Chapter 18, Problem 18.62QE

a)

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined with the given concentrations.

  $\begin{array}{l}\text{a) Zn}{\text{(s) +Fe}}^{+2}\text{(aq) }\to {\text{ Zn}}^{+2}\text{(aq)}+\text{Fe(s)}\\ \\ \left[{\text{Fe}}^{2+}\right]=\text{0}\text{.050M},\left[{\text{Zn}}^{2+}\right]=1.{\text{0×10}}^{\text{-3}}\text{M}\end{array}$

Concept Introduction:

Electrochemical cells:

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge . It allows the cations and anions to move between the two half cells.

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

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

If ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, then the reaction is predicted to be spontaneous at that direction.

If ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, then the reaction is predicted to be spontaneous at reverse direction.

Explanation of Solution

The given reaction is as follows:

  $\text{Zn}{\text{(s) +Fe}}^{+2}\text{(aq) }\to {\text{ Zn}}^{+2}\text{(aq)}+\text{Fe(s)}$

The half-cell reaction can be written as below:

  $\begin{array}{l}\text{At anode:}\\ \text{Oxidation: Zn (s)}\to {\text{ Zn}}^{2+}{\text{(aq) +2e}}^{\text{-}},{\text{E}}_{\text{cell}}^{\text{o}}=+0.76\text{V}\\ \\ \text{At cathode:}\\ {\text{Reduction: Fe}}^{\text{+2}}{\text{(aq)+2e}}^{\text{-}}\to \text{ Fe}\text{(s)},{\text{E}}_{\text{cell}}^{\text{o}}=-0.44\text{V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{+E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= }-\text{0}\text{.44 V}+0.76\text{V}\\ \\ \text{= }+0.32\text{ V}\end{array}$

Voltage of the cell is calculated as given below:

  $\begin{array}{c}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0591}}{\text{n}}\text{ logQ }\\ \text{E = 0}\text{.32}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{\left[{\text{Zn}}^{\text{+2}}\right]}{\left[{\text{Fe}}^{+2}\right]}\\ \text{E }=\text{0}\text{.32}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{1.{\text{0×10}}^{\text{-3}}\text{M}}{\text{0}\text{.050M}}\\ \text{E }=\text{0}\text{.32}\text{V - }\left(-0.0502\right)\\ \text{E }=0.37\text{V}\end{array}$

Potential is calculated to be $0.37\text{V}$.

b)

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined with the given concentrations.

  $\begin{array}{l}\text{b) AgCl}\text{(s) }+{\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\to \text{ Ag(s)}+{\text{Fe}}^{\text{+3}}\left(\text{aq}\right){\text{+Cl}}^{-}\text{(aq) }\\ \\ \begin{array}{l}\left[{\text{Fe}}^{2+}\right]=0.20\text{M},\left[{\text{Fe}}^{3+}\right]=0.010\text{M},\left[{\text{Cl}}^{-}\right]=4.0\times {10}^{-3}\text{M}\hfill \\ \hfill \end{array}\end{array}$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  $\text{AgCl}\text{(s) }+{\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\to \text{ Ag(s)}+{\text{Fe}}^{\text{+3}}\left(\text{aq}\right){\text{+Cl}}^{-}\text{(aq) }$

The half-cell reaction can be written as below:

  $\begin{array}{l}\text{At anode:}\\ \text{Oxidation: }{\text{Fe}}^{\text{+2}}\text{ (aq)}\to {\text{ Fe}}^{\text{+3}}\left(\text{aq}\right){\text{ +e}}^{\text{-}},{\text{E}}_{\text{cell}}^{\text{o}}=-0.771\text{V}\\ \\ \text{At cathode:}\\ \text{Reduction: }\text{AgCl }\left(\text{s}\right){\text{+e}}^{\text{-}}\to \text{ Ag}\left(s\right){\text{+Cl}}^{-}\left(\text{aq}\right),{\text{E}}_{\text{cell}}^{\text{o}}=+0.222\text{V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{+E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= 0}\text{.222 V}+\left(-0.771\text{V}\right)\\ \\ \text{=- 0}\text{.549 V}\end{array}$

Voltage of the cell is calculated as given below:

  $\begin{array}{c}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0591}}{\text{n}}\text{ logQ }\\ \text{E = }-\text{0}\text{.549}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{\left[{\text{Fe}}^{\text{+3}}\right]\left[{\text{Cl}}^{-}\right]}{\left[{\text{Fe}}^{+2}\right]}\\ \text{E }=-\text{0}\text{.549}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{0.010\times 4\times {10}^{-3}}{0.20}\\ \text{E }=-\text{0}\text{.549}\text{V - }\left(-0.1093\right)\\ \text{E }=-0.434\text{V}\end{array}$

Potential is calculated to be $-0.434\text{V}$.

c)

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined with the given concentrations.

  $\begin{array}{l}{\text{c) Br}}_2\text{(l) +}\text{2}{\text{Cl}}^{\text{-}}\text{(aq) }\to {\text{ Cl}}_2\text{(g) + 2}{\text{Br}}^{\text{-}}\text{(aq)}\\ \\ \begin{array}{l}\left[{\text{Br}}^{-}\right]=3.5\times {10}^{-3}\text{M},\left[{\text{Cl}}^{-}\right]=0.10M,P_{{\text{Cl}}_2}=0.50\text{atm}\hfill \\ \hfill \end{array}\end{array}$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  ${\text{Br}}_2\text{(l) +}\text{2}{\text{Cl}}^{\text{-}}\text{(aq) }\to {\text{ Cl}}_2\text{(g) + 2}{\text{Br}}^{\text{-}}\text{(aq)}$

The half-cell reaction can be written as below:

  $\begin{array}{l}\text{At anode:}\\ {\text{Oxidation: 2Cl}}^{-}\text{ (aq)}\to {\text{ Cl}}_2{\text{(g) +2e}}^{\text{-}},{\text{E}}_{\text{cell}}^{\text{o}}=-1.36\text{V}\\ \\ \text{At cathode:}\\ {\text{Reduction: Br}}_2{\text{(l)+2e}}^{\text{-}}\to \text{ 2}{\text{Br}}^{\text{-}}\text{(aq)},{\text{E}}_{\text{cell}}^{\text{o}}=+1.06\text{V}\end{array}$

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

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{+E}}_{\text{anode}}^{\text{o}}\\ \\ \text{= 1}\text{.06 V}+\left(-1.36\text{V}\right)\\ \\ \text{= -0}\text{.3 V}\end{array}$

Voltage of the cell is calculated as given below:

  $\begin{array}{c}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0591}}{\text{n}}\text{ logQ }\\ \text{E = }-\text{0}\text{.3}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{{\left[{\text{Br}}^{\text{-}}\right]}^2\text{p}\left[{\text{Cl}}_2\right]}{{\left[{\text{Cl}}^{\text{-}}\right]}^2}\\ \text{E }=-\text{0}\text{.3}\text{V - }\frac{\text{0}\text{.0591}}{2}\text{ log}\frac{{\left(3.5\times {10}^{-3}\right)}^2\times 0.50}{3.5\times {10}^{-3}}\\ \text{E }=-\text{0}\text{.3}\text{V - }\left(-0.08147\right)\\ \text{E }=-0.219\text{V}\end{array}$

Potential is calculated to be $-0.219\text{V}$.