underset{}{\rightleftharpoons }{\text{ Zn}}^{\text{2+}}\left(aq\right)+{\text{ 2e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[Zn|{\text{Zn}}^{\text{2+}}\right]\text{=}-{\text{E}}_{red}^o\left[{\text{Zn}}^{\text{2+}}|Zn\right]\\ \\ \text{=}-\left(-\text{0}\text{.762 V}\right)\text{ = 0}\text{.762 V}\end{array}$

Calculate $E_{cell}^o$ for the following reaction as follows,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{2}\text{.866 V}\right)+\left(\text{0}\text{.762 V}\right)\\ \\ =\text{3}\text{.63 V}\end{array}$