Chapter 11, Problem 19E

(a)

Interpretation Introduction

Interpretation: The galvanic cells based on given overall reaction needs to be sketched and the value of ${\text{E}}^{\text{0}}$ needs to be calculated.

  ${\text{Cr}}^{\text{3+}}\left(aq\right)+{\text{Cl}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{Cr}}_{\text{2}}{\text{O}}_7{}^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)$

Concept Introduction: A galvanic cell is an electro chemical cell that drives electrical energy from chemical reactions taking place within a particular cell.

In a galvanic cell the oxidation and the reduction half cells are separated by a connecting wire so that electrons can flow through the wire.

Answer to Problem 19E

Sketch of a galvanic cell is as follows:

  

Values of ${\text{E}}^{\text{0}}$ is 0.03.

Explanation of Solution

Below is a sketch of a galvanic cell where electrons flow from anode to cathode with a salt bridge between two to supply the required ions and complete the circuit. Cations will always flow to the anode and anions will always flow to the cathode.

  

By examining oxidation number of atoms in the reaction determine which compound is oxidized and which being reduced.

  ${\text{Cr}}^{\text{3+}}\left(aq\right)+{\text{Cl}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{Cr}}_{\text{2}}{\text{O}}_7{}^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)$

In above reaction Cr has +3 oxidation state and Cl has 0 oxidation state and in products Cr has +6 oxidation state and Cl has -1 oxidation state. Oxidation number increased for Cr, it is oxidized. This means that ${\text{Cr}}^{\text{3+}}$ is at the anode. Because a metal solid is no made, platinum metal anode is used and ${\text{Cr}}^{\text{3+}}$ exists in solution. For Cl oxidation number decreased, it was reduced. This means that ${\text{Cl}}_{\text{2}}$ is at cathode. Reduced metal is not a metal solid thus a platinum metal cathode is used and ${\text{Cl}}_{\text{2}}$ is bubbled into solution.

Determine actual half reactions that have occurred. ${\text{Cr}}^{\text{3+}}$ has taken on O which must have come from ${\text{H}}_{\text{2}}\text{O}$ . To add enough ${\text{H}}_{\text{2}}\text{O}$ to balance out O on each side. H is converted to ${\text{H}}^{\text{+}}$ .

  ${\text{2Cr}}^{\text{3+}}\text{+}{\text{7H}}_{\text{2}}\text{O}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}+14{\text{H}}^{+}$

Cr has lost three electrons because Cr atom has gone from +3 to +6. Add six electrons to the products. The half reaction is now change balanced as well.

  ${\text{2Cr}}^{\text{3+}}\text{+}{\text{7H}}_{\text{2}}\text{O}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}+14{\text{H}}^{+}+6e^{-}$

It has been converted to ${\text{Cl}}^{-}$ for ${\text{Cl}}_{\text{2}}$ .

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

Each Cl atom required one electron because they have gone from 0 to -1, thus add two electrons in reactants for a change balance.

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

Add two half reactions together and cancel out the electrons to determine overall process. Because Cr half reaction has six electrons multiply Cl half reaction by three before adding two together.

  $\begin{array}{l}{\text{2Cr}}^{\text{3+}}\text{+}{\text{7H}}_{\text{2}}\text{O}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}+14{\text{H}}^{+}+6e^{-}\\ {\text{3Cl}}_{\text{2}}+6e^{-}\to 6{\text{Cl}}^{-}\\ {\text{2Cr}}^{\text{3+}}{\text{+3Cl}}_{\text{2}}+{\text{7H}}_{\text{2}}\text{O}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}+6{\text{Cl}}^{-}+14{\text{H}}^{+}\end{array}$

Determine the standard reduction potential from reference table to calculate ${\text{E}}^{\text{0}}$ . Because Cr half reaction runs reverse as an oxidation, reverse sign of its potential.

  $\begin{array}{l}{\text{2Cr}}^{\text{3+}}\text{+}{\text{7H}}_{\text{2}}\text{O}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}+14{\text{H}}^{+}+6e^{-}-{\text{E}}^0{}_{\text{anode}}=-1.33\text{V}\\ {\text{Cl}}_{\text{2}}+2e^{-}\to 2{\text{Cl}}^{-}{\text{E}}^0{}_{\text{Cathode}}=1.36\text{V}\end{array}$

Add these values together to give ${\text{E}}^{\text{0}}$ .

  $\begin{array}{c}{\text{E}}^{\text{0}}={\text{E}}^{\text{0}}{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ =1.36-1.33\\ =0.03\end{array}$

(b)

Interpretation Introduction

Interpretation: The galvanic cells based on given overall reaction needs to be sketched and the value of ${\text{E}}^{\text{0}}$ needs to be calculated.

  ${\text{Cu}}^{\text{2+}}\left(aq\right)+\text{Mg}\left(g\right)\rightleftharpoons {\text{Mg}}^{2+}\left(aq\right)+\text{Cu}\left(aq\right)$

Concept Introduction: A galvanic cell is an electro chemical cell that drives electrical energy from chemical reactions taking place within a particular cell.

In a galvanic cell the oxidation and the reduction half cells are separated by a connecting wire so that electrons can flow through the wire.

Answer to Problem 19E

Sketch of a galvanic cell is as follows:

  

Values of ${\text{E}}^{\text{0}}$ is 2.17V.

Explanation of Solution

By examining oxidation number of atoms in the reaction determine which compound is oxidized and which being reduced.

  ${\text{Cu}}^{\text{2+}}\left(aq\right)+\text{Mg}\left(g\right)\rightleftharpoons {\text{Mg}}^{2+}\left(aq\right)+\text{Cu}\left(aq\right)$

In above reaction Cu has +2 oxidation state and Mg has 0 oxidation state and in products Cu has 0 oxidation state and Mg has +2oxidation state. Oxidation number increased for Mg, it is oxidized. This means that Mg is at the anode. For Cu oxidation number decreased, it was reduced. This means that ${\text{Cu}}^{\text{2+}}$ is at cathode. A metal formed Cu can be used as electrode.

Add two half reactions together and cancel out electrons determine overall process.

  $\begin{array}{l}\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}\\ {\text{Cu}}^{2+}+2e^{-}\to \text{Cu}\\ {\text{Cu}}^{\text{2+}}+\text{Mg}\to {\text{Mg}}^{\text{2+}}+\text{Cu}\end{array}$

Determine the standard reduction potential from reference table to calculate ${\text{E}}^{\text{0}}$ . Because Mg half reaction runs reverse as an oxidation, reverse sign of its potential.

  $\begin{array}{l}\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}{\text{E}}^0{}_{\text{anode}}=2.37\text{V}\\ {\text{Cu}}^{\text{2+}}+2e^{-}\to 2\text{Cu}-{\text{E}}^0{}_{\text{Cathode}}=0.34\text{V}\end{array}$

Add these values together to give ${\text{E}}^{\text{0}}$ .

  $\begin{array}{c}{\text{E}}^{\text{0}}={\text{E}}^{\text{0}}{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ =0.34+2.37\text{V}\\ =2.71\text{V}\end{array}$

(c)

Interpretation Introduction

Interpretation: The galvanic cells based on given overall reaction needs to be sketched and the value of ${\text{E}}^{\text{0}}$ needs to be calculated.

  ${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+{\text{Fe}}^{\text{2+}}\left(aq\right)\rightleftharpoons {\text{Fe}}^{3+}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$

Concept Introduction: A galvanic cell is an electro chemical cell that drives electrical energy from chemical reactions taking place within a particular cell.

In a galvanic cell the oxidation and the reduction half cells are separated by a connecting wire so that electrons can flow through the wire.

Answer to Problem 19E

Sketch of a galvanic cell is as follows:

  

Values of ${\text{E}}^{\text{0}}$ is 0.43V.

Explanation of Solution

By examining oxidation number of atoms in the reaction determine which compound is oxidized and which being reduced.

  ${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+{\text{Fe}}^{\text{2+}}\left(aq\right)\rightleftharpoons {\text{Fe}}^{3+}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$

In above reaction I has +5 oxidation state and Fe has +2 oxidation state and in products I has 0 oxidation state and Fe has +3oxidation state. Oxidation number increased for Fe, it is oxidized. This means that ${\text{Fe}}^{\text{2+}}$ is at the anode. Because a metal solid is not made, a platinum metal anode is used and ${\text{Fe}}^{\text{2+}}$ exists in solution.

Determine actual half reactions that have occurred ${\text{IO}}_{\text{3}}{}^{-}$ has lost O, which must be made into ${\text{H}}_{\text{2}}\text{O}$ . Add enough ${\text{H}}^{\text{+}}$ to balance out the O on each side.

  ${\text{2IO}}_{\text{3}}{}^{-}+12{\text{H}}^{\text{+}}\to {\text{I}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$

Each I atom has gone from +5 to 0 oxidation state, thus it has gained five electrons. Add ten electrons to reactants. Half-reaction is now charge balanced as well.

  ${\text{Fe}}^{\text{2+}}$ has been converted to ${\text{Fe}}^{\text{3+}}$ .

  ${\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}$

  ${\text{Fe}}^{\text{2+}}$ lose one electron thus add one electron to products for a charge balance.

  ${\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}+e^{-}$

To determine overall process add two half reactions together and cancel out electrons. Because I half-reaction has ten electrons, multiply Fe half-reaction by ten before adding two together.

  $\begin{array}{l}{\text{2IO}}_{\text{3}}{}^{-}+12{\text{H}}^{\text{+}}+10e^{-}\to {\text{I}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}\\ {\text{10Fe}}^{\text{2+}}\to 10{\text{Fe}}^{3+}+10e^{-}\\ {\text{2IO}}_{\text{3}}{}^{-}+10{\text{H}}^{\text{+}}+12{\text{H}}^{+}\to {\text{I}}_{\text{2}}+10{\text{Fe}}^{\text{3+}}+6{\text{H}}_{\text{2}}{\text{O}}^{\text{+}}\end{array}$

Determine standard reduction potentials from reference table to calculate ${\text{E}}^{\text{0}}$ . Because Fe half-reaction runs in reverse as an oxidation, reverse sign of its potential.

  $\begin{array}{l}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}+e^{-}-{\text{E}}^0{}_{\text{anode}}=-0.77\text{V}\\ {\text{2IO}}_{\text{3}}{}^{-}+12{\text{H}}^{+}+10e^{-}\to {\text{I}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}{\text{E}}^0{}_{\text{Cathode}}=1.20\text{V}\end{array}$

Add these values together to give ${\text{E}}^{\text{0}}$ .

  $\begin{array}{c}{\text{E}}^{\text{0}}={\text{E}}^{\text{0}}{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ =1.20-0.77\text{V}\\ =0.43\text{V}\end{array}$

(d)

Interpretation Introduction

Interpretation: The galvanic cells based on given overall reaction needs to be sketched and the value of ${\text{E}}^{\text{0}}$ needs to be calculated.

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

Concept Introduction: A galvanic cell is an electro chemical cell that drives electrical energy from chemical reactions taking place within a particular cell.

In a galvanic cell the oxidation and the reduction half cells are separated by a connecting wire so that electrons can flow through the wire.

Answer to Problem 19E

Sketch of a galvanic cell is as follows:

  

Values of ${\text{E}}^{\text{0}}$ is 1.56V.

Explanation of Solution

By examining oxidation number of atoms in the reaction determines which compound is oxidized and which being reduced.

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

In above reaction Ag has +1 oxidation state and Zn has 0 oxidation state and in products Ag has 0 oxidation state and Zn has +2 oxidation state. This means Zn is at anode. Because oxidation number decrease for Ag. It is reduced. This means that ${\text{Ag}}^{\text{+}}$ is cathode. Because a metal is formed Ag can be used as electrode.

Determine actual-half reactions that have occurred. Zn has only been converted to ${\text{Zn}}^{\text{2+}}$ .

  $\text{Zn}\to {\text{Zn}}^{\text{2+}}$

Zn loses two electrons so add two electrons to products to charge balance half-reaction.

  $\text{Zn}\to {\text{Zn}}^{\text{2+}}+2e^{-}$

  ${\text{Ag}}^{\text{+}}$ has only been converted to Ag.

  ${\text{Ag}}^{+}\to \text{Ag}$

  ${\text{Ag}}^{\text{+}}$ gains one electron, thus add one electron to reactants to charge balance half-reaction.

  ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$

Determine standard reduction potentials from reference table to calculate ${\text{E}}^{\text{0}}$ . Because Zn half-reaction runs in reverse as an oxidation, reverse sign of its potential.

  $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{2+}}+2e^{-}-{\text{E}}^0{}_{\text{anode}}=-0.76\text{V}\\ {\text{Ag}}^{+}+e^{-}\to \text{Ag}{\text{E}}^0{}_{\text{Cathode}}=0.80\text{V}\end{array}$

Add these values together to give ${\text{E}}^{\text{0}}$ .

  $\begin{array}{c}{\text{E}}^{\text{0}}={\text{E}}^{\text{0}}{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ =0.80+0.76\text{V}\\ =1.56\text{V}\end{array}$