Chapter 11, Problem 40E

(a)

Interpretation Introduction

Interpretation: The reaction that will occur if crystals of ${\text{I}}_{\text{2}}$ are added to a solution of $\text{NaCl}$ needs to be determined and if the reaction takes place, the balanced equation, value of $E^0,\Delta G^0\text{and}K$ needs to be determined.

Concept Introduction : The standard cell potential for the reaction between iodine and NaCl should be greater than zero for the reaction to occur. Iodine ${\text{I}}_{\text{2}}$ should be reduced to ${\text{I}}^{-}$ ions and ${\text{Cl}}^{-}$ ions in the NaCl solution should be oxidized to ${\text{Cl}}_2$ gas for the reaction to occur.

Answer to Problem 40E

The reaction is non-spontaneous.

Explanation of Solution

The half reactions are as follows:

  $\begin{array}{l}{\text{I}}_{\text{2}}+2e^{-}\to 2{\text{I}}^{-}{E}^0(\text{cathode})=0.54\text{V}\\ {\text{2Cl}}^{-}\to {\text{Cl}}_{\text{2}}+2e^{-}-E^0(\text{anode})=-1.36\text{V}\end{array}$

  ${\text{I}}_{\text{2}}+2{\text{Cl}}^{-}\to 2{\text{I}}^{-}+{\text{Cl}}_{\text{2}}{E}^0=-0.82\text{V}$

So, no reaction occurs when solid iodine is added to $\text{NaCl}$ solution as the standard cell potential is less than zero. The reaction is non-spontaneous.

(b)

Interpretation Introduction

Interpretation: The reaction that will occur when ${\text{Cl}}_{\text{2}}$ gas is bubbled into a solution of $\text{NaI}$ needs to be explained and if the reaction takes place, balanced equation, value of $E^0,\Delta G^0\text{and}K$ needs to be determined.

Concept Introduction : The standard cell potential for the reaction between ${\text{Cl}}_{\text{2}}$ gas and $\text{NaI}$ solution should be greater than zero for the reaction to occur. Chlorine gas ${\text{Cl}}_{\text{2}}$ should be reduced to ${\text{Cl}}^{-}$ ions and ${\text{I}}^{-}$ ions in the NaI solution should be oxidized to ${\text{I}}_{\text{2}}$ gas for the reaction to occur.

Answer to Problem 40E

The reaction is spontaneous and balanced equation for the reaction is,

  ${\text{Cl}}_{\text{2}}+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_{\text{2}}$

Value of $E^0\text{=0}\text{.82}\text{V}$ , $\Delta G^0$ is $-1.58\times {10}^2\text{kJ}{\text{mol}}^{-1}$ and K is $5.46\times {10}^{27}$ .

Explanation of Solution

The half cell reactions are as follows:

  $\begin{array}{l}{\text{Cl}}_{\text{2}}+2e^{-}\to 2{\text{Cl}}^{-}{E}^0(\text{cathode})=1.36\text{V}\\ {\text{2I}}^{-}\to {\text{I}}_{\text{2}}+2e^{-}{E}^0(\text{anode})=-0.54\text{V}\end{array}$

  ${\text{Cl}}_{\text{2}}+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_{\text{2}}{E}^0=1.36\text{V+(}-\text{0}\text{.54}\text{V)=0}\text{.82}\text{V}$

Chlorine gas will dissolve in NaI solution as the standard cell potential is greater than zero. Hence, standard cell potential of the reaction is $\text{0}\text{.82}\text{V}$ .

Now calculate $\Delta G^0$ .

  $\Delta G^0=-nF{E}^0{}_{\text{cell}}$........ (1)

For the reaction, n =2 and $E^0\text{=0}\text{.82}\text{V}$

Put the above values in equation (1).

  $\begin{array}{c}\Delta G^0=-2\times 96485\text{C}{\text{mol}}^{-1}\times 0.82\text{V}\\ \text{=}-\text{158235}\text{J}{\text{mol}}^{-1}\\ =-1.58\times {10}^2\text{kJ}{\text{mol}}^{-1}\end{array}$

Therefore, value of $\Delta G^0$ is $-1.58\times {10}^2\text{kJ}{\text{mol}}^{-1}$ .

Now calculate K.

  $\begin{array}{c}\Delta G^0=-RT\ln K\\ -158235\text{J}{\text{mol}}^{-1}=8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\ln K\\ \ln K=\frac{-158235\text{J}{\text{mol}}^{-1}}{-(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})(298\text{K})}\\ =63.867\\ =5.46\times {10}^{27}\end{array}$

Hence, value of K is $5.46\times {10}^{27}$ .

(c)

Interpretation Introduction

Interpretation: The reaction that will occur if a silver wire is placed in a solution of ${\text{CuCl}}_{\text{2}}$ needs to be determined and if the reaction takes place find the balanced equation, $E^0,\Delta G^0\text{and}K$ .

Concept Introduction : The standard cell potential for the reaction between silver and ${\text{CuCl}}_{\text{2}}$ solution should be greater than zero for the reaction to occur.

Answer to Problem 40E

The reaction is non-spontaneous.

Explanation of Solution

The half reactions are as follows:

  $\begin{array}{l}{\text{Cu}}^{2+}+2e^{-}\to \text{Cu}{E}^0(\text{cathode})=0.34\text{V}\\ \text{2Ag}\to 2{\text{Ag}}^{+}+2e^{-}-E^0(\text{anode})=-0.80\text{V}\end{array}$

  ${\text{Cu}}^{2+}+2\text{Ag}\to \text{Cu}+2{\text{Ag}}^{+}{E}^0=-0.46\text{V}$

So, no reaction occurs when silver wire is placed in a solution of ${\text{CuCl}}_2$ solution as the standard cell potential is less than zero. The reaction is non-spontaneous.

(d)

Interpretation Introduction

Interpretation: The reaction that will occur if an acidic solution of ${\text{FeSO}}_{\text{4}}$ is exposed to air needs to be determined and if the reaction takes place, the balanced equation, value of $E^0,\Delta G^0\text{and}K$ needs to be determined.

Concept Introduction: The standard cell potential for the reaction between acidic solution of ${\text{FeSO}}_{\text{4}}$ and air should be greater than zero for the reaction to occur. ${\text{Fe}}^{2+}$ ion is oxidized to ${\text{Fe}}^{3+}$ ions and oxygen in the air should be reduced to water for the reaction to occur.

Answer to Problem 40E

The reaction is spontaneous. The balanced reaction is,

  ${\text{O}}_2+4{\text{H}}^{+}+4{\text{Fe}}^{\text{2+}}\to 2{\text{H}}_{\text{2}}\text{O}+4{\text{Fe}}^{3+}$

Value of $E^0$ is $\text{0}\text{.46}\text{V}$ , $\Delta G^0$ is $-1.78\times {10}^2\text{kJ}{\text{mol}}^{-1}$ and K is $1.32\times {10}^{31}$ .

Explanation of Solution

The half cell reactions are as follows:

  $\begin{array}{l}{\text{O}}_2+4{\text{H}}^{+}+4e^{-}\to 2{\text{H}}_{\text{2}}\text{O}{E}^0(\text{cathode})=1.23\text{V}\\ {\text{4Fe}}^{2+}\to 4{\text{Fe}}^{\text{3+}}+4e^{-}{E}^0(\text{anode})=-0.77\text{V}\end{array}$

  ${\text{O}}_2+4{\text{H}}^{+}+4{\text{Fe}}^{\text{2+}}\to 2{\text{H}}_{\text{2}}\text{O}+4{\text{Fe}}^{3+}{E}^0=1.23\text{V+(}-\text{0}\text{.77}\text{V)=0}\text{.46}\text{V}$

Reaction occur between acidic ${\text{FeSO}}_{\text{4}}$ solution and air as the standard cell potential is greater than zero.

Now calculate $\Delta G^0$ .

  $\Delta G^0=-nF{E}^0{}_{\text{cell}}$........ (1)

For the reaction, n =4 and $E^0\text{=0}\text{.46}\text{V}$

Put the above values in equation (1).

  $\begin{array}{c}\Delta G^0=-4\times 96485\text{C}{\text{mol}}^{-1}\times 0.46\text{V}\\ \text{=}-\text{177532}\text{J}{\text{mol}}^{-1}\\ =-1.78\times {10}^2\text{kJ}{\text{mol}}^{-1}\end{array}$

Therefore, value of $\Delta G^0$ is $-1.78\times {10}^2\text{kJ}{\text{mol}}^{-1}$ .

Now calculate K.

  $\begin{array}{c}\Delta G^0=-RT\ln K\\ -177532\text{J}{\text{mol}}^{-1}=8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\ln K\\ \ln K=\frac{-177532\text{J}{\text{mol}}^{-1}}{-(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})(298\text{K})}\\ =71.656\\ =1.32\times {10}^{31}\end{array}$

Hence, value of K is $1.32\times {10}^{31}$ .