Chapter 18, Problem 18.95QE

(a)

Interpretation Introduction

Interpretation:

The overall chemical equation and the standard potential has to be given when ${\text{Fe}}^{\text{3+}}$ is added to ${\text{H}}_2{\text{SO}}_3\text{(aq)}$.

Concept Introduction:

Voltaic cell or Galvanic cell:

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

Cell diagram of voltaic or galvanic cell is as follows.

  $\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 18.95QE

The chemical equation and standard potential is given below:

  $\begin{array}{l}{\text{H}}_2{\text{SO}}_3{\text{(aq) + H}}_2{\text{O(l)+2Fe}}^{+3}\left(\text{aq}\right)\to {\text{HSO}}_4{}^{\text{-}}\text{(aq)}{\text{+3H}}^{+}\left(\text{aq}\right)+2{\text{Fe}}^{+2}\left(\text{aq}\right)\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.60\text{V}\end{array}$

Explanation of Solution

Half- cell reactions are given below:

  $\begin{array}{l}{\text{H}}_2{\text{SO}}_3{\text{(s) + H}}_2\text{O(l)}\to {\text{HSO}}_4{}^{\text{-}}\text{(aq)}{\text{+4H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.17\text{ V}\\ \\ {\text{2Fe}}^{+3}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\to 2{\text{Fe}}^{+2}\left(\text{aq}\right){\text{ E}}^{\text{0}}=0.771\text{ V}\end{array}$

Overall reaction is considered as a total reaction occurs at anode and cathode.

Half- cell reaction and overall reaction can be written as below:

  $\begin{array}{l}{\text{H}}_2{\text{SO}}_3{\text{(aq) + H}}_2\text{O(l)}\to {\text{HSO}}_4{}^{\text{-}}\text{(aq)}{\text{+3H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.17\text{ V}\\ \\ {\text{2Fe}}^{+3}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\to 2{\text{Fe}}^{+2}\left(\text{aq}\right){\text{ E}}^{\text{0}}=0.771\text{ V}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ }\\ {\text{Overall:H}}_2{\text{SO}}_3{\text{(aq) + H}}_2{\text{O(l)+2Fe}}^{+3}\left(\text{aq}\right)\to {\text{HSO}}_4{}^{\text{-}}\text{(aq)}{\text{+3H}}^{+}\left(\text{aq}\right)+2{\text{Fe}}^{+2}\left(\text{aq}\right)\end{array}$

The calculation of the ${\text{E}}_{\text{cell}}^{\text{o}}$ value is given below:

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}+{\text{E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= 0}\text{.771V}+\left(-0.17\text{V}\right)\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.60\text{V}\end{array}$

b)

Interpretation Introduction

Interpretation:

The overall chemical equation and the standard potential has to be given when ${\text{Fe}}^{+2}$ is titrated with ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$.

Concept Introduction:

Refer to (a)

Answer to Problem 18.95QE

The chemical equation and standard potential is given below:

  $\begin{array}{l}6{\text{Fe}}^{+2}\left(\text{aq}\right){\text{ + Cr}}_2{\text{O}}_7^{2-}{\text{(aq) +14H}}^{+}\left(\text{aq}\right)\to {\text{6Fe}}^{+3}\left(\text{aq}\right)\text{+}2{\text{Cr}}^{+3}\text{(aq)}{\text{+ 7H}}_2\text{O(l) }\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.56\text{V}\end{array}$

Explanation of Solution

Half- cell reactions are given below:

  $\begin{array}{l}6{\text{Fe}}^{+2}\left(\text{aq}\right)\to {\text{6Fe}}^{+3}\left(\text{aq}\right)\text{+}6{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.771\text{ V}\\ \\ {\text{Cr}}_2{\text{O}}_7^{2-}{\text{(aq) +14H}}^{+}\left(\text{aq}\right)+6{\text{e}}^{\text{-}}\to 2{\text{Cr}}^{+3}\text{(aq)}{\text{+ 7H}}_2{\text{O(l) E}}^{\text{0}}=+1.33\text{ V}\end{array}$

Overall reaction is considered as a total reaction occurs at anode and cathode.

Half- cell reaction and overall reaction can be written as below:

  $\begin{array}{l}6{\text{Fe}}^{+2}\left(\text{aq}\right)\to {\text{6Fe}}^{+3}\left(\text{aq}\right)\text{+}6{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.771\text{ V}\\ \\ {\text{Cr}}_2{\text{O}}_7^{2-}{\text{(aq) +14H}}^{+}\left(\text{aq}\right)+6{\text{e}}^{\text{-}}\to 2{\text{Cr}}^{+3}\text{(aq)}{\text{+ 7H}}_2{\text{O(l) E}}^{\text{0}}=+1.33\text{ V}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ }\\ \text{Overall:}6{\text{Fe}}^{+2}\left(\text{aq}\right){\text{ + Cr}}_2{\text{O}}_7^{2-}{\text{(aq) +14H}}^{+}\left(\text{aq}\right)\to {\text{6Fe}}^{+3}\left(\text{aq}\right)\text{+}2{\text{Cr}}^{+3}\text{(aq)}{\text{+ 7H}}_2\text{O(l) }\end{array}$

The calculation of the ${\text{E}}_{\text{cell}}^{\text{o}}$ value is given below:

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}+{\text{E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= 1}\text{.33V}+\left(-0.771\text{V}\right)\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.56\text{V}\end{array}$

b)

Interpretation Introduction

Interpretation:

The overall chemical equation and the standard potential has to be given when ${\text{HNO}}_2$ is added to ${\text{Fe}}^{\text{+2}}\left(\text{aq}\right)$.

Concept Introduction:

Refer to (a)

Answer to Problem 18.95QE

The chemical equation and standard potential is given below:

  $\begin{array}{l}{\text{HNO}}_2{\text{(aq) +H}}^{+}\left(\text{aq}\right){\text{ +Fe}}^{+2}\left(\text{aq}\right)\to \text{NO(g)}{\text{+ H}}_2\text{O(l) }{\text{+ Fe}}^{+3}\left(\text{aq}\right)\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.212\text{V}\end{array}$

Explanation of Solution

Half- cell reactions are given below:

  $\begin{array}{l}{\text{Fe}}^{+2}\left(\text{aq}\right)\to {\text{Fe}}^{+3}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.771\text{ V}\\ \\ {\text{HNO}}_2{\text{(aq) +H}}^{+}\left(\text{aq}\right)+{\text{e}}^{\text{-}}\to \text{NO(g)}{\text{+ H}}_2{\text{O(l) E}}^{\text{0}}=+0.983\text{ V}\end{array}$

Overall reaction is considered as a total reaction occurs at anode and cathode.

Half- cell reaction and overall reaction can be written as below:

  $\begin{array}{l}{\text{Fe}}^{+2}\left(\text{aq}\right)\to {\text{Fe}}^{+3}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{-}}{\text{ E}}^{\text{0}}=-0.771\text{ V}\\ \\ {\text{HNO}}_2{\text{(aq) +H}}^{+}\left(\text{aq}\right)+{\text{e}}^{\text{-}}\to \text{NO(g)}{\text{+ H}}_2{\text{O(l) E}}^{\text{0}}=+0.983\text{ V}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ }\\ {\text{Overall:HNO}}_2{\text{(aq) +H}}^{+}\left(\text{aq}\right){\text{ +Fe}}^{+2}\left(\text{aq}\right)\to \text{NO(g)}{\text{+ H}}_2\text{O(l) }{\text{+ Fe}}^{+3}\left(\text{aq}\right)\end{array}$

The calculation of the ${\text{E}}_{\text{cell}}^{\text{o}}$ value is given below:

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}+{\text{E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= 0}\text{.983V}+\left(-0.771\text{V}\right)\\ \\ {\text{E}}_{\text{cell}}^{\text{o}}\text{= }0.212\text{V}\end{array}$