Chapter 18, Problem 18.24QP

Calculate ΔG° and K_c for the following reactions at 25°C.

Interpretation Introduction

Interpretation: The following reaction at ${25}^{0}C$ should be calculated the value of $\Delta {\text{G}}^0$ and ${\text{K}}_{\text{C}}$ respectively.

Concept Introduction:

Standard reduction potential

• The standard reduction potential is the tendency for a chemical species to be reduced, and is measured in volts at standard conditions. The more positive the potential is the more likely it will be reduced.
• Redox reaction are those in which oxidation number changes.
• Half reaction are the separated oxidation and reduction reaction that make up the overall redox reaction.
• Redox equation can be balanced via the half –reaction methods, which allows for the addition of ${\text{H}}_{\text{2}}\text{O}$ to balance O, ${\text{H}}^{+}$ to balance H, and ${\text{OH}}^{-}$ for taking place in basic solution.
• Standard EMF:
• The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic cell.
• This quantity is related to the tendency for an element.

  i.e- It has a great tendency to acquire or gain the compound.

• Half-cell Reaction: A half-cell is one of the two electrodes in a galvanic cell
• Example, the Zn-Cu battery, the two half cells make an oxidizing couple.
• Electrochemical cell:
• An electrochemical cell in which a spontaneous chemical reaction generates a flow of electrons through a wire is called a galvanic cell

Answer to Problem 18.24QP

By using Standard reduction potential

First break the equation into half-reaction

$\begin{array}{l}\text{Mg(s)}\stackrel{\text{oxidation(anode)}}{\to }{\text{Mg}}^{{}_{{}^{2+}}}{\text{(aq)+2e}}^{-}{\text{ E}}^0{}_{\text{anode}}\text{ = -2}\text{.37 V}\\ {\text{Pb}}^{\text{2+}}{\text{(aq)+ 2}}^{\text{e-}}\stackrel{\text{reduction (cathode)}}{\to }{\text{Pb(s) E}}^0{}_{\text{cathode}}\text{ = -0}\text{.13 V}\end{array}$

The standard emf is given by :

${\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ = -0}\text{.13 V - (-2}\text{.37 V) = 2}\text{.24 V}$

We can calculate $\Delta {\text{G}}^0$ from the standard emf  by using the equation 18.3

$\begin{array}{l}\Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \\ \Delta {\text{G}}^0={\text{-(2)(96,500 J/Vmol}}^{\text{-}}\text{(2}\text{.24V)=-432kj/mol}\\ \\ {\text{E}}^0{}_{\text{cell}}\text{=}\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ logK}\\ \text{or}\\ \text{logK}=\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}\\ \text{and}\\ K={10}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ \\ K={10}^{\text{(2)(2}\text{.24 V)/0}\text{.0592 V}}\\ \\ {\text{K=10}}^{\text{75}\text{.7}}\text{ =5}\times {\text{10}}^{75}\end{array}$

Explanation of Solution

$\text{Mg(s)}\stackrel{\text{oxidation(anode)}}{\to }{\text{Mg}}^{{}_{{}^{2+}}}{\text{(aq)+2e}}^{-}{\text{ E}}^0{}_{\text{anode}}\text{ = -2}\text{.37 V}$

When magnesium molecule under goes oxidation give the magnesium (II) ion with liberation or losses of two electron.

${\text{Pb}}^{\text{2+}}{\text{(aq)+ 2}}^{\text{e-}}\stackrel{\text{reduction (cathode)}}{\to }{\text{Pb(s) E}}^0{}_{\text{cathode}}\text{ = -0}\text{.13 V}$

In case of lead ion, two electron are taken but the lead ion to convert itself neutral lead atom,

The standard emf is given by:

${\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ = -0}\text{.13 V - (-2}\text{.37 V) = 2}\text{.24 V}$

We can calculate $\Delta {\text{G}}^0$ from the standard emf by using the equation 18.3

$\begin{array}{l}\Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \Delta {\text{G}}^0={\text{-(2)(96,500 J/Vmol}}^{\text{-}}\text{(2}\text{.24V)=-432kj/mol}\\ {\text{E}}^0{}_{\text{cell}}\text{=}\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ logK}\\ \text{or}\\ \text{logK}=\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}\\ \text{and}\\ K={10}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ \\ K={10}^{\text{(2)(2}\text{.24 V)/0}\text{.0592 V}}\\ \\ {\text{K=10}}^{\text{75}\text{.7}}\text{ =5}\times {\text{10}}^{75}\end{array}$

Interpretation Introduction

Interpretation: The following reaction at ${25}^{0}C$ should be calculated the value of $\Delta {\text{G}}^0$ and ${\text{K}}_{\text{C}}$ respectively.

Concept Introduction:

Standard reduction potential

• The standard reduction potential is the tendency for a chemical species to be reduced, and is measured in volts at standard conditions. The more positive the potential is the more likely it will be reduced.
• Redox reaction are those in which oxidation number changes.
• Half reaction are the separated oxidation and reduction reaction that make up the overall redox reaction.
• Redox equation can be balanced via the half –reaction methods, which allows for the addition of ${\text{H}}_{\text{2}}\text{O}$ to balance O, ${\text{H}}^{+}$ to balance H, and ${\text{OH}}^{-}$ for taking place in basic solution.
• Standard EMF:
• The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic cell.
• This quantity is related to the tendency for an element.

  i.e- It has a great tendency to acquire or gain the compound.

• Half-cell Reaction: A half-cell is one of the two electrodes in a galvanic cell
• Example, the Zn-Cu battery, the two half cells make an oxidizing couple.
• Electrochemical cell:
• An electrochemical cell in which a spontaneous chemical reaction generates a flow of electrons through a wire is called a galvanic cell

Answer to Problem 18.24QP

By using Standard reduction potential

$\begin{array}{l}{\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ = 1}\text{.23 V - 0}\text{.77 V = 0}\text{.46 V}\\ \Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \Delta {\text{G}}^0\text{ = -(4)(96,500 J/V}\text{.mole})\text{(0}\text{.46V) = -180KJ/mol}\\ \text{logK}={\text{10}}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ \\ {\text{ K=10}}^{\text{(4)(0}\text{.46V)}/\text{0}\text{.0592 V}}\\ \\ {\text{K=10}}^{\text{31}\text{.1}}\text{ = 1}\times {\text{10}}^{31}\end{array}$

Explanation of Solution

We can calculate $\Delta {\text{G}}^0$ from the standard emf  by using the equation 18.3

$\begin{array}{l}{\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ = 1}\text{.23 V - 0}\text{.77 V = 0}\text{.46 V}\\ \Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \Delta {\text{G}}^0\text{ = -(4)(96,500 J/V}\text{.mole})\text{(0}\text{.46V) = -180KJ/mol}\\ \text{logK}={\text{10}}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ {\text{ K=10}}^{\text{(4)(0}\text{.46V)}/\text{0}\text{.0592 V}}\\ {\text{K=10}}^{\text{31}\text{.1}}\text{ = 1}\times {\text{10}}^{31}\end{array}$

Interpretation Introduction

Interpretation: The following reaction at ${25}^{0}C$ should be calculated the value of $\Delta {\text{G}}^0$ and ${\text{K}}_{\text{C}}$ respectively.

Concept Introduction:

Standard reduction potential

• The standard reduction potential is the tendency for a chemical species to be reduced, and is measured in volts at standard conditions. The more positive the potential is the more likely it will be reduced.
• Redox reaction are those in which oxidation number changes.
• Half reaction are the separated oxidation and reduction reaction that make up the overall redox reaction.
• Redox equation can be balanced via the half –reaction methods, which allows for the addition of ${\text{H}}_{\text{2}}\text{O}$ to balance O, ${\text{H}}^{+}$ to balance H, and ${\text{OH}}^{-}$ for taking place in basic solution.
• Standard EMF:
• The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic cell.
• This quantity is related to the tendency for an element.

  i.e- It has a great tendency to acquire or gain the compound.

• Half-cell Reaction: A half-cell is one of the two electrodes in a galvanic cell
• Example, the Zn-Cu battery, the two half cells make an oxidizing couple.
• Electrochemical cell:
• An electrochemical cell in which a spontaneous chemical reaction generates a flow of electrons through a wire is called a galvanic cell

Answer to Problem 18.24QP

By using Standard reduction potential

${\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ =0}\text{.53 V - (-1}\text{.66 V) = 2}\text{.19 V }$

We can calculate $\Delta {\text{G}}^0$ from the standard emf  by using the equation 18.3

$\begin{array}{l}\Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \Delta {\text{G}}^0\text{ = -(6)(96,500 J/V}\text{.mole})\text{(2}\text{.19V) = -1}\text{.27 }\times {\text{10}}^3\text{KJ/mol}\\ {\text{E}}^0{}_{\text{cell}}\text{=}\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ logK}\\ \text{or}\\ \text{logK}=\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}\\ \text{and}\\ K={10}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ \\ K={10}^{\text{-(6)(2}\text{.19V)/0}\text{.0592V}}\\ \\ {\text{K=10}}^{221.96}\text{ =9}\text{.1}\times {\text{10}}^{221}\end{array}$

Explanation of Solution

We can calculate $\Delta {\text{G}}^0$ from the standard emf  by using the equation 18.3s

$\begin{array}{l}{\text{E}}^0{}_{\text{cell}}{\text{ = E}}^0{}_{\text{cathode}}{\text{- E}}^0{}_{\text{anode}}\text{ =0}\text{.53 V - (-1}\text{.66 V) = 2}\text{.19 V }\\ \Delta {\text{G}}^0={\text{-nFE}}^{\text{0}}{}_{\text{Cell}}\\ \Delta {\text{G}}^0\text{ = -(6)(96,500 J/V}\text{.mole})\text{(2}\text{.19V) = -1}\text{.27 }\times {\text{10}}^3\text{KJ/mol}\\ {\text{E}}^0{}_{\text{cell}}\text{=}\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ logK}\\ \text{or}\\ \text{logK}=\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}\\ \text{and}\\ K={10}^{\frac{{\text{nE}}^0{}_{\text{cell}}}{\text{0}\text{.0592 V}}}\\ \\ K={10}^{\text{-(6)(2}\text{.19V)/0}\text{.0592V}}\\ \\ {\text{K=10}}^{221.96}\text{ =9}\text{.1}\times {\text{10}}^{221}\end{array}$