Chapter 25, Problem 25.36P

(a)

Interpretation Introduction

Interpretation:

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

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 25.36P

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ is $\text{1}\text{.88 V}$.

Explanation of Solution

The given reaction is shown below,

  $\text{2 Na}\left(s\right)\text{ + 2}{\text{H}}_{\text{2}}O\left(l\right)\underset{}{\rightleftharpoons }{\text{ 2 Na}}^{\text{+}}\left(aq\right){\text{ + 2 OH}}^{-}\left(aq\right){\text{ + H}}_2\left(g\right)$

Reduction half-cell is given below,

  $\text{2}{\text{H}}_{\text{2}}O\left(l\right){\text{+ 2e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ 2 OH}}^{-}\left(aq\right){\text{ + H}}_2\left(g\right)E_{red}^o=-\text{0}\text{.828 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{Na}\left(s\right)\underset{}{\to }{\text{ Na}}^{\text{+}}\left(aq\right)+{\text{ 1e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[\text{Na}|{\text{Na}}^{+}\right]\text{=}-{\text{E}}_{red}^o\left[{\text{Na}}^{+}|\text{Na}\right]\text{=}-\left(-\text{2}\text{.71 V}\right)\\ \\ =\text{2}\text{.71 V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(-\text{0}\text{.828 V}\right)+\left(\text{2}\text{.71 V}\right)\\ \\ =\text{1}\text{.88 V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to (a).

Answer to Problem 25.36P

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

Explanation of Solution

The given reaction is shown below,

  ${\text{2 H}}^{\text{+}}\left(aq\right)\text{ + Pd}\left(s\right)\underset{}{\rightleftharpoons }{\text{ Pd}}^{\text{2+}}\left(aq\right){\text{ + H}}_2\left(g\right)$

Reduction half-cell is given below,

  ${\text{2 H}}^{\text{+}}\left(aq\right){\text{+ 2e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ H}}_2\left(g\right)E_{red}^o=\text{0}\text{.00 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{Pd}\left(s\right)\underset{}{\rightleftharpoons }{\text{ Pd}}^{\text{2+}}\left(aq\right)+{\text{ 2e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[\text{Pd}|{\text{Pd}}^{\text{2+}}\right]\text{=}-{\text{E}}_{red}^o\left[{\text{Pd}}^{\text{2+}}|\text{Pd}\right]\text{=}-\text{0}\text{.951 V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(0.00\text{ V}\right)+\left(-\text{0}\text{.951 V}\right)\\ \\ =-\text{0}\text{.951 V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ has to be calculated for the given reaction $\text{4 Fe}{\left(OH\right)}_{\text{2}}\left(s\right){\text{ + O}}_2\left(g\right){\text{ + H}}_{2}O\left(l\right)\underset{}{\rightleftharpoons }\text{ 4 Fe}{\left(OH\right)}_{\text{3}}\left(s\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 25.36P

The value of ${\text{E}}_{\text{cell}}^{\text{o}}$ is $\text{0}\text{.962 V}$.

Explanation of Solution

The given reaction is shown below,

  $\text{4 Fe}{\left(OH\right)}_{\text{2}}\left(s\right){\text{ + O}}_2\left(g\right){\text{ + H}}_{2}O\left(l\right)\underset{}{\rightleftharpoons }\text{ 4 Fe}{\left(OH\right)}_{\text{3}}\left(s\right)$

Reduction half-cell is given below,

  ${\text{O}}_2\left(g\right){\text{ + H}}_{2}O\left(l\right){\text{+ 4e}}^{-}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ 4 OH}}^{-}\left(aq\right)E_{red}^o=\text{0}\text{.401 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{4 Fe}{\left(OH\right)}_{\text{2}}\left(s\right)\text{ + }O{H}^{-}\left(aq\right)\underset{}{\rightleftharpoons }\text{ 4 Fe}{\left(OH\right)}_{\text{3}}\left(s\right)+{\text{ e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[\text{Fe}{\left(OH\right)}_{\text{3}}|\text{Fe}{\left(OH\right)}_{\text{2}}\right]\text{=}-{\text{E}}_{red}^o\left[\text{Fe}{\left(OH\right)}_{\text{2}}|\text{Fe}{\left(OH\right)}_{\text{3}}\right]\\ \\ \text{=}-\left(-\text{0}\text{.56 V}\right)\text{ = 0}\text{.56 V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{0}\text{.401 V}\right)+\left(\text{0}\text{.56 V}\right)\\ \\ =\text{0}\text{.962 V}\end{array}$